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I have the following definition of a compact space

A topological space is a compact space if every filter $\mathcal B$ on $X$ has an accumulation point. That is there exists $x \in X$ such that

"for all $N \in \mathcal N(x)$, for all $A \in \mathcal B$, $N\cap A \neq \varnothing$"

I can't understand this definition. I am trying to show that using this definition of compact space

  1. $(a,b) \subset\mathbb R$ is not a compact space
  2. $[a,b] \subset\mathbb R$ is a compact space

By Heine-Borel theorem it is obvious that $[a,b]$ is a compact space as it is bounded and closed.

But how to proceed starting with the quoted definition?

Asaf Karagila
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Galilean
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1 Answers1

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While it's true that this filter property characterises compactness and it's an important fact, it's not "meant" for proving or disproving spaces compact. The "real" definition is that every open cover has a finite subcover. And it's fairly straightforward (if you know about filters already) to show that these notions are equivalent.

For $(a,b)$ the open cover definition is just fine to disprove compactness, e.g. take $\mathcal{U}= \{(a+\frac1n, b): n \in \Bbb N\}$ as an open cover that has no finite subcover. We could easily turn it into a filter without an accumulation point (the recipe is in the proof of this characterisation) but why do that? It's just more work and doesn't add insight.

To see specifically that $[a,b]$ is compact, my personally favourite way is to apply Alexander's subbase theorem and the order-completeness of $\Bbb R$. Here again, using filters is not the easy way to do it. It would just be a translation of the cover definition again.

Henno Brandsma
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