# Tychonoff’s theorem implies AC

###### Proof.

Let $C$ be a non-empty collection  of non-empty sets. Let $Y$ be the generalized cartesian product of all the elements in $C$. Our objective is show that $Y$ is non-empty.

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. 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 well-defined, since $h(X_{A})\in A$. Now, it is easy to see that the function $\phi:Y\to Z$ given by $\phi(g)=g^{*}$ is a bijection, whose inverse    $\phi^{-1}:Z\to Y$ is given by $\phi^{-1}(h)=h^{\prime}$. This shows that $Y$ and $Z$ are equipollent.

2. 2.

Next, we topologize each $X_{A}$ in such a way that $X_{A}$ is compact  .

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 non-empty element $S\cup\{A\}$ in $\mathcal{D}$, so that $A-S\neq\varnothing$, and is finite. By assumption  , each element in $A-S$ belongs to some open set in $\mathcal{D}$. So to cover $A-S$, 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. 3.

Finally, we prove that $Z$, and therefore $Y$, is non-empty.

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 non-empty, we employ a characterization  of compact space: $X$ is compact iff every collection of closed sets  in $X$ having FIP has non-empty intersection  (http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty). Let us look at the collection $\mathcal{S}:=\{p_{A}^{-1}(A)\mid A\in C\}$. Given $A_{1},\ldots,A_{n}\in C$, pick an element $a_{i}\in A_{i}$, since $A_{i}\neq\varnothing$ by assumption. Note that this is possible, since there are only a finite number of sets. Define $f:D\to\bigcup D$ as follows:

 $f(X_{A}):=\left\{\begin{array}[]{ll}a_{i}&\textrm{if }A=A_{i}\textrm{ for some% }i=1,\ldots,n,\\ A&\textrm{otherwise.}\end{array}\right.$

Since $f(X_{A_{i}})=a_{i}\in A_{i}$, $f\in p_{A_{i}}^{-1}(A_{i})$ for each $i=1,\ldots,n$. Therefore,

 $f\in p_{A_{1}}^{-1}(A_{1})\cap\cdots\cap p_{A_{n}}^{-1}(A_{n}).$

Since $p_{A_{1}}^{-1}(A_{1}),\ldots,p_{A_{n}}^{-1}(A_{n})$ are arbitrarily picked from $\mathcal{S}$, the collection $\mathcal{S}$ has finite intersection property, and since $X$ is compact, $Z=\bigcap S$ must be non-empty.

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

Title Tychonoff’s theorem implies AC TychonoffsTheoremImpliesAC 2013-03-22 18:45:38 2013-03-22 18:45:38 CWoo (3771) CWoo (3771) 7 CWoo (3771) Proof msc 54D30 msc 03E25