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?


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.

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.