# Compact condition for base elements

Suppose $$X$$ is a topological space and $$\{B_i\}$$ form a base for the topology on $$X$$, where the $$i$$ run over some index set $$J$$.

$$X$$ is said to be compact if every open cover of $$X$$ contains a finite subcover of $$X$$.

Suppose you know that for every covering of $$X$$ by base elements, $$X=\bigcup_{i\in I}B_i$$, there exists a finite subcover $$X=\bigcup_{i\in S}B_i$$ where $$S$$ is a finite subset of $$I$$.

Does this then imply that $$X$$ is compact? If we have an arbitrary open covering of $$X$$, say $$X=\bigcup_{j\in J} U_j$$, then for each $$j\in J$$, there exists a covering of $$U_j$$ by some base elements $$B_{j_i}$$. Putting these together, we have $$X=\bigcup_{i\in I,j\in J}B_{i_j}$$, for which we know there is a finite subcover.

But does this then imply that our original cover of arbitrary open sets $$U_i$$ has a finite subcover?

Let $$\mathscr{U}$$ be an open cover of $$X$$. For each $$x\in X$$ and each $$U\in\mathscr{U}$$, choose $$i(x,U)\in J$$ such that $$x\in B_{i(x,U)}\subset U$$. Then $$X\subset \bigcup_{x\in X,U\in\mathscr{U}}B_{i(x,U)}$$ By assumption, there are finitely many $$x_1,\dots,x_n\in X$$ and $$U_1,\dots,U_m\in \mathscr{U}$$ such that $$X\subset \bigcup_{k=1}^n\bigcup_{l=1}^mB_{i(x_k,U_l)}$$ Now use that $$B_{i(x,U)}\subset U$$, so that $$X\subset \bigcup_{i=1}^n U_i$$ and we're done.
• The previous proof seems to rely on the axiom of choice. This is not necessary, since we can define, for every $$x\in X$$ and $$U\in\mathscr{U}$$, the set $$J(x,U)=\{i\in J:x\in B_j\subset U\}$$. Then $$X\subset \bigcup_{x\in X,U\in\mathscr{U}}\bigcup_{i\in J(x,U)}B_j$$ The rest of the argument is identical.