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A basis of a topological space $X$ is a family of open sets ${B_i : i \in I}$ for some indexing set $I$, where any open set in $X$ can be written as the union of two or more members of ${B_i : i \in I}$.

An open cover of topological space $X$ is a family of open sets ${B_i : i \in I}$ such that $X \subset \cup_{i \in I}U_{i}$, and each $B_i$ is open.

Considering that (by definition using open sets), $X$ is open, it seems to me that any basis of a topological space is an open cover, because if every open set in $X$ can be written as the union of two basis elements, then certainly the whole space can be (since the space is certainly open). However, an open cover isn't always going to be a basis, since not every open subset of $X$ can be written as the union of the open cover elements.

But say if every open cover is indeed a basis for $X$, and suppose further that every open cover admits a finite subcover. Does this mean that every open subset of our space $X$ is compact? But what if $X$ is Hausdorff? I recall a theorem that says if a space is Hausdorff, then every compact subset is closed and bounded. So then our space $X$ would have to be zero-dimensional (clopen basis), right?

I do apologize if my post is confusing or maybe even pointless. I am new here and I am trying to wrap my head around the differences between an open cover for a space, and a basis for a space.

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  • $\begingroup$ If every open cover is a basis, considering the open cover $\mathscr{C} = \{X\}$ tells you all about the topology. $\endgroup$ – Daniel Fischer Apr 21 '14 at 11:34
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If every open cover of a topological space $X$ is a basis for $X$, then the open cover $\{ X \}$ is a basis for $X$, and it follows that $X$ has the trivial topology: only $\varnothing$ and $X$ are open. And, yes, this space is compact (or quasi-compact if your definition of compactness includes the Hausdorff condition).

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