Topology: A Categorical Approach by Tai-Danae Bradley & Tyler Bryson & John Terilla

Author:Tai-Danae Bradley & Tyler Bryson & John Terilla [Bradley, Tai-Danae & Bryson, Tyler & Terilla, John]
Language: eng
Format: epub
Tags: point-set topology; category theory; category; functor; natural transformation; connectedness; Hausdorff; compactness; convergence; filters; limits; colimits; adjunctions; fibrations; fundamental groupoid
ISBN: 9780262539357
Google: AL74DwAAQBAJ
Publisher: MIT Press
Published: 2020-07-15T00:34:04.886036+00:00

Proof.  We used Zorn’s lemma to prove Tychonoff’s theorem. Although we don’t prove it, the axiom of choice implies Zorn’s lemma (see exercise 3.4.3 at the end of the chapter), from which it follows that Tychonoff’s theorem is implied by the axiom of choice.

To prove that Tychonoff’s theorem implies the axiom of choice, let {Xα}α∈A be a collection of nonempty sets. We need to make a bunch of compact spaces so we can apply the Tychonoff theorem. First, add a new element to Xα called “∞α,” letting Yα = Xα ∪{∞α}. Each set Yα can be made into a space by defining the topology to be {∅, {∞α}, Xα, Yα}. Note that Yα is compact—there are only finitely many open sets so every open cover is finite. Thus by Tychonoff’s theorem, Y:= ∏α∈AY α is compact.

Now consider a collection of open sets {Uβ}β∈A of Y where Uβ is the basic open set in Y obtained by taking the product of all Yαs for α ≠ β and putting the open set {∞β} in the βth factor. Notice that any finite subcollection {Uβ1, …, Uβn} cannot cover Y, for the function f defined as follows is not in Choose a partial function which is possible without the axiom of choice since the product is finite. Then extend to a function f ∈ Y by setting f(α) = ∞α for all α ≠ β1, …, βn, which is possible since we’re not making any choices.

Therefore, the collection {Uβ} cannot cover Y. So there is a function f ∈ Y not in the ∪α∈AUα. This says that for no α ∈ A does fα = ∞α. Therefore, fα ∈ Xα for each α, which is a desired choice function.

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