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36

When you move from $\exists s\in S$, to specifying "Let $s$ be an element of $S$" you are using what is known as existential instantiation. This is an inference rule of the underlying logic, stating that if there are objects satisfying some property, we can add a new symbol to the language with the statement that this symbol satisfies our property. So you ...


11

This is a misunderstanding about the essence of axiomatic set theory. In axiomatic set theory, you don't assume that a set exists because you can think of it; in a sense, the entire point of axiomatic set theory is not to do that, to decouple the notion of existence of sets from such pre-existing (pun intended) notions. When you say "Define $f(S)=s$", you'...


9

Good question! This is really subtle stuff; its impossible to give a proper answer without picking an actual formal system for first-order logic and trying to formalize your argument inside. After a lot of struggle, you'll notice that it can't be done. An informal explanation, however, is that you're implicitly using the "logicians" axiom of choice: AC....


8

This is a confusing matter, mainly because the kind of reasoning you use in your proof is usually taken to be valid. However, in order to formalize that reasoning in axiomatic set theory, we need to reduce it to particular symbolic formulas in a formal logic system. And it turns out that the rules of symbolic logic and set theory that are sufficient to ...


6

Think about it this way: you're trying to write down instructions for a robot to pick an element from each set. The instructions have to be completely unambiguous. If you just say, "OK, each set $A_i$ has an element $a_i$; so let $f(i)$ be that element," this clearly doesn't unambiguously tell the robot how to pick some element from each $A_i$. Suppose $A_i=...


5

No. For two main reasons: Even in the absence of choice $\aleph_\omega$ cannot be a power set of anything. The proof is the same usual proof that relies on Koenig's theorem, which may require choice, but it does not require choice if the power set is well-orderable. So in any case it is provable you will have uncountable cardinals which are not power ...


3

I claim that more than ZF is needed for this result. Specifically, I claim that, in the basic Fraenkel model, the set $A$ of atoms with the discrete topology is a counterexample. (I'm using a permutation model here, but the result transfers to ZF by the Jech-Sochor theorem.) Of course, $A$ isn't compact because the cover by singletons has no finite ...


3

The main issue here is defining concretely, from the existing universe, a choice function (or whatever object). I have two white socks, plain, new, entirely indistinguishable. I hold one sock in my left hand, and the other in my right. Now I turn around for a few seconds and turn back to you. Did I switch the socks? You can't tell. Your power of ...


3

First of all note that the proof of the Axiom of Choice from the Well Ordering Principle is not valid unless you admit the latter as an axiom; it is not a proof of AC per se (which is why ZFC contains AC as axiom), but shows that one can take WOP as axiom instead of AC, and then derive AC from it. Second, note that in spite of appearances you are not trying ...


2

A good visual way to 'see' the problem is this: For the first method, imagine each well-order of $\cup_{i\in I}A_i$ is a marble, all of which are put into a hat which is the hat of well-orders of that set. We then invoke the well-ordering principle to show that there is at least one marble in that hat, and then reach into that hat and take out a marble (any ...


2

Going from Hausdorff to Tikhonov gains you nothing. Theorem $4.70$ in Horst Herrlich, Axiom of Choice, Springer Lecture Notes in Mathematics $1876$: Equivalent are: Products of compact Hausdorff spaces are compact. Products of finite discrete spaces are compact. Products of finite spaces are compact. Hilbert cubes $[0,1]^I$ are ...


1

This axiom is controversial because although it seems like a relatively intuitive idea, there are still some issues. One of the best ways of understanding it is in this way: Take the set of all pairs of shoes and find a way to pick one shoe from each pair in order to form a new set. Easy! Just let our choice function be take every right shoe from each pair ...


1

Your "functions with finite support" approach is the standard approach to take to this, and is very likely what any author who considers the question trivial has in mind. The correct notion of "finite" to take when defining monomials is the usual one, as you have done (this is necessary for the polynomial ring you get to be the free commutative $k$-algebra ...


1

Here is an option which takes place between your two suggestions, without appealing to choice: Direct limit of finitely generated subrings. We define $k[X]$ to be the direct limit of the system $\{ k[Y]\mid Y\in[X]^{<\omega}\}$ with inclusions as the embeddings. Now to check that the operations are well-defined we only need to check they restrict ...


1

Let me elaborate on Asaf's claim that $\aleph_\omega$ cannot be a power set. Suppose $|P(X)|=\aleph_\omega$ for some set $X$. Since $|X|<|P(X)|$, $|X|<\aleph_\omega$, and so $|X|=\aleph_n$ for some $n<\omega$. It would then follow that $$\aleph_\omega^{\aleph_0}\leq\aleph_\omega^{\aleph_n}=(2^{\aleph_n})^{\aleph_n}=2^{\aleph_n^2}=2^{\aleph_n}=\...


1

Below is an excerpt from an old answer of mine that I think addresses the difference between making one choice and making an infinity of choices well. Note that the kind of reasoning you use in your suggested proof of AC is actually allowed in mathematics in general. But in order to formalize that reasoning in axiomatic set theory we need the axiom of ...


1

For why we can "pick from one" but not from infinitely many: There is a blob. ​ Let x be one. ​ [further reasoning] ​ Thus Q. is short for There exists a blob. For all blobs x: [further reasoning] Q. For all blobs x, Q. ​ Thus Q. , using the rule For formulas Q in which x does not occur free, from ​ ​ (∀x)(P(x) implies Q) ​ ...



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