Prove that $X$ is compact if and only if every cover of $X$ by members of $B$ has finite subcover.

Let $X,\tau$ be a space and let $B$ be a basis for $\tau$. Prove that $X$ is compact if and only if every cover of $X$ by members of $B$ has finite subcover.

here is what I got

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Assume that every cover of $X$ by members of $B$ has finite subcover. Since $B$ be a basis for $\tau$, $B$ is open cover for $X$. From our assumption, $B$ contain finitely many subcover of $X$, so $X$ is compact.

I'm not sure I know how to argue the converse.

• Compact = every open cover of $X$ has a finite subcover. The condition says: there is a base $\mathcal{B}$ for $X$ such that every open cover of $X$ that only uses members of $\mathcal{B}$, has a finite subcover. So one direction is clear. Mar 19, 2014 at 14:03
• To show compactness from the condition, start with an arbitrary open cover $O_i$ ($i \in I$) for $X$. Now use the fact that $\mathcal{B}$ is a base, and apply to condition to another related cover.. Mar 19, 2014 at 14:07

First assume $X$ is compact then take a covering of $X$ by members of $B$ since each member of $B$ is open we have a open covering of $X$, then because $X$ is compact the covering by members of $B$ have a finite subcovering.
Second assume every covering by members of $B$ have a finite subcovering. Take a open covering of $X$ call it $\{U_{\alpha},\alpha \in A\}$, by definition each $U_\alpha$ is union of elements of $B$, then take set formed for all the elements of $B$ which are contained in some $U_\alpha$. By definition this set is a open covering of $X$ by elements of $B$, then it have a finite subcovering $B_1,B_2,...,B_n$ each one contained in at least one element of $\{U_{\alpha},\alpha \in A\}$, for each $B_i,i=1..n$ pick $U_i,i=1..n$ with the property $B_i \subset U_i$, then $U_1,U_2,...,U_n$ is a finite subcovering of $X$ (because $B_1,B_2,...,B_n$ it is)