Q:

Tychonov's theorem converse

I just want to validate my proof about the converse of Tychonov's theorem that says:

If the product topology $\prod_{x \in I}X$ is compact, then each $X_i$ is compact

I define the projection function that maps $\prod_{x \in I}X$ to $X_i$ such as:
$P: \prod_{x \in I}X \rightarrow X_i$ 
We know that $P$ is continuous and since the image of a compact set is compact under continuous function then $X_i$ must be compact
Please let me know if it's a valid argument.
Thanks

A:

The projections are only surjective when all the factors are nonempty. For example $\emptyset \times \mathbb{R} = \emptyset$ is compact, but $\mathbb{R}$ isn't. You have to remember that it's the image of a compact space under a continuous function that is compact, not necessarily the whole codomain.
You're also right to say that you need the axiom of choice when the index set is infinite, but that's in addition to requiring that the factors are nonempty. I'm not sure what happens if you don't assume the axiom of choice (which is equivalent to Tychonoff's theorem anyway...) -- you could potentially get an empty product of an infinite number of noncompact spaces, I imagine.
Otherwise, barring the weird notation (I guess you meant something like $\prod_{x \in I} X_x$ instead of $\prod_{x \in X} X$), your argument is fine.

