# Definition of Compact Set: every open cover of basic open sets has finite subcover

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?

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 $$\{V_{\alpha}|\alpha \in I\}$$ .
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 $$P$$ .