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let $X$ be a topological space; $X$ is said quasi-compact if for every open cover of $X$ there is a finite subcover. I need help to prove the following proposition:

Let $B$ be a basis of $X$; $X$ is quasi-compact if and only if every cover of $X$ done with elements of $B$ admits a subcover.

Clearly one implication is trivial.

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    $\begingroup$ Hint: Any open subset of $X$ can be covered using elements from $B$ (also, you forgot to write finite in the problem). $\endgroup$ – Tobias Kildetoft Feb 28 '13 at 15:44
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Suppose that every cover with basic open sets admits a finite subcover.

Let $\cal U$ be an open cover of $X$, for every $U\in\cal U$ replace it with all the basic open sets contained in $U$. So $\mathcal V=\{V\in B\mid\exists U\in\mathcal U: V\subseteq U\}$ is an open cover with basic open sets. It admits a finite subcover, $V_1,\ldots,V_n$.

For every $i$ there is some $U_i\in\cal U$ such that $V_i\subseteq U_i$. One can show that $\{U_1,\ldots,U_n\}$ is a finite subcover of $\cal U$.

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