Is the converse of Tychonoff's theorem true?

Question

Tychonoff's theorem states that the product of any collection of compact topological spaces is compact with respect to the product topology.

I would like to know that if the product of some topological spaces is compact with respect to the product topology, does it imply that these spaces are compact?

Answer 1

Consider the map $\phi_j:\Pi_i X_i\to X_j$ with respect to the product topology. These are continous, and we know the range of a continous map on a compact space is compact.

Answer 2

Suppose there is $j$ such that $X_j$ is non-compact. Then, there exists an open covering of X, say $\{U_i\}_{i \in J}$, that has no finite subset covering $X_j$. Then $\{\pi_j^{-1}(U_i) \hphantom{/}/ \hphantom{/} i \in J\}$ is an open covering of $\Pi X_i$, which has no finite subset covering it.