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Here's an exercise in normed spaces that I can't get my head around. It reads as follows: "Let X be a compact space equipped with norm d. If X is countable, then the set of isolated points in X is both open and dense." Just point me in the right direction. Thanks a bunch!

Gearclose
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  • You meant "metric", not "norm", right? Norms live on vector spaces - there are no compact or countable vector spaces. (Everyone else: Yes, I know that last statement is false. Context...) – David C. Ullrich Jan 01 '16 at 19:01
  • Yes that's right, X is only a metric. Thanks! – Gearclose Jan 01 '16 at 19:15
  • Hint for the "open" part: What's the definition of "isolated point"? Any union of open sets is open... – David C. Ullrich Jan 01 '16 at 19:17
  • So, if J is the set of isolated points, then for x in J we have a>0 such that B(x,a) is only {x}, which is contained in J. Taking the union for all x (which are at most countable, which doesn't really matter) we get an open set! Thanks! – Gearclose Jan 01 '16 at 19:51
  • The density part eludes me still... – Gearclose Jan 01 '16 at 20:23

1 Answers1

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You've already figured out why the set of isolated points $I$ is open.

To show $I$ is dense: Because your space is countable, you can enumerate the non-isolated points $x_1, x_2,$ and so on. Define $U_0 = X$ and for all natural numbers $n$, let $U_{n+1} = U_n \setminus \{x_n\}$. Use induction to show $U_n$ is open and dense for all $n$. Then use the Baire Category theorem.

Dabbler
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