Tychonoff's Theorem and Axiom of Choice I know that Tychonoff's Theorem requires the Axiom of Choice, but I am struggling to see where I have used it in the proof below. I suspect it lies somewhere in the step of going from basis elements to elements of the open cover, but it seems that only finitely many choices are needed there.
Let $X_i$, $i \in I$, be compact spaces. We wish to show $X:=\prod_{i \in I}X_i$ is compact. Let $\{U_j~|~j \in J\}$ be an open cover of $X$. Each $U_j$ is of the form $U_j=\cup_k U_{j,k}$ for some basis sets $U_{j,k}$. That is, each $U_{j,k}$ is of the form $\prod_{i \in I} V_i$ where $V_i$ is open in $X_i$ and $V_i=X_i$ for all but finitely many $i$.
Now consider the collection of all the $U_{j,k}$'s. Choose one such $U_{j,k}=\prod_{i \in I}V_i$. Since $V_i=X_i$ for all but finitely many $i$, we just need to "fill in" the coordinates for which $V_i \neq X_i$. For each such coordinate $i$, the projections of the $U_{j,k}$'s onto the $i$th coordinate cover $X_i$. Since $X_i$ is compact, finitely many of the projects cover it, and from those projections we recover finitely many of the $U_{j,k}$. Note this only requires finitely many choices. We do this for finitely many coordinates and find that finitely many basis elements cover $X$.
Finally, from each of these finitely many basis elements we recover a set $U_j$ from the open cover containing it. Again this only requires finitely many choices, and these $U_j$'s must cover $X$.
 A: You have not used the axiom of choice (in any essential way), but your proof is incorrect.
You have described covering $X$ in terms of covering each coordinate $X_i$.  It seems that what you have in mind here is that sets $U_j$ cover each coordinate if their projections onto $X_i$ cover $X_i$ for all $i$.  However, this condition does not imply the sets $U_j$ actually cover all of $X$.  For instance, let's take an example with just two factors.  Say $X=[0,1]\times[0,1]$.  Let $U_1=[0,2/3)\times[0,2/3)$ and $U_2=(1/3,1]\times(1/3,1]$.  These two open sets cover each coordinate, since $[0,2/3)$ and $(1/3,1]$ cover $[0,1]$.  However, they do not cover $X$, since (for instance) $(1/4,3/4)\not\in U_1\cup U_2$.
You have described a way to find a finite subcollection of your open cover which covers each coordinate.  However, such a finite subcollection need not cover $X$, so this does not prove $X$ is compact.
A: To see the flaw in your argument,it's instructive to study the standard proof that the Tychonov theorem implies AC:
Assume Tychonov holds.
Suppose $S_i, i \in I$ is some family of non-empty sets. We want to show that $\prod_{i \in I} S_i \neq \emptyset$, i.e. there exists some choice function
$f: I \rightarrow \cup_{i \in I} S_i$ such that $\forall i \in I: f(i) \in S_i$.
To do this, take a different point $p_i \notin S_i$ (exists by regularity, no choice is needed here!), and define a space $X_i = S_i \cup \{p_i\}$ with the topology $\mathcal{T}_i = \{\emptyset, \{p_i\}, S_i, X_i\}$. The topology being finite, each $X_i$ is compact. So $\prod_i X_i$ is compact.
Now define a subbasic family of open sets of $X$ by defining for each $j \in I$: $U_j = \{x \in X: x_j = p_j\} = \pi_j^{-1}[\{p_j\}]$. (depending on one coordinate only)
Claim: no finite subfamily of $\{U_j : j \in I\}$ covers $X$:
Let $U_{j_1},\ldots U_{j_n}$ be a finite subfamily. Then for each of the $j_i$ (So finitely may choices!) pick $q_{j_i} \in S_{j_i}$ and define $x \in X$ by $x_i = p_i $ if $i \notin \{j_1,\ldots, j_n\}$ and $x_{j_i} = s_{i_j}$ otherwise. Then it's clear that $x \notin \cup_{i=1}^n U_{j_i}$
But as $X$ is compact, this family $\{U_j: j \in I\}$ cannot be an open cover of $X$, so there must be some $y \in X\setminus_{j \in I} U_j$. For this $y$ we must have $y_j \in X_j$ and $y_j \neq p_j$, so $y_j \in S_j$ for any $j$.
So $y \in \prod_j S_j \neq \emptyset$ as required.
This shows that in your proof you cannot construct a finite subcover in the way you say, as it would fail for the above family (when you cannot invoke choice). In your sketch you never use that your family is a cover, and there are no choices involved in the coordinate open sets (as you thought the issue was there).
