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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?

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Yes, checking a basis is enough. Perhaps the following is clearer:

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.

Some remarks:

  • 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.
  • The statement is still true when dealing with a subbasis, not just a basis. This is a non-trivial result, usually called the Alexander Subbasis Theorem. See here for a proof.
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