Tychonoff’s theorem implies AC
In this entry, we prove that Tychonoff^{}’s theorem^{} implies that product of nonempty set of nonempty sets is nonempty, which is equivalent^{} to the axiom of choice^{} (AC). This fact, together with the fact that AC implies Tychonoff’s theorem, shows that Tychonoff’s theorem is equivalent to AC (under ZF). The proof was first discovered by John Kelley in 1950, and is now an exercise in axiomatic set theory.
Proof.
Let $C$ be a nonempty collection^{} of nonempty sets. Let $Y$ be the generalized cartesian product of all the elements in $C$. Our objective is show that $Y$ is nonempty.
First, some notations: for each $A\in C$, set ${X}_{A}:=A\cup \{A\}$, $D:=\{{X}_{A}\mid A\in C\}$, $X$ the generalized cartesian product of all the ${X}_{A}$’s, and ${p}_{A}$ the projection from $X$ onto ${X}_{A}$.
We break down the proof into several steps:

1.
$Y$ is equipollent^{} to $Z:=\bigcap \{{p}_{A}^{1}(A)\mid A\in C\}$.
An element of $X$ is a function $f:D\to \bigcup D$, such that $f({X}_{A})\in {X}_{A}$ for each $A\in C$. In other words, either $f({X}_{A})\in A$, or $f({X}_{A})=A$. An element of $Y$ is a function $g:C\to \bigcup C$ such that $g(A)\in A$ for each $A\in C$. Finally, $h\in {p}_{A}^{1}(A)$ iff $h({X}_{A})\in A$.
Given $g\in Y$, define ${g}^{*}\in X$ by ${g}^{*}({X}_{A}):=g(A)\in A$. Since $A$ is arbitrary, ${g}^{*}\in Z$. Conversely, given $h\in Z$, define ${h}^{\prime}\in Y$ by ${h}^{\prime}(A):=h({X}_{A})$, which is welldefined, since $h({X}_{A})\in A$. Now, it is easy to see that the function $\varphi :Y\to Z$ given by $\varphi (g)={g}^{*}$ is a bijection, whose inverse^{} ${\varphi}^{1}:Z\to Y$ is given by ${\varphi}^{1}(h)={h}^{\prime}$. This shows that $Y$ and $Z$ are equipollent.

2.
Next, we topologize each ${X}_{A}$ in such a way that ${X}_{A}$ is compact^{}.
Let ${\mathcal{T}}_{A}$ be the coarsest topology^{} containing the cofinite topology^{} on ${X}_{A}$ and the singleton $\{A\}$. A typical open set of ${X}_{A}$ is either the empty set^{}, or has the form $S\cup \{A\}$, where $S$ is cofinite in $A$.
To show that ${X}_{A}$ is compact under ${\mathcal{T}}_{A}$, let $\mathcal{D}$ be an open cover for ${X}_{A}$. We want to show that there is a finite subset of $\mathcal{D}$ covering ${X}_{A}$. If ${X}_{A}\in \mathcal{D}$, then we are done. Otherwise, pick a nonempty element $S\cup \{A\}$ in $\mathcal{D}$, so that $AS\ne \mathrm{\varnothing}$, and is finite. By assumption^{}, each element in $AS$ belongs to some open set in $\mathcal{D}$. So to cover $AS$, only a finite number of open sets in $\mathcal{D}$ are needed. These open sets, together with $S\cup \{A\}$, cover ${X}_{A}$. Hence ${X}_{A}$ is compact.

3.
Finally, we prove that $Z$, and therefore $Y$, is nonempty.
Apply Tychonoff’s theorem, $X$ is compact under the product topology. Furthermore, ${\pi}_{A}$ is continuous^{} for each $A\in C$. Since $\{A\}$ is open in ${X}_{A}$, and $A={X}_{A}\{A\}$, $A$ is closed in ${X}_{A}$, and thus so is ${p}_{A}^{1}(A)$ closed in $X$.
To show that $Z$ is nonempty, we employ a characterization^{} of compact space: $X$ is compact iff every collection of closed sets^{} in $X$ having FIP has nonempty intersection^{} (http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty). Let us look at the collection $\mathcal{S}:=\{{p}_{A}^{1}(A)\mid A\in C\}$. Given ${A}_{1},\mathrm{\dots},{A}_{n}\in C$, pick an element ${a}_{i}\in {A}_{i}$, since ${A}_{i}\ne \mathrm{\varnothing}$ by assumption. Note that this is possible, since there are only a finite number of sets. Define $f:D\to \bigcup D$ as follows:
$$f({X}_{A}):=\{\begin{array}{cc}{a}_{i}\hfill & \text{if}A={A}_{i}\text{for some}i=1,\mathrm{\dots},n,\hfill \\ A\hfill & \text{otherwise.}\hfill \end{array}$$ Since $f({X}_{{A}_{i}})={a}_{i}\in {A}_{i}$, $f\in {p}_{{A}_{i}}^{1}({A}_{i})$ for each $i=1,\mathrm{\dots},n$. Therefore,
$$f\in {p}_{{A}_{1}}^{1}({A}_{1})\cap \mathrm{\cdots}\cap {p}_{{A}_{n}}^{1}({A}_{n}).$$ Since ${p}_{{A}_{1}}^{1}({A}_{1}),\mathrm{\dots},{p}_{{A}_{n}}^{1}({A}_{n})$ are arbitrarily picked from $\mathcal{S}$, the collection $\mathcal{S}$ has finite intersection property, and since $X$ is compact, $Z=\bigcap S$ must be nonempty.
This completes^{} the proof. ∎
Remark. In the proof, we see that the trick is to adjoin the set $\{A\}$ to each set $A\in C$. Instead of $\{A\}$, we could have picked some arbitrary, but fixed singleton $\{B\}$, as long as $B\notin A$ for each $A\in C$, and the proof follows essentially the same way.
References
 1 T. J. Jech, The Axiom of Choice. NorthHolland Pub. Co., Amsterdam, 1973.
 2 J. L. Kelley, The Tychonoff’s product^{} theorem implies the axiom of choice. Fund. Math. 37, pp. 7576, 1950.
Title  Tychonoff’s theorem implies AC 

Canonical name  TychonoffsTheoremImpliesAC 
Date of creation  20130322 18:45:38 
Last modified on  20130322 18:45:38 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  7 
Author  CWoo (3771) 
Entry type  Proof 
Classification  msc 54D30 
Classification  msc 03E25 