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<?xml version="1.0"?>

<!DOCTYPE TEI.2 SYSTEM "base.dtd">




<title>The Joy of Sets</title></titleStmt>

<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

upon written application to any of the holding bodies.

The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any


<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

British Academy and the Arts and Humanities Research Board. </p></availability>




<recording dur="00:45:42" n="5582">


<respStmt><name>BASE team</name>



<langUsage><language id="en">English</language>



<person id="nm0863" role="main speaker" n="n" sex="m"><p>nm0863, main speaker, non-student, male</p></person>

<person id="sm0864" role="participant" n="s" sex="m"><p>sm0864, participant, student, male</p></person>

<person id="sm0865" role="participant" n="s" sex="m"><p>sm0865, participant, student, male</p></person>

<person id="sm0866" role="participant" n="s" sex="m"><p>sm0866, participant, student, male</p></person>

<personGrp id="ss" role="audience" size="l"><p>ss, audience, large group </p></personGrp>

<personGrp id="sl" role="all" size="l"><p>sl, all, large group</p></personGrp>

<personGrp role="speakers" size="6"><p>number of speakers: 6</p></personGrp>





<item n="speechevent">Lecture</item>

<item n="acaddept">Mathematics</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG</item>

<item n="module">unknown</item>





<u who="nm0863"> # a number of you went to the first year discussion evening yesterday <pause dur="1.4"/> and <pause dur="3.1"/> i wasn't able to be there myself but if any of you here <pause dur="0.4"/> have any feedback about <pause dur="0.7"/> this particular course that you'd like to relay to me <pause dur="0.8"/> that came up at that meeting i'd be very grateful to hear it so come and see me <pause dur="0.9"/> after the session if there are any special points that i can address <pause dur="2.1"/> let's just cast our mind back over the last two weeks what have we been doing in this course <pause dur="3.4"/> we've been <pause dur="0.4"/> looking at arithmetic from <pause dur="0.8"/> a new point of view <pause dur="0.9"/> we've taken familiar ideas like <pause dur="0.2"/> adding subtracting multiplying dividing <pause dur="0.5"/> ordinary numbers <pause dur="1.6"/> and <pause dur="0.6"/> looking at them from <pause dur="0.8"/> different structural viewpoints through division with remainder Euclid's algorithm <pause dur="0.9"/> using that to <pause dur="3.3"/> prove formally <pause dur="0.6"/> the fundamental theorem of arithmetic which says every <pause dur="0.2"/> natural number can be <pause dur="0.2"/> uniquely <pause dur="0.3"/> expressed as a product of prime powers <pause dur="1.4"/> we've also taken time out to <pause dur="0.6"/> study very quickly <pause dur="0.2"/> the complex numbers those of you who hadn't <pause dur="0.4"/> met them before <pause dur="0.3"/> are strongly advised to <pause dur="1.2"/> get plenty of practice at

using them and if you have trouble with the ideas to consult <pause dur="0.9"/> a mate <pause dur="0.4"/> a buddy who does know about them <pause dur="2.3"/> and # <pause dur="2.8"/> we also # <pause dur="0.9"/> took time out to study <pause dur="2.4"/> mathematical induction <pause dur="2.2"/><kinesic desc="puts on transparency" iterated="n"/> and # <pause dur="0.4"/> i thought you'd be amused to see this <pause dur="0.5"/><vocal desc="laughter" iterated="y" n="ss" dur="1"/> alternative <pause dur="0.2"/> image <pause dur="0.8"/> for the <pause dur="0.4"/> falling dominoes <pause dur="0.8"/> as an indication of how induction works <pause dur="4.4"/> so basically <pause dur="0.5"/> we've been doing two things we've been bridging the gap <pause dur="1.0"/> looking at <pause dur="0.2"/> arithmetic in a <pause dur="0.9"/> a more professional way <pause dur="0.6"/> and we've been <pause dur="1.1"/> filling a gap <pause dur="0.4"/> of knowledge in <pause dur="1.0"/> complex numbers <pause dur="0.3"/> and we've been studying <pause dur="0.7"/> at least one method of proof namely <pause dur="0.5"/> mathematical induction <pause dur="0.4"/> well today i want to focus on the <pause dur="1.3"/> foundations aspect of the course <pause dur="1.7"/><kinesic desc="changes transparency" iterated="y" dur="3"/> and i want to go <pause dur="1.4"/> right back to the very beginning <pause dur="6.3"/> do you <pause dur="0.2"/> read these notes i produced do you actually read them or do you <pause dur="1.0"/> i mean in your own time <pause dur="0.3"/> they are meant to be read i don't have <pause dur="0.6"/> time within the lecture to read out every word <pause dur="0.5"/> and i hope you do <pause dur="2.2"/> look for the # <pause dur="0.4"/> the details when you <pause dur="0.2"/> take them away <pause dur="1.1"/> i hope the <pause dur="0.2"/> assignments i set give you the

motivation to go away and read them <pause dur="1.1"/> # <pause dur="0.8"/> today we're <pause dur="1.4"/> talking about the fact that mathematics could be entirely based <pause dur="0.2"/> on set theory <pause dur="0.4"/> and that the notion of a set <pause dur="1.5"/> is <pause dur="0.3"/> a primitive <pause dur="0.2"/> undefined <pause dur="0.4"/> concept <pause dur="2.0"/> and i'd like to make the point that <pause dur="0.6"/> however good your dictionary is <pause dur="0.7"/> it cannot define <pause dur="0.4"/> everything <pause dur="0.4"/> without <pause dur="0.3"/> some kind of <pause dur="0.6"/> circular definition <pause dur="1.3"/> and <pause dur="0.4"/> to check this out i took a couple of words at random <pause dur="0.9"/> # and i pursued them <pause dur="0.2"/> through a list of definitions <pause dur="0.6"/> mathematics <pause dur="0.3"/> the science of magnitude and number <pause dur="0.8"/> i looked up science <vocal desc="clears throat" iterated="n"/><pause dur="0.6"/> <reading>knowledge ascertained by observation <pause dur="0.5"/> and experiment critically tested systematized brought under general principals <pause dur="0.9"/> knowledge <pause dur="0.2"/> that which is known <pause dur="0.6"/> to know is to be informed of</reading> and then we get our first circular definition <pause dur="0.5"/> <reading>to inform is to pass on knowledge</reading> so at some point <pause dur="0.4"/> you must reach words in the dictionary which cannot be <pause dur="0.5"/> defined <pause dur="1.3"/> without making a circular argument <pause dur="2.6"/> i looked up another word at random i don't know <kinesic desc="reveals covered part of transparency" iterated="n"/> why i chose cow <pause dur="0.5"/> but it very

quickly became <pause dur="0.7"/> circular <pause dur="1.3"/> cow <reading>the female <pause dur="0.3"/> bovine animals <pause dur="0.4"/> bovine <pause dur="0.3"/> of or pertaining to cattle <pause dur="1.4"/> # cattle bovine animals <pause dur="0.7"/> of the genus Bos especially <pause dur="0.2"/> oxen bulls and cows</reading> and we're back in the circle again <pause dur="9.0"/><kinesic desc="reveals covered part of transparency" iterated="n"/> so i say there are only a finite number of English words so it's impossible to avoid circular definitions unless <pause dur="0.5"/> you accept certain basic words as <pause dur="0.2"/> primitive undefined concepts <pause dur="0.4"/> which we all know the meaning of but don't need or not even able to define <pause dur="3.2"/> and <pause dur="0.6"/> mathematicians of course are bound by the same rules <pause dur="2.3"/> we're very big on definitions but we can't define absolutely everything <pause dur="1.2"/> however <pause dur="0.5"/> we do rather well <pause dur="0.9"/> we make just one <pause dur="0.9"/> we make do with just one <pause dur="0.2"/> primitive idea <pause dur="0.8"/> the concept of a set so i'm not going to define the concept of a set <pause dur="1.0"/> but i'm going to talk about it <pause dur="0.7"/> and hope that your idea of what a set is <pause dur="0.8"/><kinesic desc="reveals covered part of transparency" iterated="n"/> coincides closely <pause dur="0.7"/> with my idea <pause dur="0.7"/> and that of most mathematicians <pause dur="2.0"/> <reading>it turns out that the whole of mathematics can be founded on this one simple notion <pause dur="0.9"/> and a subtext of this

course of lectures a not so hidden agenda <pause dur="0.3"/> is to show how this can be achieved</reading> <pause dur="2.7"/> so we we can't offer a formal definition of a set because that's our primitive undefined object but we can <pause dur="0.9"/> make an informal <pause dur="0.6"/> definition as to what we mean by it <pause dur="17.2"/><kinesic desc="changes transparency" iterated="y" dur="10"/> and we now come to <pause dur="1.1"/> A-point-one definitely not a definition <pause dur="0.3"/> a set is just a collection of things <pause dur="0.6"/> and the things in the set are called elements <pause dur="0.4"/> whoops <pause dur="0.8"/> or <pause dur="0.6"/><vocal desc="laughter" iterated="y" n="ss" dur="1"/> the members <pause dur="1.9"/> of that set <pause dur="0.8"/> and we have a bit of standard notation <pause dur="1.4"/> typically we use <pause dur="0.2"/> upper case letters for the sets themselves <pause dur="0.8"/> not always but <pause dur="1.9"/> very often <pause dur="0.3"/> and we use lower case <pause dur="0.8"/> letters often related closely to the upper case letter in the alphabet <pause dur="0.3"/> to denote the members <pause dur="0.4"/> or the elements of that set so <pause dur="0.4"/> here's a bit of <pause dur="0.2"/> standard <pause dur="0.5"/> notation <pause dur="1.0"/> X <pause dur="1.6"/> this funny epsilon sign <pause dur="0.3"/> big-X <pause dur="1.3"/> this should be read out loud as <pause dur="0.6"/> the element X <pause dur="0.4"/> is a member of <pause dur="0.2"/> or belongs to <pause dur="0.4"/> or is in <pause dur="0.6"/> the set <pause dur="0.7"/> of capital-X <pause dur="2.9"/> if we have a finite set we can <pause dur="0.8"/> list its elements <pause dur="0.2"/> X-one X-two up to X-N <pause dur="1.4"/> if we know what N is we can write them all out <pause dur="1.7"/> but we

don't always deal with just finite sets <pause dur="0.9"/> # <pause dur="0.8"/> the integers themselves are <pause dur="1.2"/> an example of an infinite set and even integers you can describe <pause dur="0.2"/> in this way <pause dur="1.3"/> we use these braces a set of <pause dur="1.5"/> integers N elements <pause dur="0.2"/> little-N in Z the set of all integers <pause dur="0.4"/> two-divides-N that's a way of describing <pause dur="0.6"/> in set theoretical notation the even integers <pause dur="0.6"/> or sometimes we <pause dur="1.1"/> suggest <pause dur="0.5"/> what <pause dur="0.4"/> the set looks like <pause dur="0.8"/> this is meant to indicate the set of all <pause dur="0.4"/> even integers <pause dur="4.7"/> but this sort of notation depends <pause dur="0.5"/> on <pause dur="1.1"/> the tacit understanding <pause dur="0.3"/> that i've given enough information <pause dur="1.4"/> for you to see what's going on <pause dur="0.2"/> and that you've cottoned on to what i mean <pause dur="7.7"/><kinesic desc="reveals covered part of transparency" iterated="n"/> someone who is bloody-minded might say well why isn't the next one seven there's no reason why it shouldn't be <pause dur="0.5"/> but i've meant to indicate <pause dur="0.3"/> what i'm up up to there <pause dur="2.0"/> so what do we expect of a set these are <pause dur="0.2"/> important <pause dur="1.4"/> properties <pause dur="3.4"/> <reading>a set <pause dur="0.2"/> is determined by its elements</reading> and that sounds rather trite <pause dur="1.4"/> but a lot of people have trouble deciding when two sets are equal <pause dur="1.3"/> so let's be very

clear about this <pause dur="1.8"/> <reading>two sets are equal <pause dur="0.2"/> if and only if <pause dur="0.5"/> they contain <pause dur="0.5"/> the same <pause dur="0.2"/> elements</reading> <pause dur="1.0"/> so we have the notion of equality which we use the usual equal sign for <pause dur="0.4"/> between <pause dur="0.2"/> sets <pause dur="5.9"/> <reading>the elements of a set <pause dur="0.4"/> must be distinguishable <pause dur="0.4"/> from each other</reading> <pause dur="0.4"/> we only count each element once <pause dur="0.5"/> so if we wrote down one and one <pause dur="1.1"/> it's really only got one element and we should <pause dur="0.4"/> accept the convention of writing it <pause dur="0.6"/> as a single <pause dur="0.2"/> one <pause dur="0.2"/> so we think of all the elements <pause dur="0.4"/> of the set when we list them <pause dur="0.7"/> as being different <pause dur="4.3"/> now this is an important one <pause dur="0.4"/> when you write out a set <pause dur="0.6"/> the order <pause dur="0.7"/> the way you write them out the <trunc>l</trunc> the order in which you write them out <pause dur="0.3"/> or label them <pause dur="1.0"/> is <pause dur="0.4"/> unimportant so <pause dur="0.3"/> the set containing <pause dur="0.3"/> X and Y can be written <kinesic desc="indicates point on screen" iterated="n"/> that way <pause dur="0.7"/> or <kinesic desc="indicates point on screen" iterated="n"/> that way those two sets are the X and Y have the same <pause dur="0.3"/> elements therefore they are equal <pause dur="0.4"/> even though <pause dur="0.5"/> they're written out in different order so <pause dur="1.8"/> <trunc>th</trunc> the property of being a set takes no account of the ordering or the <trunc>labeli</trunc> or the labelling of the elements <pause dur="2.7"/> if you've got a set <pause dur="1.4"/> it's not well defined unless we

have some method of deciding whether <pause dur="0.7"/> anything <pause dur="0.3"/> is in the set <pause dur="0.3"/> or not <pause dur="0.9"/> so there's got to be some well defined <pause dur="0.6"/> procedure <pause dur="0.4"/> that tells you <pause dur="0.3"/> if i give you this is it in your set or not if you can always do that then your set <pause dur="0.4"/> is a genuine set and well defined <pause dur="1.8"/> now this last one <pause dur="0.9"/> <trunc>i</trunc> is # <pause dur="0.8"/> is just perhaps important to stress a lot of people think the objects of a set <pause dur="0.4"/> should be somehow <pause dur="0.2"/> related to each other like a set of numbers or a set of <pause dur="0.5"/> # students in a room or whatever <pause dur="0.4"/> but they can be totally <pause dur="0.6"/> # <pause dur="0.3"/> different have <trunc>n</trunc> no relation to each other no uniformity no common property so <pause dur="0.3"/> for instance <pause dur="0.5"/> # <pause dur="0.3"/> a particular cat <pause dur="0.4"/> and a particular fiddle <pause dur="0.4"/> and a particular cow <pause dur="0.6"/> and the moon <pause dur="0.2"/> and a certain dog <pause dur="0.4"/> and the word laughter can all be <pause dur="1.0"/> elements of this set with one <pause dur="0.5"/> two <pause dur="0.2"/> three <pause dur="0.2"/> four five six elements <pause dur="6.1"/><kinesic desc="reveals covered part of transparency" iterated="n"/> if you've got any questions about sets <pause dur="0.4"/> i'll give you a break in the middle <pause dur="0.6"/> to <pause dur="2.3"/> have you got one now i'm happy to take questions at any point but if you <pause dur="0.3"/> it's <pause dur="0.2"/> often hard to ask questions in a big audience if you've got some questions <pause dur="0.4"/> i'll let you

have a chance to talk to your neighbours about it and then <pause dur="0.6"/> you might feel more confident about asking <pause dur="1.6"/> so here are some examples of <trunc>s</trunc> <pause dur="0.4"/> sets the students attending the first <pause dur="0.5"/> foundations lecture <pause dur="1.5"/> you here today i presume are a subset <pause dur="0.3"/> of that set probably a proper subset it being now nine o'clock in the morning <pause dur="0.2"/> although the first one was at nine too <pause dur="0.4"/> but you were <pause dur="0.7"/> # <pause dur="0.7"/> bit less <pause dur="2.0"/> # <pause dur="0.9"/> inured to the ways of the world on the first day on the first Wednesday of term <pause dur="0.6"/> # <pause dur="0.2"/> here's an interesting set the set of <pause dur="0.4"/> real numbers that satisfy this equation <pause dur="0.8"/> if you draw the graph so the graph lies the X axis completely so <pause dur="0.6"/> there are no real numbers satisfying that so that's another way of defining the empty set <pause dur="0.5"/> however <pause dur="0.3"/> if we allow <vocal desc="clears throat" iterated="n"/><pause dur="0.3"/> the <pause dur="0.7"/> variable X to be in the field of complex numbers rather than the field of real numbers <pause dur="1.0"/> there are indeed <pause dur="0.4"/> two numbers <pause dur="0.3"/> in that set so be <pause dur="0.3"/> very <pause dur="0.3"/> specific <pause dur="0.2"/> where the elements come from <pause dur="0.3"/> if you are using <kinesic desc="indicates point on screen" iterated="n"/> this notation <pause dur="0.3"/> the set of <pause dur="0.4"/> certain things with <pause dur="0.4"/> a straight line and then <pause dur="0.3"/> a certain

property <pause dur="0.4"/> the set of all X in here <pause dur="0.4"/><kinesic desc="indicates point on screen" iterated="n"/> satisfying that property <pause dur="5.5"/> some non-examples <pause dur="2.8"/> there are a set of people <pause dur="0.3"/> in the first foundations lecture <pause dur="0.4"/> who will be alive on January the first <pause dur="0.6"/> two-thousand we hope that is <pause dur="0.4"/> the same set but we won't know <pause dur="0.5"/> until the first of January two-thousand-and-fifteen <pause dur="0.6"/> # but i certainly shan't be around and so that's one <pause dur="0.5"/> less <trunc>peop</trunc> one less person <pause dur="0.3"/> we won't know until two-thousand-and-fifteen <pause dur="0.4"/> # whether what what set is who belongs to it <pause dur="1.4"/> # the set of water <pause dur="0.6"/> in <pause dur="0.4"/> the Atlantic Ocean <pause dur="0.5"/> well there are lots of problems with that set <pause dur="0.9"/> water <pause dur="0.3"/> doesn't have any <pause dur="0.3"/> discreteness so <pause dur="0.5"/> you could talk about the molecules of water perhaps but water <pause dur="0.3"/> is <pause dur="0.4"/> # # an amorphous <pause dur="0.5"/> structure <pause dur="1.2"/> # <pause dur="1.5"/> you could <pause dur="0.5"/> talk about the molecules of water in the Atlantic Ocean you'd still have trouble <pause dur="0.4"/> # because <pause dur="0.6"/> the Atlantic Ocean isn't terribly well defined where does it stop and where does it begin <pause dur="0.6"/> and molecules of water are constantly dancing <pause dur="0.4"/> # above the surface of the water are they in or are they out <pause dur="0.2"/> and if someone gave you <vocal desc="laugh" iterated="n"/><pause dur="0.4"/> a molecule how

could you ever tell <pause dur="0.2"/> whether it came from the Atlantic Ocean or not because all molecules of water are the same <pause dur="0.2"/> so i wouldn't accept <kinesic desc="indicates point on screen" iterated="n"/> this as a set <pause dur="0.5"/> and then <pause dur="0.4"/> # there are also problems <pause dur="0.7"/> in logic <pause dur="0.5"/> with making sets too big <pause dur="0.9"/> and in particular the set of all sets <pause dur="1.0"/> is not allowed <pause dur="0.4"/> for reasons that we shall see later <pause dur="2.7"/> okay <pause dur="0.9"/> everyone happy about sets anybody got any questions <pause dur="2.4"/> right the first important idea <pause dur="0.7"/> in sets is to talk about subsets </u><gap reason="break in recording" extent="uncertain"/> <u who="nm0863" trans="pause"> <kinesic desc="changes transparency" iterated="y" dur="6"/> i don't think it's a very difficult idea <pause dur="1.1"/> a subset <pause dur="0.2"/> of a set <pause dur="1.0"/> is a set consisting of <pause dur="0.6"/> some <pause dur="0.3"/> possibly all the elements i haven't done that <pause dur="0.7"/> very clearly <pause dur="22.6"/><kinesic desc="adjusts transparency" iterated="n"/> # the <trunc>sub</trunc> # B is a subset of A <pause dur="0.2"/> if any element of B <pause dur="0.8"/> is in A <pause dur="0.6"/> that's all it says <pause dur="1.0"/> # <pause dur="0.2"/> the notation is important however <pause dur="0.9"/> # <pause dur="0.2"/> for any old subset we use this <pause dur="1.9"/> curly less than or equal to sign <pause dur="0.3"/> that allows the possibility that B <pause dur="0.2"/> is equal to A note from the definition that A <pause dur="1.0"/> is one of the subsets of A A itself's a whole set <pause dur="0.7"/> and that the empty set of course is also <pause dur="0.4"/> a subset <pause dur="1.2"/> if we want to <pause dur="0.4"/> highlight the fact that we have a <pause dur="0.8"/> a proper subset we use

the slightly different <pause dur="0.3"/> notation <pause dur="0.6"/> if we remove the bottom line there <pause dur="0.4"/> and <pause dur="0.3"/><kinesic desc="indicates point on screen" iterated="n"/> this notation means B <pause dur="0.2"/> is a subset of A <pause dur="0.3"/> and B <pause dur="0.2"/> is not equal to A <pause dur="0.5"/> that means there's something <pause dur="0.2"/> in A <pause dur="0.3"/> but not in B <pause dur="12.3"/><kinesic desc="reveals covered part of transparency" iterated="n"/> # <pause dur="0.5"/> so the remarks i've already made <pause dur="0.2"/> don't forget that A <pause dur="0.2"/> is one of the subsets of A <pause dur="0.8"/> note that every set contains the empty set as a subset because the condition <pause dur="0.4"/> is vacuously satisfied <pause dur="0.8"/> now this is very important <pause dur="1.0"/> you're often asked to <pause dur="2.3"/> show that two sets <pause dur="0.2"/> described in different ways <pause dur="1.0"/> are actually equal <pause dur="0.8"/> and there's <pause dur="0.8"/> one standard <pause dur="0.4"/> and infallible technique for doing that <pause dur="2.1"/> what does it mean to say two sets are equal <pause dur="0.4"/> means they've got you've got to show they have the same <pause dur="0.2"/> elements <pause dur="0.6"/> and the way you do it is to show that if you've got anything in <pause dur="0.7"/> B <pause dur="1.5"/> it's in C if you got anything in C <pause dur="0.4"/> it's in B <pause dur="1.1"/> those two together <pause dur="0.7"/> tell me that the elements in B are the same as the elements in C and therefore the two sets are equal <pause dur="0.5"/> so the routine is this is what i call a double inclusion argument <pause dur="0.5"/> to show that B equals C <pause dur="0.9"/> you show first that B is a subset

of C <pause dur="0.3"/> then you show C as a subset of B <pause dur="0.7"/> # and you've got a safe <pause dur="0.2"/> way <pause dur="0.5"/> of checking that <pause dur="0.2"/> the two sets have the same elements <pause dur="1.9"/> and finally the observation i already made if B is a proper subset of A <pause dur="0.5"/> there's at least one element of A which is not in B <pause dur="3.1"/><kinesic desc="reveals covered part of transparency" iterated="n"/> let's have some <pause dur="0.8"/> examples <pause dur="1.9"/> every non-empty set <pause dur="0.4"/> has a least two subsets mainly the whole set <pause dur="0.9"/> and the empty subset <pause dur="5.8"/> for each <pause dur="0.6"/> X in a non-empty set A there's always a one element subset <pause dur="1.3"/> and we use the notation <pause dur="0.8"/> X with <pause dur="1.3"/> braces round it <pause dur="2.3"/> and it's important to distinguish <pause dur="0.5"/> X <pause dur="0.2"/> as an element of <pause dur="0.6"/> X <pause dur="1.0"/> and this <pause dur="0.5"/><kinesic desc="indicates point on screen" iterated="n"/> subset <pause dur="1.2"/> or sometimes a singleton <pause dur="0.2"/><vocal desc="clears throat" iterated="n"/><pause dur="0.7"/> # <pause dur="0.7"/> is a subset <pause dur="1.3"/> and therefore it's known to the set of all subsets <pause dur="0.4"/> of A so <pause dur="1.1"/> X <pause dur="0.6"/> is distinct from it's different from <pause dur="0.4"/> the set containing it <pause dur="2.9"/> and this is an excuse to remind you of some <pause dur="0.2"/> important subsets of the real numbers <pause dur="0.3"/> that <pause dur="0.2"/> crop up time and time again <pause dur="0.6"/> in <pause dur="0.7"/> analysis i don't know if you've met them yet but <pause dur="0.2"/> if not <pause dur="0.5"/> you soon will <pause dur="1.1"/> # <pause dur="3.7"/> round brackets <pause dur="1.3"/> A comma B <pause dur="0.2"/> the assumption there is that A and B are

real numbers <pause dur="0.4"/> and that A is less than B <pause dur="2.7"/> and this set stands for all the X <pause dur="0.7"/> that lie <pause dur="0.2"/> between A and B <pause dur="3.8"/><kinesic desc="writes on board" iterated="y" dur="5"/> so if this is your real line <pause dur="1.2"/> if you're assuming that A is less than B <pause dur="3.2"/> then <pause dur="5.2"/><kinesic desc="writes on board" iterated="y" dur="6"/> all the points in there are denoted by the open interval <pause dur="1.6"/> A-B <pause dur="1.6"/><vocal desc="clears throat" iterated="n"/><pause dur="1.2"/> and if you stick in <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="11"/> both the end points as well <pause dur="3.8"/> you use the square bracket notation <pause dur="3.5"/><vocal desc="clears throat" iterated="n"/><pause dur="1.5"/> so <pause dur="0.3"/> you get <kinesic desc="indicates point on board" iterated="n"/> this set by adding the two end points to <kinesic desc="indicates point on board" iterated="n"/> that set <pause dur="1.1"/> notice that these intervals are always infinite however close <pause dur="0.4"/> A is to B <pause dur="1.5"/> and then you've got the choice of <pause dur="1.5"/> a half open interval this is called a closed interval because you've closed it off at the ends <pause dur="0.9"/> it can have half open intervals with <pause dur="0.8"/> A missing <pause dur="0.3"/> or with B missing <pause dur="0.6"/> and this notation speaks for itself <pause dur="0.6"/> and you can even use the symbol <pause dur="0.2"/> infinity <pause dur="0.3"/> symbolically <pause dur="0.8"/> to be <pause dur="1.3"/> A <pause dur="0.2"/> to infinity it's all real numbers strictly bigger than <pause dur="0.6"/> A <pause dur="0.7"/> # minus-infinity <pause dur="0.3"/> A in square brackets is all real numbers <pause dur="0.4"/> less than or equal to A <pause dur="0.6"/> and in <kinesic desc="indicates point on screen" iterated="n"/> this notation minus-infinity plus-infinity would be another way <pause dur="0.2"/> of describing the whole real line <pause dur="0.6"/>

that's just the symbolic <pause dur="0.6"/> notation which is useful </u><gap reason="break in recording" extent="uncertain"/> <u who="nm0863" trans="pause"> <kinesic desc="reveals covered part of transparency" iterated="n"/> so we've talked about sets <pause dur="1.0"/> we've talked about subsets <pause dur="0.8"/> and now we come to <pause dur="2.6"/> some useful notation <pause dur="0.2"/> in making <pause dur="0.4"/> new sets from given sets <pause dur="0.5"/> we defined three important binary operations a binary operation is <pause dur="0.6"/> something that <pause dur="1.3"/> has <pause dur="0.3"/> an input of two things and an output of one <pause dur="0.4"/> so when you feed in <pause dur="0.4"/> two sets A and B <pause dur="0.5"/> the intersection <pause dur="0.4"/> is a subset consisting of <pause dur="0.3"/> all elements that are both in <pause dur="0.8"/> A <pause dur="0.4"/> and B <pause dur="1.8"/> if the union again a binary operation feed in two sets A and B <pause dur="0.6"/> the union just consists of all elements that are <pause dur="1.0"/> in X <pause dur="1.4"/> or in B or possibly in both <pause dur="0.2"/> it's not <pause dur="0.8"/> # <pause dur="1.0"/> <trunc>i</trunc> <trunc>i</trunc> it's not <pause dur="0.5"/> the exclusive or it's the inclusive or <pause dur="0.3"/> so <pause dur="0.2"/> A-union-B contains everything in A everything in B <pause dur="0.2"/> together with everything in both that's the union intersection union <pause dur="0.3"/> and the set theoretical difference <pause dur="0.4"/> everything in A take away <pause dur="0.7"/> from A <pause dur="0.2"/> everything that lies in B <pause dur="0.2"/> that's called the set theoretical <trunc>diff</trunc> difference # difference <pause dur="0.3"/> the set of elements that lie in A <pause dur="0.5"/> but not in B <pause dur="21.6"/><kinesic desc="changes transparency" iterated="y" dur="17"/> and <pause dur="1.2"/> one of my

standard <pause dur="0.8"/> bits of advice to anyone studying mathematics if you don't understand it draw a picture <pause dur="0.4"/> and that's what John Venn did <pause dur="1.9"/> he <pause dur="0.7"/> had this rather nice <pause dur="1.5"/> # <pause dur="0.7"/> idea of representing a set as an area <pause dur="0.8"/> and then <pause dur="0.2"/> visually the intersection <pause dur="0.5"/> is everything in #<pause dur="0.6"/> common <pause dur="1.8"/> the union is everything in both <pause dur="0.7"/> and the difference <pause dur="0.6"/> is everything in A but not in B do you find those pictures helpful i do <pause dur="0.2"/> i've always found them helpful <pause dur="0.4"/> so thanks to <pause dur="0.2"/> John Venn for <pause dur="0.5"/> being brave enough to draw a picture and letting us all know about it <pause dur="4.5"/> so <pause dur="0.4"/> <reading>the diagrams above show two sets A and B in general position with separate regions <pause dur="0.3"/> for each of the three subsets <pause dur="0.8"/> # <pause dur="0.5"/> A-minus-B A-intersection-B and B-minus-A</reading> there are three <pause dur="0.2"/> different <pause dur="0.6"/> connected regions here <pause dur="2.3"/><kinesic desc="indicates point on screen" iterated="n"/> and these <pause dur="1.2"/> shaded regions are made up <pause dur="0.5"/> in different <pause dur="0.7"/> ways of the three <pause dur="0.3"/> connected regions <pause dur="5.8"/><kinesic desc="reveals covered part of transparency" iterated="n"/> <reading>Venn diagrams can be used to show <pause dur="0.8"/> special relationships between sets such as <pause dur="0.3"/> inclusion</reading> <pause dur="3.5"/><kinesic desc="indicates point on screen" iterated="n"/> this picture suggests that <pause dur="0.3"/> A is entirely contained in B so A is a subset of B <pause dur="5.5"/>

now <pause dur="3.9"/> one important thing in studying <pause dur="0.6"/> sets <pause dur="0.2"/> is to have a safe space in which to do your set theory <pause dur="1.0"/> if you allow <pause dur="0.8"/> too much extraneous possibility you get into logical difficulties which we will see <pause dur="1.1"/> # so what we tend to do <trunc>i</trunc> when we're working with set theory is to fix some <pause dur="1.1"/> genuine <pause dur="0.3"/> set some bona fide set U <pause dur="0.3"/> that's big enough to contain <pause dur="0.2"/> all the sets that we're <pause dur="0.4"/> likely to consider <pause dur="0.3"/> in a given <pause dur="0.2"/> setting <pause dur="2.0"/> and # <pause dur="4.6"/> to call that the universe <pause dur="0.3"/> the the set everything happens in so in this case <pause dur="0.4"/> i've drawn the universe as a sort of square <pause dur="0.2"/> containing box <pause dur="0.2"/> and the assumption is that <pause dur="0.6"/> while i'm doing set theory <pause dur="0.8"/> everything will happen inside that box <pause dur="2.2"/> if you find the universe you chose is to small <pause dur="0.4"/> to begin with you can obviously enlarge it <pause dur="0.9"/> # so that # <pause dur="0.7"/> you include new sets if you have to <pause dur="0.7"/> so if we <pause dur="0.4"/> started out with the real numbers as the universe <pause dur="0.4"/> and we came upon an equation which had complex roots <pause dur="0.3"/> it might be convenient just to enlarge the universe <pause dur="0.4"/> to U <pause dur="0.5"/> and then you'd include those roots <pause dur="2.0"/> # the only

important thing is that once you've chosen your universe <pause dur="0.2"/> for a given <pause dur="0.5"/> discussion <pause dur="0.5"/> you'd better stick to it <pause dur="4.8"/> and once we've got a universe <pause dur="1.0"/> we can define <pause dur="4.4"/> a complement <pause dur="3.9"/><kinesic desc="puts on transparency" iterated="n"/> if we've got a universe then <pause dur="1.1"/> the complement denoted by <kinesic desc="indicates point on screen" iterated="n"/> this little exponent C <pause dur="0.8"/> is everything in the universe that's not in A <pause dur="0.5"/> # it's denoted by the <pause dur="0.2"/> shaded region <pause dur="0.4"/> and obviously <pause dur="0.7"/> what the complement is depends entirely on which universe you're working in <pause dur="1.8"/> so you could define <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="7"/> A-to-the-C <pause dur="1.1"/> with our other notation it's U-<pause dur="0.4"/>minus-A <pause dur="2.9"/> if you change your universe you obviously <pause dur="0.3"/> change your complement <pause dur="6.1"/><kinesic desc="reveals covered part of transparency" iterated="n"/> so i've defined the complement here <kinesic desc="indicates point on screen" iterated="n"/> to it's the shaded region in the Venn diagram <pause dur="2.9"/> and # <pause dur="4.7"/><kinesic desc="reveals covered part of transparency" iterated="n"/> i'm going to have <pause dur="1.4"/> a little break there to focus on <pause dur="0.6"/> a problem that's of <pause dur="0.3"/> some interest <pause dur="5.9"/> typically <pause dur="1.7"/> # <pause dur="1.3"/> people draw circles for their sets when they're using Venn diagrams <pause dur="1.0"/> so let <kinesic desc="indicates board" iterated="n"/> this blackboard be the universe <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="2"/> if i draw one set <pause dur="1.9"/> it divides the universe into two parts the <pause dur="0.2"/> inside <pause dur="0.9"/> and the complement outside <pause dur="3.9"/><kinesic desc="writes on board" iterated="y" dur="1"/> if i draw <pause dur="0.7"/> two sets <pause dur="2.2"/><kinesic desc="writes on board" iterated="y" dur="4"/> circles <pause dur="1.4"/> how many regions do i get <pause dur="2.3"/>

one <pause dur="0.3"/> two <pause dur="0.2"/> three <pause dur="0.4"/> and everything not in those four <pause dur="0.6"/> so with two sets i get four <pause dur="3.7"/><kinesic desc="writes on board" iterated="y" dur="1"/> regions in the <pause dur="0.3"/> plane <pause dur="5.8"/> how many do i get if i <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="3"/> have three sets <pause dur="3.0"/> i'm looking at all possible <pause dur="0.6"/> intersections <pause dur="0.9"/> of these various sets <pause dur="1.5"/> and i'm looking at the connecting regions i get <pause dur="1.2"/> let's do a count the outside one two three four <pause dur="0.2"/> five six <pause dur="1.5"/> have i left some out the outside one <pause dur="0.2"/> two three four five <pause dur="0.3"/> six <pause dur="0.2"/> seven <pause dur="0.7"/> and the outside was eight <unclear>side two</unclear><pause dur="3.8"/><vocal desc="clears throat" iterated="n"/><pause dur="0.9"/> what do you think i get in general <pause dur="1.9"/> well <pause dur="1.7"/> you've guessed it you should get <pause dur="0.6"/> two-to-the-N <pause dur="0.8"/> sets because every time you <pause dur="0.2"/> draw a new set <pause dur="0.6"/> you divide in half <pause dur="0.4"/> each of the existing sets <pause dur="2.3"/> my question is if you stick to circles <pause dur="1.1"/> do you <pause dur="0.2"/> always get <pause dur="0.8"/> two-to-the-N regions <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="1"/> so i want you to look at the region <pause dur="2.3"/> for four sets <pause dur="0.8"/> and decide whether now you get <pause dur="0.7"/> get your paper out get your pen out <pause dur="0.3"/> and see if you can get <pause dur="1.6"/> sixteen regions with four circles <pause dur="23.7"/><event desc="attempts task set" iterated="y" n="ss" dur="uncertain"/><kinesic desc="writes on board" iterated="y" dur="2"/></u><gap reason="break in recording" extent="uncertain"/> <u who="nm0863" trans="pause"> that's one thing i didn't make clear <pause dur="2.7"/> was that i was talking about what i call simply connected regions <pause dur="0.6"/> those are the regions <pause dur="0.4"/> that if you were a fly <pause dur="0.3"/> you could move

around in you get chambers here <pause dur="0.3"/> without crossing any of the <pause dur="1.1"/> boundaries and the general consensus here is that there are fourteen <pause dur="0.5"/> rather than sixteen does anyone know why </u><pause dur="2.1"/> <u who="sm0864" trans="pause"> <gap reason="inaudible" extent="5 secs"/> </u><u who="nm0863" trans="overlap"> yeah <pause dur="0.2"/> i mean the reason is that two circles intersect <pause dur="0.4"/> in how many points </u><pause dur="0.9"/> <u who="ss" trans="pause"> two </u><pause dur="1.5"/> <u who="nm0863" trans="pause"> two exactly two no more than two <pause dur="1.0"/> and therefore <pause dur="0.2"/> by the same argument <pause dur="0.6"/> as the pizza problem <pause dur="0.6"/> the number of new regions you <pause dur="0.5"/> can introduce by <pause dur="0.3"/> drawing a new circle <pause dur="0.4"/> is strictly limited so you can't <pause dur="0.4"/> using circles <pause dur="0.4"/> get all the regions you'd like to <pause dur="0.3"/> to show <pause dur="0.2"/> the disposition of N sets in its greatest generality <pause dur="0.8"/> so <pause dur="0.8"/> if you allowed yourself the <pause dur="0.7"/> luxury of rectangles <pause dur="0.3"/> rather than <pause dur="0.7"/> circles <pause dur="0.6"/> could you then do it now <pause dur="0.2"/> i'm <pause dur="0.4"/> offering a small prize <pause dur="0.5"/> to a written solution to the first one handed in which is <pause dur="0.4"/> lucid <pause dur="0.6"/> mathematically correct <pause dur="0.4"/> and is either a proof <pause dur="0.8"/> or a counter-example <pause dur="0.8"/> the object is <pause dur="0.6"/> to get <pause dur="0.6"/> two-to-the-N <pause dur="0.4"/> connected regions <pause dur="1.6"/> and sets <pause dur="1.4"/> okay and for your delectation <pause dur="0.9"/> we have <pause dur="0.3"/> a number <pause dur="0.3"/> and your <pause dur="0.2"/> object here <pause dur="0.3"/> is to come up with a plausible reason <pause dur="0.2"/> why <pause dur="0.2"/> that number of the

week is an interesting number <pause dur="1.9"/> and i will <pause dur="0.4"/> tell you next week <pause dur="1.0"/> okay <pause dur="1.0"/> any questions at this point </u><pause dur="4.3"/> <u who="sm0865" trans="pause"> now i don't see how you can <pause dur="0.5"/> <gap reason="inaudible" extent="3 secs"/></u><pause dur="1.1"/> <u who="nm0863" trans="pause"> you don't see how the </u><u who="sm0865" trans="overlap"> # are you saying that all rectangles <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.3"/> <u who="nm0863" trans="pause"> if instead of using circles you use rectangles could you represent <pause dur="0.8"/> the two-to-the-N connected regions that you get <pause dur="0.7"/> by intersecting </u><pause dur="1.1"/> <u who="sm0865" trans="pause"> so how can you <gap reason="inaudible" extent="2 secs"/> </u><pause dur="0.5"/> <u who="nm0863" trans="pause"> well <pause dur="0.2"/> this was a counter-example with circles <pause dur="0.9"/> here you you can actually prove that you can't get more than fourteen <pause dur="0.4"/> by counting the intersections <pause dur="0.9"/> with the existing circles </u><u who="sm0866" trans="overlap"> <gap reason="inaudible" extent="1 sec"/> </u><pause dur="0.5"/> <u who="nm0863" trans="pause"> sorry </u><pause dur="0.2"/> <u who="sm0866" trans="pause"> <gap reason="inaudible" extent="2 secs"/> </u><pause dur="0.4"/> <u who="nm0863" trans="pause"> it means that you can never get sixteen <pause dur="0.2"/> i can </u><u who="sm0866" trans="overlap"><gap reason="inaudible" extent="1 sec"/> </u><u who="nm0863" trans="latching"> prove that you can never get sixteen regions which you should get <pause dur="0.6"/> if you're using <pause dur="0.4"/> circles <pause dur="0.8"/> so it may be <pause dur="0.5"/> that you can never get sixteen regions <pause dur="1.0"/> using <pause dur="0.6"/> rectangles <pause dur="0.4"/> or even if you can get sixteen it may fail <pause dur="0.2"/> at the next level thirty-two <pause dur="0.4"/> or it may fail at sixty-four so that's your problem <pause dur="0.3"/> can you always do it for all N <pause dur="1.3"/> or <pause dur="0.4"/> does it fail at some point with rectangles <pause dur="8.4"/> now <pause dur="1.1"/><kinesic desc="reveals covered part of transparency" iterated="n"/> there are lots of <pause dur="2.5"/> rules about intersections <pause dur="0.7"/> and <pause dur="0.2"/><kinesic desc="indicates point on screen" iterated="n"/> here are <pause dur="0.8"/> # <pause dur="0.2"/> two examples of these <pause dur="0.3"/> so-called identities for sets <pause dur="1.5"/> De Morgan's Laws <pause dur="5.6"/> when you fix some

universe as usual <pause dur="1.2"/> you take two sets <pause dur="0.5"/> and then one of De Morgan's Laws says the complement of the intersection is the union of the <trunc>comp</trunc> <pause dur="0.2"/> complement so <pause dur="0.2"/> the rule there is <pause dur="0.7"/> whenever you <pause dur="1.8"/> apply complementation to a <pause dur="0.2"/> <unclear>case</unclear> of two things <pause dur="0.3"/> with an intersection <pause dur="0.4"/> to <pause dur="1.1"/> complement each of the constituents and put a union <pause dur="0.3"/> and vice versa <pause dur="0.6"/> if you had a union row <pause dur="1.3"/> this gives you this gives you an algorithm <pause dur="0.4"/> for converting <pause dur="0.6"/> # <pause dur="0.9"/> the complement of a complicated expression using intersections and unions <pause dur="0.4"/> into <pause dur="0.4"/> # <pause dur="0.6"/> the intersection and union as complements <pause dur="2.5"/> so i'll just give you an idea <pause dur="0.4"/> of how <pause dur="1.0"/> you prove <pause dur="0.8"/> that <pause dur="1.9"/> two sets are equal <pause dur="0.9"/> so <pause dur="0.2"/> i'm going to prove the left-hand log <pause dur="0.8"/> and the point about that it holds <pause dur="0.6"/> for all sets A and B <pause dur="0.8"/> it's not dependent on their being a particular set <pause dur="0.4"/> and when something holds for all sets <pause dur="0.6"/> like that we call it an identity <pause dur="1.0"/> set theoretical identity so we've got to go <pause dur="0.6"/> we've got to do the double inclusion argument so we start out with the left-hand set <pause dur="0.4"/> and we take X <pause dur="0.2"/> in the complement <pause dur="0.4"/> of A-intersect-B <pause dur="3.0"/>

so X is <pause dur="1.0"/> not in <pause dur="1.5"/> the intersection so it's not in both A and B <pause dur="0.3"/> so X is not in at least <pause dur="0.5"/> one <pause dur="0.2"/> of A and B so it's either not in A <pause dur="0.4"/> or not in B so X is either in the complement of A or the complement of B or possibly both <pause dur="0.4"/> and therefore it's in the union <pause dur="1.1"/> of <pause dur="0.7"/> A-complement B-complement <pause dur="0.9"/> that proves <pause dur="0.4"/> that inclusion <pause dur="5.1"/> now you take the right-hand side you take an element of the right-hand side <pause dur="2.1"/> # the right-hand side of the first De Morgan Law will let us <pause dur="0.6"/> but if it's in the left-hand that's the left-hand side of this <kinesic desc="indicates point on screen" iterated="n"/> <pause dur="1.5"/> inclusion <pause dur="0.4"/> if X is in there <kinesic desc="indicates point on screen" iterated="n"/> <pause dur="0.3"/> then it's either not in A <pause dur="1.4"/> or it's not in B <pause dur="0.8"/> it's therefore never in both <pause dur="1.3"/> and hence it's <pause dur="0.3"/> not in the intersection <pause dur="0.9"/> so <kinesic desc="indicates point on screen" iterated="n"/> this is in <kinesic desc="indicates point on screen" iterated="n"/> there we've shown the inclusion <pause dur="0.3"/> both ways <pause dur="0.3"/> and that's <pause dur="0.2"/> the way we show two sets are equal <pause dur="1.8"/> if in doubt <pause dur="2.2"/> you <pause dur="6.3"/> # <pause dur="0.2"/> draw a picture <pause dur="8.2"/><kinesic desc="writes on board" iterated="y" dur="3"/> and # <pause dur="0.7"/> so the first the left-hand side of the first De Morgan Law <pause dur="0.5"/> is the complement <pause dur="3.7"/><kinesic desc="writes on board" iterated="y" dur="9"/> of # <pause dur="0.3"/> the intersection there's the intersection <pause dur="2.9"/> and the complement of that <pause dur="1.0"/> is everything <pause dur="1.7"/><kinesic desc="writes on board" iterated="y" dur="12"/> not <pause dur="5.3"/> # it's everything not inside there <pause dur="2.5"/><kinesic desc="indicates point on board" iterated="n"/> that's

the only <pause dur="0.3"/> bit that's not in the set <pause dur="0.7"/> if you take the right-hand side <pause dur="3.3"/><kinesic desc="writes on board" iterated="y" dur="2"/> you have to take the <pause dur="0.5"/> complement of A <pause dur="7.3"/><kinesic desc="writes on board" iterated="y" dur="1"/> the complement of A <pause dur="0.4"/> is everything <pause dur="4.3"/><kinesic desc="writes on board" iterated="y" dur="5"/> outside A <pause dur="8.1"/><kinesic desc="writes on board" iterated="y" dur="10"/> and the complement of B <pause dur="0.9"/> is everything <pause dur="1.1"/> outside B <pause dur="2.6"/> and anything that's not <pause dur="0.8"/> in the union of <kinesic desc="indicates point on board" iterated="n"/> those two the pink and the white <pause dur="0.3"/> is again <kinesic desc="indicates point on board" iterated="n"/> the intersection <unclear>of the two</unclear> so <pause dur="0.7"/> the two sets are equal <pause dur="0.6"/> in Venn terms as well <pause dur="1.3"/> but <pause dur="0.9"/> because you can't fully represent <pause dur="1.5"/> the <pause dur="0.8"/> sets using <pause dur="0.9"/> traditional Venn diagrams with circles you have to be very cautious about saying <pause dur="0.4"/> the Venn diagram <pause dur="0.3"/> is a proof <pause dur="0.5"/> it's okay up to three sets <pause dur="0.5"/> but you get into difficulty <pause dur="0.6"/> if you have more than <pause dur="0.4"/> three <pause dur="3.5"/><kinesic desc="reveals covered part of transparency" iterated="n"/> so <reading>De Morgan's Laws are examples of set theoretical identities <pause dur="1.3"/> they're equations that hold <pause dur="0.4"/> whatever sets are substituted for A and B <pause dur="1.4"/> we can think of the symbols <pause dur="0.3"/> A and B as variables that range over all possible sets <pause dur="0.4"/> an identity <pause dur="0.3"/> is an equation between two set theoretical expressions <pause dur="0.3"/> that hold for all values of the variable</reading> <pause dur="3.9"/><kinesic desc="changes transparency" iterated="y" dur="15"/> there are lots and lots of identities <pause dur="1.6"/> involving two sets <pause dur="8.3"/> # there are

some fairly obvious ones like the <pause dur="0.3"/> <trunc>u</trunc> the intersection of A and B is the same as the intersection of B and A the same with unions <pause dur="0.6"/> # not quite so obvious are these distributed laws <pause dur="0.5"/> A intersect the union of B and C <pause dur="0.3"/> is the union of A-intersect-B and A-intersect-C <pause dur="0.3"/> and similar one with <pause dur="0.4"/> unions and <pause dur="1.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> intersections interchange <pause dur="3.3"/> and this <pause dur="0.5"/> raises the principle of duality <pause dur="0.4"/> and if you have any <pause dur="0.6"/> identity <pause dur="1.3"/> # <pause dur="0.7"/> involving sets <pause dur="0.8"/> intersections and unions <pause dur="0.7"/> you will automatically get another <pause dur="0.2"/> identity <pause dur="1.1"/> interchanging all the <pause dur="0.7"/> unions and intersections <pause dur="0.9"/> so it's all <pause dur="0.2"/> from the first <pause dur="0.7"/><kinesic desc="indicates point on screen" iterated="n"/> distributed law there i swap over <pause dur="0.9"/> the unions and the intersections and i get another <pause dur="0.3"/> true identity <pause dur="0.3"/> and the reason for it <kinesic desc="indicates point on screen" iterated="n"/> here <pause dur="0.4"/> is spelled out <pause dur="0.4"/> using De Morgan's Laws <pause dur="1.4"/> i'll leave you to read that one in your own time <pause dur="1.2"/> because in the last <vocal desc="clears throat" iterated="n"/><pause dur="0.3"/> few minutes <pause dur="1.4"/> i want to come on to <pause dur="0.7"/><kinesic desc="reveals covered part of transparency" iterated="n"/> some <pause dur="0.5"/> logical problems that you <pause dur="0.3"/> get if you <pause dur="0.6"/> allow <pause dur="1.2"/> too many sets to <pause dur="0.2"/> arise <pause dur="2.6"/> it's it's worth knowing that # <pause dur="0.6"/><kinesic desc="indicates point on screen" iterated="n"/> this book which i will <pause dur="0.8"/> circulate on a list of recommended books for the

course in due course <pause dur="0.7"/> it's a <trunc>di</trunc> it's <pause dur="0.2"/> rather large book called Discrete Mathematics with Application <pause dur="0.3"/> it's pretty strong on set theory <pause dur="0.3"/> and it has a very comprehensive list <pause dur="0.4"/> of all the basic identities for sets <pause dur="0.4"/> of this kind <pause dur="3.9"/><kinesic desc="reveals covered part of transparency" iterated="n"/> yep <pause dur="0.5"/> there is a snake <pause dur="0.6"/> lurking in the grass <pause dur="1.3"/> # <pause dur="1.1"/> incidentally if you like this kind of <pause dur="0.7"/><kinesic desc="reveals covered part of transparency" iterated="n"/> there was a splendid book <pause dur="0.8"/> # <pause dur="0.5"/> by <pause dur="1.0"/> Raymond Smullyan <pause dur="0.6"/> the title of the book is What is the Name of this Book <pause dur="0.7"/> and # <pause dur="1.5"/> it has lots of # <pause dur="0.6"/> splendid examples of <pause dur="1.2"/> logical and verbal paradoxes <pause dur="0.2"/> anyway consider this particular one you've got a certain town there's a barber <pause dur="0.7"/> who shaves <pause dur="0.5"/> all the men <pause dur="0.5"/> who do not shave themselves <pause dur="0.3"/> I go there <pause dur="1.2"/> but no one else <pause dur="0.4"/> if they shave themselves the barber doesn't shave <pause dur="1.3"/> so <pause dur="0.7"/> let the universe U <pause dur="0.8"/><kinesic desc="changes transparency" iterated="y" dur="16"/> consist of all the men in that town <pause dur="5.8"/> and let the set A <pause dur="2.6"/> consist of all the men <pause dur="0.9"/> who shave themselves <pause dur="2.4"/> by definition of the universe <pause dur="2.1"/> either someone shaves themselves <pause dur="1.5"/> or they don't <pause dur="2.3"/> and the question is <pause dur="0.5"/> does the barber belong to A <pause dur="0.6"/> or A-complement if he belongs to A and he shaves himself <pause dur="0.4"/><vocal desc="clears throat" iterated="n"/><pause dur="0.4"/> but that's not

possible because he only shaves men not in A <pause dur="0.3"/> on the other hand if he belongs to the complement of A <pause dur="0.2"/> by definition of <pause dur="0.4"/> A <pause dur="0.7"/> he shaves himself and so he belongs to A <pause dur="0.5"/> so you have the contradiction that the barber is in <pause dur="0.4"/> the set and its complement <pause dur="0.9"/> which is the set to the empty set <pause dur="1.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> # <pause dur="1.7"/> there are two ways to escape <pause dur="0.7"/> from this paradox <pause dur="0.7"/> you can either assume the barber had a beard <pause dur="1.2"/> or <pause dur="0.5"/> you can point out that # <pause dur="0.7"/> # <pause dur="0.9"/> the paradox is just a logical <pause dur="0.4"/> booby trap <pause dur="0.5"/> # because it's not possible <pause dur="0.8"/> for the statement <pause dur="0.7"/> the barber shaves all men who do not shave themselves <pause dur="0.6"/> and shaves no one else it's not possible for that be a true statement <pause dur="0.7"/> if you allow <pause dur="0.7"/> the barber in the universe <pause dur="0.4"/><kinesic desc="reveals covered part of transparency" iterated="n"/> so that was <pause dur="1.1"/> # a trick <pause dur="1.3"/> but a more serious problem <pause dur="0.2"/> # was pointed out by Bertrand Russell who <pause dur="0.2"/> spent years <pause dur="0.5"/> # trying to <pause dur="1.6"/> # flog himself to death in <pause dur="0.6"/> using <pause dur="0.6"/> set theory <pause dur="0.3"/> mathematical logic to <pause dur="0.8"/> prove all mathematics <pause dur="1.4"/> # <pause dur="1.5"/> he's sometimes defined <pause dur="0.2"/> as the <pause dur="1.5"/> philosopher who shaves all people who do not shave themselves <pause dur="0.9"/><kinesic desc="reveals covered part of transparency" iterated="n"/> # <pause dur="0.2"/> but his <pause dur="1.1"/> problem was the following <pause dur="0.3"/> his paradox <pause dur="1.6"/> take the

universe to be the set of all sets <pause dur="1.2"/> and then you can define the subset of that universe <pause dur="0.5"/> # to consist of all sets that don't contain themselves <pause dur="1.3"/> i point out that # <pause dur="1.1"/> the set of integers that's not an integer so <pause dur="0.8"/> Z is not in Z <pause dur="0.3"/> Z does not contain itself <pause dur="0.3"/> so Z <pause dur="0.2"/> that set <pause dur="0.6"/> is in our set A <pause dur="0.3"/> on the other hand the set of all sets is a set <pause dur="0.6"/> so it belongs to itself <pause dur="0.5"/> and so <pause dur="1.1"/> the complement of A contains U so there are both kinds of sets those that contain themselves <pause dur="0.4"/> those that don't <pause dur="6.6"/><kinesic desc="reveals covered part of transparency" iterated="n"/> now the question <pause dur="1.2"/> # <pause dur="0.2"/> we have to ask is <pause dur="0.8"/> does A <pause dur="1.0"/> the set of all sets that don't contain themselves does it contain itself <pause dur="0.3"/> or not does it belong to A or the complement of A <pause dur="1.3"/> well if A <pause dur="1.4"/> belongs to itself <pause dur="1.3"/> then <pause dur="0.8"/> what does A define <pause dur="0.9"/> it <reading>must be one of those sets that do not belong to itself in other words A is not in A</reading> <pause dur="1.1"/> so it's in the complement <pause dur="0.4"/> <reading>on the other hand if A

belongs to the complement then A belongs to itself <pause dur="0.5"/> or it symbols A is in A</reading> either way we get A in A-<pause dur="0.5"/>intersect A-complement which is the empty set so that's a serious <pause dur="0.3"/> objection <pause dur="0.6"/> if we allow the set of all sets <pause dur="0.6"/> to be a set <pause dur="1.8"/><kinesic desc="reveals covered part of transparency" iterated="n"/> so we don't <pause dur="0.6"/> so we have to be much stricter about what sets are allowed in the scheme of things <pause dur="0.8"/> and # there are ways of <pause dur="1.0"/> for getting round this <pause dur="0.6"/> basically you start with the <trunc>em</trunc> <pause dur="0.2"/> the empty set and build up <pause dur="0.7"/> and <pause dur="0.2"/> what you can get by that process <pause dur="0.9"/> is what's allowed and you don't end up with a set of all sets <pause dur="0.9"/><vocal desc="clears throat" iterated="n"/><pause dur="2.8"/> # <pause dur="0.6"/> poor old Russell <pause dur="0.6"/> got his comeuppance because <pause dur="0.7"/> in nineteen-thirty-one <pause dur="1.1"/> <reading>Gödel showed it wasn't possible to prove rigorously that mathematics is free from contradictions</reading> <pause dur="0.8"/> so <pause dur="0.8"/> we always have to live with a cloud of doubt over our heads