Pairing: Given any two sets, you can make a set containing those two sets. (You can also take set a, pair it with itself {a,a} which (via Extensionality) gives you {a}. So by pairing you can make a set containing one set too.) Union: Given a set of sets, you can make a set containing all of the elements' elements. {{1,2}, {2,3}} unionized is {1,2,3} (Union is necessary to construct Vω.) 2:30 Yay, Pairing is unnecessary in ZF! If you want to make a pair {a,b}, just define a function from the Natural numbers: 1 -> a 2 -> b 3 -> b 4 -> b ... Then take its image (Replacement Axiom) which will be {a,b,b...}={a,b}. Pairing is needed only in weaker set theories. (E.g. without Powerset and Infinity, you cannot even show there is any set with more than 1 element, so there you need Pairing.) 4:00 "Is Union needed?" Yes, e.g. for V2ω. Let us define K = |V2ω| Borcherds says Vω has the cardinality of ω, but it is ℵ0, not ω! |Vω+1| = ℵ1 I assume... Or rather Beth1 because for some reason Alephs were not enough for the mathematicians... 5:00 Now, take all sets A such that |A|

Can someone explain how the model of ZF without union works, because it seems like he defined a descending chain of sets which violates the axiom of foundation

There are two unfortunate terminology errors in the presentation--- you just said the "transitive closure of a" without adding "has cardinality less than kappa", as that defines the set of allowed a. The second error is that a singular cardinal kappa IS the limit of less than kappa cardinals less than kappa, in this case the V_\omega+k for all k>0, you misspoke and said "is NOT a limit of less than kappa cardinals", guaranteed to be seriously confusing to students.

You seem to define the axiom of pairing differently than I'm used to. I understood it to mean if one has a set A={a} and a set B={b} then one can form a set of ordered pairs AxB={(a,b)}. Is this a different convention or a mistake on my part? N.B. If it matters to conventions, I'm a mathematical physicist.

@billh17

2 жыл бұрын

What do you mean by the notation (a, b)? I suspect that you mean it is an order pair. But, every object in ZFC must be a set. How do you express (a, b) in terms of being a set? The definition that I can up with is the following (which I don't think is the one that you have): Def. Let a, b, c be sets. Then, c is an order pair of a and b means: 1) a in c 2) (Ez)(z in c and (Ax)(x in z if and only if x = a or x = b)) But, the main problem is how do you even know that if you have a set a that you also have a set A such that A consists of the single element a? As pointed out in the video, this follows from the axiom of pairing because if a is a set then { a, a} = { a } is a set. How does your understanding justifies the existence of singleton sets?

yeeeeeeeeee

I do not understand the argument used to say that the consistancy of ZFC does not imply that of ZFC-foundation. Syntaxically, if you cannot derive a contradiction from a larger set of axioms you cannot derive on from a smaller one, and there is nothing more to it. It just mean that consistency of ZFC with and without foundation are equivalent, and by no means that ZFC prove its own consistency

@marcelolynch8943

2 жыл бұрын

The claim is not that "the consistency of ZFC does not imply the consistency of ZFC-foundation" rather that ZFC ⊬ Con(ZFC-), that is, that the formula Con(ZFC-) which expresses "ZFC-foundation is consistent" is not a theorem in ZFC. Note that Prof. Borcherds says "the *axioms* of ZFC do not imply the consistency of ZFC-" and not "the consistency of ZFC..."

@An-ht8so

2 жыл бұрын

@@marcelolynch8943 Right, thank you.

SUMMA!!!!

But isn't the axiom of pairing needed to make the axiom of union work?