Let $X$ be a topological space and $K$ be a subset of $X$. Let $B$ be a basis of topology on $X$ (i.e. every open subset of $X$ is a union of some members in $B$).

Claim: If every open cover of $K$ by elements of $B$ has a finite sub-cover then $K$ is compact.

Proof: Let $\{ U_{\alpha}:\alpha\in I\}$ be an arbitrary open cover of $K$. We want to find finite sub-cover of it. Now each $U_{\alpha}$ is equal to union of some members of $B$. Thus $K$ is contained in a union of members of $B$ and by hopothesis, it has a finite sub-cover. Thus, if $K\subseteq B_1 \cup B_2 \cup \cdots \cup B_n$, then for each $i$ in $1,2,\ldots, n$, pick-up one member $U_{\alpha_i}$ such that $B_i\subseteq U_{\alpha_i}$. It is then clear that $K$ is covered by $U_{\alpha_1}, U_{\alpha_2},\cdots,U_{\alpha_n}$.

Q.1 Is this argument correct?

Q.2 Using this fact, I think we can easily prove that if $K_1$ and $K_2$ are compact subsets of $X_1$ and $X_2$ then $K_1\times K_2$ is compact subset of $X_1\times X_2$. Is this correct?


Yes, the proof you provided is exactly correct .

Now, for the $2nd$ problem, this typical fact is not enough as because if $I$ be an index set, then obviously $\cup_{\alpha \in I} (U_{\alpha} × V_{\alpha}) \neq (\cup_{\alpha} U_{\alpha}) × (\cup_{\alpha} V_{\alpha})$; in general, for some family of sets $\{U_{\alpha}|\alpha \in I\}$ and $\{V_{\alpha}|\alpha \in I\}$ .

The crucial fact required here is the somewhat called "The tube lemma".

If $X$ and $Y$ be two topological spaces with $Y$ being compact, then if $S$ be a non-empty subset of $X$ and if $W$ be a neighborhood of $(S×Y)$, then then there exists a point $s \in S$ and a neighborhood $P \subseteq W$ with $s \in P$ such that the set $\{s\} × Y$ is totally contained in $P$ .

Consider any book on basic general topology , rather Munkres and you can find the proof there .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.