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  1. At the beginning of proof of problem 2, The author choose arbitrary a in (0,1) and

Claimed that there exists some closed interval [a1,a2] containing a.

I know this is right, but I'm curious that this is always guranteed or is just specific case.

Is it possible for some open sets not to have a compact subset (except singleton) ?

  1. The author showed that f is continuous on [a1,a2] and this set is arbitrary.

But does it gurantee that if f is continuous on all of such compact sets, then it follows that f is continuous on open set such as (0,1)? Doesn't exist some point of (0,1) not in such compact sets?

I think this questions can be summarized as can open set be expressed as an union of compact sets?

If I missed something, could you point it out?

It will be very helpful to solidate my understading about elementary topology.

Thank you for your answer in advance.

glimpser
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  • In a Hausdorff space the property that, given a point $p$, any open set around $p$ contains a compact set containing $p$ is equivalent to the property of being "locally compact." There are Hausdorff spaces that are not locally compact.
    • – symplectomorphic Aug 10 '17 at 06:33
  • @symplectomorphic thx :D though I don't know what is Hausdorff space... then Is Euclidean space Hausdorff space? – glimpser Aug 10 '17 at 06:36
  • You missed something. The author proves that $f$ is continuous at an arbitrary point $a$ in $(0,1)$. (Here is the relevant quotation: "... and by Supplement 1 we know $f$ is continuous at $a$.") And this is exactly what it means to say that $f$ is continuous on $(0,1)$. (That is what the last sentence of the proof means.)
    • – symplectomorphic Aug 10 '17 at 06:36
  • Yes, Euclidean space is Hausdorff and locally compact. – symplectomorphic Aug 10 '17 at 06:38
  • @symplectomorphic umm. Since it is locally compact, for any a in (0,1), there exists some compact set containing a. So if I show that f is continuous on this arbitrary compact set, then it follow that f is continuous on given open set? Is it right? – glimpser Aug 10 '17 at 06:40
  • The compact set isn't arbitrary in the proof. Here is the point. Pick an arbitrary point of $(0,1)$; by local compactness there is a compact subset containing it but lying in $(0,1)$; by the author's argument, $f$ is uniformly continuous on the compact set and therefore continuous at $a$. Since the point $a$ was arbitrary, $f$ is continuous at every point of the open interval. – symplectomorphic Aug 10 '17 at 06:46
  • But yes, what you say is also true. Given an open interval $(a,b)$, if you show that $f$ is continuous on an arbitrary compact subset of the interval, then $f$ must be continuous on the interval. Why? By local compactness: every point in the interval is contained in one of the compact sets contained in the interval. – symplectomorphic Aug 10 '17 at 06:47
  • @symplectomorphic Then what "locally compact" means that the compact set we are talking about should be in the given space with locally compact property?Then if every point in the interval is contained in one of the compact set contained in the interval, it doesn't need to be unique, right?I feel like It could be many... – glimpser Aug 10 '17 at 06:55
  • I can't understand your English or your question. – symplectomorphic Aug 10 '17 at 06:58
  • Sorry, the compact set containing arbitrary point a is uniquely determined? I think it could be a lot, because , for instance, 1/2 in (0,1) it could be in [1/10,2/3] or [1/3,2/5]. – glimpser Aug 10 '17 at 07:02
  • @symplectomorphic And if you have time, could you look at my another question "Question related to connected set"? sorry, I don't know how to link to the question. – glimpser Aug 10 '17 at 07:03
  • Of course it's not uniquely determined. It doesn't matter. – symplectomorphic Aug 10 '17 at 07:05
  • @symplectomorphic Thank you really! – glimpser Aug 10 '17 at 07:11
  • Linked. – Alex Ravsky Aug 17 '17 at 05:07