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In my functional analysis homework, I had to prove something like this :

Let $D_n \subset D_{n-1} \subset \dots \subset E$ where $(E, \mathfrak T)$ is a Hausdorff topological space and the sequence of sets $D_n$ is non-empty, compact and decreasing. Prove that there is a point in the intersection of all $D_n$.

The obvious guess to begin with was to consider $x_n \in D_n$ arbitrary. I know the result is true when $(E, \mathfrak T)$ is metrizable, because then you can find a convergent subsequence which gives you a point in the intersection. If the topological space is not metrizable, all you get is a limit point. Using the fact that this topological space came up as a Banach space, I could use analysis (i.e. Hahn-Banach theorem and more properties that I had for the sets $D_n$) to complete the proof, but I was sad because I didn't find a purely topological argument.

My question is : If I am asked the question as it is written above, is it true, or are there counter-examples?

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Look for 'finite intersection property'. – Sigur Dec 16 '12 at 1:28
@Sigur : Oh. Let me give it a thought for a second. – Patrick Da Silva Dec 16 '12 at 1:34
up vote 1 down vote accepted

That's true in any topological space: it pretty much follows straight from applying DeMorgan to the basic topological definition of compactness (any open cover has a finite subcover). Lean back, and I'm sure you can write the few lines necessary in a few minutes after keeping that in mind.

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Yeah, it was pretty elementary in the end. I don't know why I couldn't see it. At least that simplifies my homework! Thanks to you two, Sigur and gnometorule. – Patrick Da Silva Dec 16 '12 at 1:36
When see a trick for the first time, it always is hard. :) – gnometorule Dec 16 '12 at 1:37
I actually thought of using the intersection property but it confused me real badly so I looked for another way (and good for me, I had found one), but it wasn't very satisfactory. Thanks for pointing in the right direction =) – Patrick Da Silva Dec 16 '12 at 1:40

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