Axioms of set Theory - Lec 02 - Frederic Schuller

[@tinski]

What a great lecturer! A very welcome relief from the more common style of "supervised reading of slides" I have to endure at uni.

[@account1307]

TIMESTAMPS 1. Axiom on the epsilon relation - 10:20 2. Axiom on the existence of an empty set - 21:45 3. Axiom on pair sets 40:30 4. Axiom on union sets 48:04 5. Axiom of replacement 59:39 5.5 The axiom of replacement implies, but is not implied by, the *principle of restricted comprehension* 1:06:14 (Note: The PRC is elevated to the status of an axiom in other treatments and is called the axiom of subsets/speification/seperation) 6. Axiom on existence of power sets 1:26:53 7. Axiom of infinity 1:29:18 8. Axiom of choice 1:36:31 9. Axiom of foundation 1:46:17

[@twwc960]

The Axiom of Infinity as shown in this video is a bit old fashioned. It is the version that was first put forward by Zermelo, but almost no one uses it anymore. A more modern formulation is this: "There exists a set which contains the empty set, for all elements y in the set y∪{y} is an element of the set." With this formulation, the natural numbers look like this: {∅, {∅}, {∅,{∅}}, {∅,{∅},{∅,{∅}}}, ... } This may appear more complicated than that shown in the video, but it is much nicer for two reasons: first, the natural number n is represented by a n element set rather than a one element set, and second, it generalizes much more naturally to the system of transfinite ordinal numbers, where the successor operation s(x)=x∪{x} applies to transfinite ordinals as well as natural numbers , and the natural numbers have the same defining property as the ordinals: they are transitive sets strictly well ordered by the ∈ relation.

[@gleisonstanlley2134]

@@matthewcory4733 I don't think you got the point.

[@musicarroll]

This was the content of Von Neumann's PhD thesis, which he wrote while simultaneously getting a master's degree in engineering to satisfy his father's wishes that he learn something more practical than mathematics.

[@pavlenikacevic4976]

The whole construction of the theory is conventionally done differently than his approach, e.g. the concept of equality is defined in the underlying predicate logic itself, and his definition of equality actually becomes the axiom of extensionality. His approach is not as formal but works just as well for the purpose of this course

[@ebsen1023]

1:00:08 "Not a functional relationship, but a functional relation. So we may assume its existence." The man even got time for jokes. Amazing

[@tobiassugandi]

I don't think that's what he meant

[@mikeCavalle]

excellent -- presentation of material, audio quality, camera work, marvelous

[@finaltheorygames1781]

Man, I love this guy and man I love this guys lectures. I can understand things clearly and you speak english and thank you for uploading.

[@samuelabreu4349]

I love this class and I watch it once in a while. Thanks, professor.

[@synaestheziac]

I’m watching it for the third time in the past year or two

[@spyphyfarnsworth6050]

l like german precision. It looks in Deutschland even explanation of Schrodinger equation is started from axiomatic set theory)

[@bighands69]

Set theorem comes from philosophy that was practiced in the classical world.

[@theunknownscientist3249]

That's just sad for us non-german mathematicians, we don't get to understand how it all connects.

[@mastershooter64]

Lmao

[@joshuameyers7962]

Mistake at 1:12:35. "There exists y in m such that P(y)" is formalized as "there exists y such that (y in m AND P(y))". Schuller mistakenly puts "IMPLIES" instead of "AND". "IMPLIES" is used in the similar context for the universal quantifier, but for the existential quantifier, "AND" is used.

[@joshuameyers7962]

Update: At 1:20:45 he fixes the error after a student points it out.

[@subhrajitdasgupta3868]

Thanks a lot professor I was searching for a detailed discussion on this topic but none of the videos I found earlier was satisfactory except this one. I'm a student of MS Statistics but I have developed interest on these parts of mathematics which may not be directly requiring in Statistics and reading books on this topic was not really being helpful to build a good concept on this, in a word your video saved me. Thanks!

[@Gabriel-mf7wh]

You said Russell used naïve set theory in his book, but that's not true. He defined an alternative foundation, called Type Theory, because he knew naive set theory (used by Frege) would lead to the paradox.

[@JK-vf1dt]

True. Naive theory was initiated by people like Canton. Russell's paradox was a important step in realizing the mistakes.

[@robertfarrell5871]

@@JK-vf1dt Yes, Dr Schuller seems to conflate Frege, whom he does not mention, but who wrote those "two volumes", with Whitehead and Russell, whose PM had three volumes. Russell stated his paradox in a letter to Frege, just as Frege was about to publish the second volume of his Grundgesetze.

[@tim-701cca]

Really thanks to this professor, though he may not see the comments. He is the first person that taught me a lot of math which I may learn some of them before. That’s exactly the way I want to learn logic and set theory. I feel I got a lot while watching two hours here. I still have some questions but it is a good start to me😊

[@PrimitiveBaroque]

I think the professor's chalk is an infinite set. lol I love these lectures they are excellent.

[@rommelborillo6216]

Hagoromo chalk

[@thinkingpi]

I have no way to describe the beauty of this lecture (short and sweet). Fabulous.....

[@jovaha]

ZFC set theory: If it looks like a duck, swims like a duck, and quacks like a duck, then it is a set.

[@mohammedkhalili1154]

jovaha why??

[@jovaha]

because when using ZFC as a foundation for math all objects - functions, relations, numbers,... - are really sets.

[@mohammedkhalili1154]

Yet a set in this theory is not well defined, one may consider strange things as a set

[@NewCalculus]

But how can you tell it is a duck when you don't know what the hell it is?!! See 2:26 you huge baboon!! "There will be no definition of e or of what is a set" If it looks like shit, smells like shit, and feels like shit, then it must be shit.

[@antonio7326]

A set is like being, you do not know what being is but you know who being is.

[@LaureanoLuna]

11:20 I wouldn't list the constraint on the membership relation as an axiom because it is rather an interpretation, the choice of a domain.

[@ChristopheDoloire]

Great talk. Very clear, instructive, and free :) Thanks a lot sir !!

[@mohammedkhalili1154]

I am enjoying your lecture, all i can to thank you is to write it, so thank you

[@addisonsmith9479]

1:60:14 "I am going to write it down now and you will recognize it" I love German mater of fact-ness

[@henrywang6931]

Regarding the left sock right sock intuition for the axiom of choice, didn't we prove earlier that {x,x}={x} ? So if left and right sock are indistinguishable then {left sock, right sock} = {sock}, there is no need to use axiom of choice. What's wrong with my understanding?

[@paulelliott9487]

His axiom that he mentions first, the first E axiom, is not really an axiom at all. It can not be expressed in the predicate calculus. It is really a meta-mathematical statement about the meaning of the intended model. In ZF we assume we are talking about "sets" whatever they are. It is not necessary to avoid Russell's paradox. That is achieved in ZF by removing the unrestricted axiom of comprehension. Such an axiom does not occur in any of my books on set theory!

[@MohandeepSingh007]

Thanks for clearing my doubt without asking! :D

[@solidsnake4022]

What do you mean by it can't be expressed in predicate calculus? From my understanding of predicate calculus the following statement "There exists a set that has no sets in it" can be expressed in predicate calculus.

[@gleisonstanlley2134]

Exactly.

[@nychan2939]

Do you think "both x and y are sets" is not a necessary condition? I can think of many examples that x is not a set and we still have x∈y or x∉y. Sorry, I'm so stupid.

[@millerfour2071]

11:52, 16:17, 26:25, 38:12, 46:28, 49:54, 1:04:32, 1:10:05, 1:16:24 (for all x exists unique y, R(x,y)) , 1:18:53 (union cannot be constructed from PRC because it restricts the elements in one set at the beginning), 1:25:14, 1:28:10, 1:34:37, 1:41:52, 1:43:53, 1:47:44

[@tim-701cca]

After reading some books about logic and Zfc, Some points noted: Set, membership symbol Equality symbol Axiom of epsilon symbol (not required) Axiom of extensionality (added) 1:17:09 also follows from axiom of empty set EE PURP IC F

[@sharafhussain1987]

I am also struggling with the idea of presenting the first axiom as he has done as opposed to including axiom of extensionality which is standard to zfc theory. What was his motivation behind it?

[@antoniomantovani3147]

Clear, essential, precise, and very usefull

[@ANSIcode]

I'm missing some kind of treatment of the "definition problem". As in: How do you go from something you know exists and is unique to having a simbol for it and being sure all its permitted uses can be made sense of?

[@fullfungo]

Search up “extension by definitions” and “conservative extension”. The basic idea is that definitions do not add any “power”, and any proof that uses the definitions can be rewritten so that it does not.

[@garywpearson1955]

In the introduction to his original paper on Gödel's incompleteness theorem, Gödel included an example related to the "Barber paradox" with respect to the logic set developed by Russell and Whitehead in Principa Mathmatecia, called PM. Now, if this example can not be proven nor disproven, it would seem to not fit the given definition of a proposition. Hence, the question arises whether or not Gödel's example in his introduction actually applies to ZFC logic. If we exclude all statements that don't fit our definition of propositions (those statements that are neither True nor False) then the example given by Gödel would seem not to apply to ZFC set theory. I know many others have looked at this and concluded that his incompleteness theorem does indeed apply to ZFC logic. So, what am I missing here?

[@worldboy9684]

So it turns out mathematics actually has a "42"... Top five most enlightening experiences ever. Very highly recommended.

[@LaureanoLuna]

It seems to me that his proof that Replacement implies Separation (i.e. what he calls PRC) at about 1:15 requires Choice in order to choose the marked member of m satisfying P(x). If so, what he proves is that Replacement plus Choice implies Separation.

[@andrewn635]

Correct you get a golden star sticker

[@6_6740]

I think you can rewrite his functional relation R so that it doesn't need y-hat anymore: R(x,y) := (P(x) ^ x = y) v (NOT P(x) ^ NOT(For all z in m: P(z) => z =/= y) Read: Either P(x) is true, then y must equal x, or P(x) is not true, than y must be a set such that it is not true that all z in m where P(z) is true are not equal to y. I think it is clear that this still implies that there has to be some y in m chosen for which P(y) is true - but at least we do not have to choose this y ourselves. We just require that- Ohhh noo i just realized that this R(x,y) isn't functional. Haha ok nevermind. At this point I would agree with you that he actually needs the Axiom of Choice for this proof. Good catch! :)

[@martinepstein9826]

You only need the axiom of choice to make infinitely many choices. By ordinary predicate logic if there exists y such that P(y) then you can set y-hat such that P(y-hat).

[@connorfrankston5548]

Thank you Dr. Schuller, these lectures are great! I wonder if you have anything more to say about the axiom of choice considering the Banach-Tarski paradox and Lebesgue immeasurable sets. Is there some major downside to using the weaker axiom of dependent choice and avoiding the Banach-Tarski paradox and immeasurable sets? Is there any analogue of the well-ordering theorem equivalent to DC instead of AC?

[@juliensorel1427]

On 1:19, one needs to identify im_R(m) with the ``set of interest" somehow. It is the ``consists" part that is missing ... and then conclude it is actually a set.

[@samuelmersha5530]

This guy knows what he is talking about.

[@callmedeno]

Anyone know of a good book for this early part of the course that goes through the axioms with rigour and deep explanation?

[@steyndewet1191]

Coolest black board and cleaning system - have not seen that.

[@twwc960]

You forgot one very important axiom: the Axiom of Extensionality. It can take two forms, depending on whether or not the notion of equality is already included in your system of logic. If it is, the axiom of Extensionality simply says that if two sets have the same elements, they are equal. Because equality is pre-existing, it has all the properties that equality should have: it is an equivalence relation (reflexive, transitive, and symmetric) and it has the substitution property: if a=b then any or all occurrences of a in a proposition can be replaced with b without changing the truth or falsity of the proposition. If equality is not included in the system of logic and is instead defined, as you have done, by saying two sets are defined to be equal if they have the same elements, you still need an axiom to guarantee the substitution property of equality. For example: consider the three sets d, e, and f, and suppose you construct two sets a = {d,e,f} and b={d,e,f}, and let c={a}. Question: is b∈c? Without an axiom of extensionality the answer is no. It may be that a=b, but = doesn't come with all the properties that it should. So we could have a=b, but a∈c while b∉c. So when equality of sets is defined this way, the axiom of extensionality takes the following form: "If two sets have the same elements (i.e. they are equal) then they are members of the same sets." This guarantees that equality has the substitution property, since terms that are sets can only occur to the left or right of an ∈ symbol in a proposition, and the definition of equality means equals can be substituted for equals on the left, and the axiom of extensionality guarantees they can be substituted on the right as well. See the Wikipedia page on the Axiom of Extensionality for a discussion of this issue.

[@gonzaklo]

"and the definition of equality means equals can be substituted for equals on the left, and the axiom of extensionality guarantees they can be substituted on the right as well" : exchange left and right.

[@twwc960]

Yes. You're absolutely right. Thank you.

[@gonzaklo]

Thank you also for a very clear explanation!

[@yongminpark5909]

After accepting the axiom of (unordered) paring, one can easily arrive at the form of extension axiom you said. The following steps are for proving it. Prop 1. For two sets x and y, x = y iff {x} = {y}. (Note. recall {x} := {x, x}. In fact, this is essential step. One can notice that this is a weaker form of the extension axiom what we now prove.) Prop. 2 For all x, z, x in z iff {x} is included in z. (Note. recall the def of inclusion relation) Prop 3. (extension axiom, another version) For all sets x, y, x=y implies r(x,y), where the predicate r is defined by r(x,y) : for all z, x in z iff y in z. pf. Assume x = y, then {x} = {y}. Also, assume x in z, then {x} is included by z. Therefore, {y} is included by z from the def of '='. Thus, we have the conclusion that y in z. Similarly, one can complete the proof. [] If one find some error in this comment, please let me know.

[@twwc960]

The problem in your proof is Proposition 1. Namely that x=y iff {x}={y}. Remember, that we're not assuming equality to be predefined, so we can't assume the substitution property of equality. When we say x=y in this situation then, we mean merely that x and y are two sets which have the same elements and nothing more. So, suppose x=y. How do we prove {x}={y}? Well, we would start by asking, do {x} and {y} have the same elements? Certainly, x∈{x}, by definition. At this point, we might be tempted to say that since x∈{x} and x=y then y∈{x}, but we cannot do so. The substitution property of equality has not been established. All we know at this point is that x and y are two sets which contain the same elements. It doesn't follow that {x}={y}, at least not without further axioms.

[@merveillevaneck5906]

lol....i actually had a nightmare about the coming of aliens after learning the definition of the intersection of a set with the universal set....and not quite understanding it

[@FadiAkil]

1:45:47 very wrong. The Axiom of Choice is still highly suspicious, especially when, e.g., applied to uncountable sets. As Émile Borel once said: "Any argument where one supposes an arbitrary choice to be made an uncountably infinite number of times ...[is] outside the domain of mathematics." For him & many other professional mathematicians, it verges on the mystical. That's why it's considered a "dark" axiom. 1:38:45 (moreover it leads to Banach-Taraski Pardox). Not due to Russell's intuition that's comparably trivial/less problematic.

[@NewCalculus]

Set theory is NOT mathematics.

[@warwolt]

I think it was maybe misstated, it's now part of standard mathematics, even if it has dubious philosophical implications

[@dineafkir5184]

Amazing lecturer... I truly enjoy it

[@조수정-w6m]

Thank you for this lecture!

[@aeroscience9834]

I don't know what elementary school he went to, but mine never came close to talking about sets

[@kockarthur7976]

I think it's actually a joke (hyperbole).

[@ZhiSiZhanShi]

I'm not German, but I remember getting some basic notions of set theory in the first or second year of elementary school, mostly illustrated using Venn diagrams. Basic concepts such as union and intersection were introduced graphically using those.

[@vitjastrum]

In my first year of elementary school, I learned how to count to 20. Are you sure you're not talking about secondary education (starting at age ~10).

[@ZhiSiZhanShi]

It really depends on the country I guess. In my country secondary education starts at 11-12, and I definitely learned it in primary school. If I had to guess, probably around the age of 9. Of course, education systems evolve over time (I'm talking over 20 years ago here), and things may be different nowadays.

[@Ranginor]

@@kockarthur7976 Actually it isnt. There was a Time in Germany where Math in school started with set Theory. Today this isnt practiced anymore but some classes did that.

[@synaestheziac]

This may have been said already but there is a small logic mistake around 1:12:50. “There exists y in m such that ...” should be formalized with a conjunction, not an implication, so that it reads “ there exists y in m AND...” . The implication is too weak because it can be made true by ex falso quodlibet.

[@synaestheziac]

Oh haha I finally got to 1:22:00 or so when a student points out the mistake. Schuller correctly switches the existential to a universal quantifier, then correctly defines the existential as the negation of the universal of the negation, but he doesn’t give the simplified version I mentioned above. Just to be clear: Ex € m : P(x) is, by definition of E, equivalent to ~Ax € m : ~P(x), which is, by definition of Ax € m, equivalent to ~Ax : (x € m => ~P(x)), which is, by definition of E, equivalent to Ex : ~(x € m => ~P(x)), which is, by the truth table for =>, equivalent to Ex : (x € m ^ ~~P(x)), which is, by the law of double negation, equivalent to Ex : (x € m ^ P(x)), which is the simplest and most straightforward equivalent to Ex € m : P(x) (Key: “E” is the existential quantifier, “€” is the element relation, and “~” is negation.)

[@Kralasaurusx]

@@synaestheziac Thanks for the explanation. That threw me off too - I paused the video there because it made me question everything I thought I knew, and spent a while pondering whether there was something I was missing, or if it should in fact be ∧ instead of ⇒. Sanity restored.

[@matthewmasood]

You are the best professor ever

[@KushagraSachanIITBHU]

What reference, if any(?), is Prof. Schuller using for these set theory lectures? This presentation is highly idiosyncratic, with the first axiom of ZFC presented here being highly non-standard (to the extent of not being present in any of the sources I could reach via searches over the web). The usual Axiom of Extensionality-which is related with the Principle of Restricted Comprehension is the one presented everywhere as being the prescription for what counts as a set-however, this principle itself appears as a consequence of the Axiom of Replacement, in Prof.'s exposition. The fact that Prof's lecture is self-contained in its logic is fine, but not being able to reconcile its (relatively topsy-turvy) presentation with standard texts present elsewhere is really throwing me off. Any comments clarifying upon my doubt(s) are extremely welcome!

[@GianlucaUK]

He also gave a weird definition of equality as a set property and not as an element of the language. The fact that two sets are equal if they are a subset of each other is not a definition, but a theorem that you prove using the axiom of extensionality. You’re right, it looks kind of self-contained but totally non-standard. Also, there is a total mixture of mathematics and metamathematics. The thing he defines as the first axiom of Zermelo Fraenkel cannot be an axiom because it’s a descriptive statement and not a sentence of the language of set theory.

[@52wtf]

Do anyone know what the first axiom presented at 10:30 is called? I have tried searching for things "membership relation axiom" or "epsilon-relation axiom", but i haven't been able to find anything. Also the axiom hasn't been in any list on the ZFC's that i have found online.

[@beppenonantola216]

Extensionality axiom

[@beppenonantola216]

True. you're right

[@musikinspace]

28:00 If "ex falso quodlibet" then it seems you can also prove that two empty sets are not equal. Wouldn't that be a contradiction?

[@scollyer.tuition]

I was thinking precisely the same thing. I'm not at all sure what rules this out - I'm no logician though, so it may be something trivial.

[@samedy00]

If we replace (y *in* x) with (y *not in* x) here, the statement will be, of course, true by "ex falso quodlibet". But then from this statement we can not conclude that x' *is not a subset of* x. Thus, we can not prove, that x' and x are not equal.

[@clivegoodman16]

Surely the intersection is not defined on the empty set but is defined on non empty sets.

[@nrrgrdn]

Very clear explanations!

[@ElizaberthUndEugen]

How is it possible we can use functions (functional relations in the video) to prove statements about set theory while we are currently constructing the axions of set theory when function are defined on sets? This... feels like circular dependency. And regrading the function R(x,y) (meaning xRy, as I understand it?) is basically a function by cases R(x) : if P(x) then x else y^, right? But... why are we allowed to do that at this point? What formal basis is there that allows this construction?

[@steliostoulis1875]

ElizaberthUndEugen functional relations have nothing to do with sets

[@Esloquees]

1:16:34 (P(x)^x=Y) [this is the part that defines the existence of such thing] or (¬P(X)^y'=y) [this part defines everything that is not R(x,y)]

[@spotted756]

In our manuals sets are described as a primitive concept which doesn't get defined. Then we start going through ways of representing sets. I dunno why Google showed me this video after 6 years. :D

[@flixerstudios1862]

What textbook/notes/problems are useful to use along this course?

[@davidwright5719]

As he presents the first (E) axiom, it’s just the tautology p | !p. As he writes third (P) axiom, he writes u=x | u=y when he means uεx | uεy.

[@Esloquees]

1:13:20 the statement on the board is valid for every contradiction Ø

[@shivamfaraday1752]

Find the lecture notes here: drive.google.com/file/d/1nchF1fRGSY3R3rP1QmjUg7fe28tAS428/view?usp=sharing

[@taghreednaser1886]

Μany thanks

[@lehu8529]

Could somebody explain to me why the axiom of foundation excludes sets that contain themselves? At first glance it looks to me like a = {a} is not allowed; but what about a = {a, {ø}}? After all a is a non-empty set which contains an element, in this case {ø} which has none of its elements in common with a (ø is not in a)

[@lehu8529]

I think I got it.. from a set X which contains itself we can always construct the power set P(X) and then by the axiom of schema specification we can write: B := {s ε P(X): X ε s}.. this B will violate the axiom of regularity, since none of B's elements can be disjoint to B.. thus a “set” containing itself can't be a set!

[@tim-701cca]

@@lehu8529 It is late but it seems not correct. Try to use axiom of pairing , and foundation. The only element of {A} is A and hence A is disjoint from {A}. So there is no set A that is an element of itself.

[@phee4174]

I'm confused, shouldn't the union of a set ( a , b) , where b's elements are just c and d, (and a,c,d are sets) be a set whose element are the sets c and d, and all the elements of set a, not all the elements of c, all the elements of d, and all the elements of a?

[@Ashm00r]

Great lecture, thanks.

[@y031962]

thanks for posting this. I have one question when y element x, is a predicate that returns false or true. So y not-element x is not a predicate. Am I right?

[@jdbrown371]

Axiom of Choice 1:36:00

[@nychan2939]

I don't know where the epsilon relation came from. Do you think "both x and y are sets" is not a necessary condition? I can think of many examples that x is not a set and we still have x∈y or x∉y.

[@jelmar35]

These axioms define what a set is. If your example contradicts these axioms, then by definition, your example is not a set. Sets in axiomatic set theory are their own thing and are not the same thing as our intuitive notion of a set or collection

[@siddean5048]

Amazing lecturer

[@KipIngram]

That is an amazing chalkboard.

[@jdbrown371]

1:10:30 proof that Axiom of Replacement implies the Principle of Restricted Comprehension. A.R. --> P.R.C.

[@KipIngram]

47:00 - You know, I can't help commenting that the reason we *have* these "old prejudices" that we need to "get rid of" is because that's how we are *taught* when we are young. Why? Why don't they just teach things to us right to start with? Then the prejudices would never develop. What is the *purpose* of teaching a corrupted version of the material first? I'm having such thoughts again and again as I continue my studies. Why do we teach high school and early college students classical physics as though it is "exact," and then force them to dismantle that when they learn quantum theory? Why do we focus on the cross product (which only works in 3D) in early physics and never mention the wedge product (which is universal)? Etc. etc...

[@fanchongseow3552]

The use of the Greek letter epsilon ϵ is confusing as it can mean "relation" or "belong to".

[@user-vq5nz6yk1h]

Could someone explain the proof of the restricted comprehension? I didn't understand the second part for there exists a y in m for which P(y) is true. But the video was really nice and everything else was well understandable for me.

[@DavidRuizTijerina]

Intuitively, what he's doing is proposing an algorithm to construct the set. If there is no y \in m such that P(y) is true, then the set is the empty set. If there is some y' \in m such that P(y') is true, then you construct the set by going over all elements x \in m with a functional relation R. When When P(x) is true, R returns x, and you put it into the set. When P(x) is not true, R returns y'. R needs to return something, and there's no problem if you already had y' (explained when he talked about the axiom on pair sets, I think). What I don't like is the way he defines R. Not from a mathematical perspective (I'm no mathematician), but aesthetically. I think of it as a mapping, where you input x and get y. But it's written in a way that it returns true or false, which seems confusing to me.

[@wroanee]

The way R is ( just a predicate of two variables ) defined, It does not return any value except true or false. and It is one kind of mapping based on true value. R maps from x to y such that R(x,y) is true. So, functional R does not construct the set here, ImR(m) does that job using R as a predicate.

[@jasonc0065]

You missed extensionality: x:a && x=y => y:a

[@alonyouval3452]

Thanks, that was really good one (:

[@lazywarrior]

very good for theoretical physicists!

[@taraspokalchuk7256]

What is the difference between (a,b) and (a,(b))? If we don't have a hierarchy of sets and elements?

[@user-vq5nz6yk1h]

There is the elementrelation. And b and (b) are different sets. One set is b and one contains b. There is kind of a hierarchy of sets and elements due to the element function and that this function holds in your example different variables. What he tried to say was in my opinion that every "element" of a set is in ZFC a set itself.

[@user-vq5nz6yk1h]

*relation not function. Sorry.

[@qxtr5853]

Does anyone know, what the pedantic representation of q1 with all theoretically necessary parentheses is? I don't get whether the first implies operator is inside or outside the for all y.

[@varun4929]

Does anyone know what reference book one can use for this lecture series

[@ansgarschaefer5411]

At kzbin.info/www/bejne/d3Ktc2yiYq10kNU, the "=" sign is used to describe identity between two variables (not sets, as it has been used before). But throughout the course, "=" has only been defined for sets, and has never been defined for "variables". How can I understand this?

[@yueteng2018]

OMG this is so muggle friendly, thank you prof!!

[@LeoHsieh]

regarding the axiom of replacement, can anyone tell me what exactly are allowed in this 'functional relation' construction??? it seems like there are sooo many arbitrary 'function relation's and it is just ridiculous. I search the wikipedia and it used the term 'definable class function' instead of 'functional relation' *The axiom schema of replacement states that if F is a definable class function, and A is any set, then the image F[A] is also a set.* and it is quite confusing what exactly class function is? can someone give me some examples?

[@antoniomantovani3147]

It's like the beginning, you are sure that there is one, and you see all the caratteristics

[@Lauschangreifer]

24:00: Wenn a und b Mengen sein sollen, müsste dann die formale Definition nicht auch Aussagen über b ∈ a enthalten?

[@ndibalemasadik4704]

how can i get more videos about set theory

[@johnhippisley9106]

Very clear!

[@samuelmersha5530]

Wait, How did he clean the board?

[@chordsequencer001]

If the empty set were a subset of every set then nothing in phi would imply that there was nothing in any selected set A.

[@bghnv123]

in the formal proof at q1, what stops him from writing ∀y : ( y∈x ⇒ x=x') proving his goal out of false assumption? or for that matter couldn't he proof that not(x=x') is true contradicting everything?

[@mannyglover]

Benjamin Rutkowski yes, this confused me

[@jimallysonnevado3973]

You cannot write that down because that is not a tautology,axiom or modus ponens. But certainly you can always write down “A => B” as one of the steps where A is a false proposition that you know and B is the one you are trying to prove, by ex falso quad libet. No problem yet here, because A=>B means “B is true if A is”, but that does not mean B on its own right is true. Now, if your set of axioms are consistent, you can never write down A on its own no matter any of axioms, tautology or modus ponens you use (because A is false). And so, you still cannot conclude B from the A=>B you wrote down. However, if the axioms you wrote down can help you arrive at the false conclusion A at some point, then B becomes True immediately and there goes the problem, everything you try to prove becomes true but this happens only in an inconsistent system of axioms.

[@bghnv123]

Thank for this answer. In the 5 years since I posted this, I've read 3 books on axiomatic set theory, finished my undergrad in mathematics, published some papers, and started grad school. Looking back on the comment that I made is really silly and is nice to see how far I have gone since posting that. Otherwise I would have forgotten about how I started out. And thanks for the very thorough answer too! I'm sure it'll help some people. @@jimallysonnevado3973

[@yakunwang7181]

I am confused by this question: is x\ in x a proposition? If is is, then by the first axiom x is a set! Therefore we can constructed any set! So x\ in x should not be a proposition. But by the last proposition we know there is no set x such that x\ in x, that is to say x \in x is false and thus it is a proposition. So what's wrong here?

[@yakunwang7181]

It seems I understand. Although we have "x is a set"⇔"x ∈ x is a proposition", we still can not constructed a set x. Since we only know "for any set x, x ∈ x is false", not "for anything x, x ∈ x is false".

[@yakunwang7181]

I missed this precondition: x is a set. Once x is a set, of course x ∈ x is a proposition.

[@MikeRosoftJH]

In set theory (ZFC), everything is a set.

[@taraspokalchuk7256]

43:30 Is U(P(N))=N? Union of the power set of N.

[@taraspokalchuk7256]

***** thank you

[@krystofsrsen8655]

"A functional relation, not a functional relationship... So we may assume its existence" 🤣🤣

[@yaminnew2953]

You do all this to specify what sets are. But then what are variables?

[@MikeRosoftJH]

This is standard predicate logic; each variable stands for some object of the universe of the theory (in case of set theory - ZF or ZFC - every object is a set).

[@pratikprajapati9133]

Great Man

[@akashraj5073]

when i looked into the example @ 1:28:42 .i found out that the power set is a topology , does this mean all power sets are a topology.

[@stijndhondt9611]

If you equip a set X with the power set P(X) you get the topological space (X, P(X)), which is a trivial topology called the discrete topology. So, yes, you can always use the power set to construct a topology, but it is a very useless one, since every subset of X is open in this topology :)

[@akashraj5073]

@@stijndhondt9611 thank you

[@henrywang6931]

Why can't we define the power set of m as the set P(m) whose elements are sets that contains elements of the m, and that the union of these sets must give the set m? Or, is this just a somewhat restatement of the definition using PUC?

[@scottwhitman9868]

The axiom is that the power set exists. What it is saying is that the collection of the subsets of any set is itself a set.

[@ibrahimn628]

@27:00 he said the preposition is true independent of what y is. How so?

[@wolfgang2205]

the proposition p=>q would be always true if p is false because of the definition of the binary operator "=>". and for all y, y is included by x is false because of the assumption which says x is an empty set. and like i said if the assumption is false, the conclusion wouldn't matter what its truth value is, the whole proposition would be true.

[@ibrahimn628]

@@wolfgang2205 got it, thank you for taking the time.

[@mikeCavalle]

omg -- i haven't seen a professor wash a slate chalkboard for more the 50 years...

[@Someone-cr8cj]

έψιλον/epsilon accent of the ε not on the ι but who am i, right?

[@cj-cs7lm]

so many chalkboards!!

[@jarrodanderson2124]

It's from the future!!!

[@whdaffer1]

Is he saying 'naive'?

[@Esloquees]

1:16:15 ach so! | richtig!

[@hamzehabuabed6333]

You have an intelligent brain

[@taraspokalchuk7256]

If x = (a,b) what is union of x?

[@taraspokalchuk7256]

so if x={2,3}, than Ux={2,3}=x, ok what about x={2,{3}}? what is Ux? Ux should consist of all elements of 2 (which doesn't make sense) and all elements of {3}, which is 3.

[@danielegarrisi3229]

U(a,b) = U{{a},{a,b}}= {a} u {a,b} = {a,b}. So, the union of the ordered pair (a,b) is the pair of a and b.

[@vincentverne7052]

By convention, capital letters are used to denote sets and letters x, y , X, Y are not used to denote sets.

[@andreemcaldas]

Not "convention". You should just say "usually". But then "usually" (usually) means people that are not doing set theory.

[@stevemuller8320]

42:50 "either x or y" - here the everyday-speech is misleading - "either x or y or both" = x v y

[@arunruthraa4559]

take a shot, each time he says ah-ha

[@Esloquees]

1:13:54 ¡Take it! ø