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A result about compactness says that A topological space is compact if every basic open cover has a finite subcover.

The proof runs as follows: Let$\{G_i\}$ be an open cover and $\{B_j\}$ an open base. Each $G_i$ is te union o certain $B_j$'s and the totality o all such $B_j$'s is clearly an open cover. Now this has a finite subcover, and for each set in this finite subcover we can find a $G_i$ which contains it. The class of $G_i$'s which arises in this way is evidently a finite subcover of the original open cover.

Doesn't the proof imply that the condition, "every basic open cover needs to have a finite subcover", can be reduced to only. "there exists at least one basic cover which has a finite subcover"? Or, does it imply that the condition, "there exists at least one basic cover which has a finite subcover" implies "every basic open cover needs to have a finite subcover"?

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  • $\begingroup$ by open base you mean?? $\endgroup$ – user87543 Feb 17 '14 at 9:45
  • $\begingroup$ Do you mean that the $B_i$ are elements of a basis for the topology? $\endgroup$ – Arthur Feb 17 '14 at 9:59
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Suppose you do something funny, and consider a base $\mathcal{B}$ of a topological space $X$ with $X \in \mathcal{B}$. Clearly there are covers of $X$ by sets in $\mathcal{B}$ admitting a finite subcover: all of them which include $X$, for example. But not all topological spaces are compact, so something must be amiss.

To see this, you really should be a bit more careful in your notation.

Fix a base $\mathcal{B}$ for $X$, and assume that all covers of $X$ by sets in $\mathcal{B}$ admit a finite subcover. Let $\mathcal{U} = \{ U_i : i \in I \}$ be an open cover of $X$. Each $U_i$ is a union of sets in $\mathcal{B}$, so for each $i \in I$ let $\mathcal{B}_i \subseteq \mathcal{B}$ be such that $\bigcup \mathcal{B}_i = U_i$. Consider now $\mathcal{V} = \bigcup_{i \in I} \mathcal{B}_i$. It is easy to show that $\mathcal{V}$ covers $X$: given $x \in X$ there is an $i \in I$ such that $x \in U_i = \bigcup \mathcal{B}_i$, so there is a $V \in \mathcal{B}_i \subseteq \mathcal{V}$ which contains $x$. Now there are $V_1 , \ldots , V_n \in \mathcal{V}$ such that $X = V_1 \cup \cdots \cup V_n$. For each $j \leq n$ fix some $i_j \in I$ such that $V_j \in \mathcal{B}_{i_j}$ (and so $V_i \subseteq U_{i_j}$). Now consider $U_{i_1} \cup \cdots \cup U_{i_n}$.

Note that we picked a subfamily of $\mathcal{B}$ which has the property that each is a subset of some set from the given open cover. It may be (as in the examples hinted at above) that finite subcovers of covers by basic open sets must include some very large sets, which in general are not subsets of any set in a given open cover.

So to conclude that $X$ is compact, we need to say that any cover by basic open sets arising from a construction as above has itself a finite subcover, and the simplest way to phrase this is to simply assume that every cover by basic open sets has a finite subcover.

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