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I have a question I my metric spaces course book I cannot solve:

Show that the following subset is closed in $C[0,1]$: $\{ f \in C[0,1] \mid f(a)=0 \textrm{ for all }a \in A \}$, where $C[0,1]$ is the space of continuous real-valued functions on $[0,1]$ with the sup metric and $A \subseteq [0,1]$.

Stahl
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Pim
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  • It might be easier to show the complement is open. You'll probably want to use the fact that $[0,1]$ is compact along the way. – MartianInvader Jun 25 '13 at 18:19
  • i must confess that i think my answer is "higher level", but showing the complement of the set is open is far more direct for a beginner (and a more likely line of inquiry). hopefully you can work out both approaches :) – citedcorpse Jun 25 '13 at 18:29
  • Correct me if I'm wrong, but I don't see where the fact that $[0,1]$ is compact is necessary to solve this problem. – dc2814 Jun 25 '13 at 19:19
  • @dc2814 it's just so the $\sup$ norm is always defined, otherwise you have to restrict to the functions where it is defined – citedcorpse Jun 26 '13 at 08:32
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    @exitingcorpse Ahh, got it. Thanks! – dc2814 Jun 26 '13 at 12:06

3 Answers3

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If $A$ was just one point, could you see why it's true? Can you see how this proves it for general $A$?

citedcorpse
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I think the way to approach this is to use the following characterization for closed sets.

Let $X$ be a metric space and $B\subset X$. Then $B$ is closed in $X$ iff $B$ contains all of its limit points. $y$ is a limit point for $B$ if there is a sequence $\{b_n\}\subset B$ such that $b_n \to b$.

So suppose $g$ is a limit point of $B=\{f\in C[0,1] | f(a) = 0\; \forall a\in A\}$, and $\{g_n\} \subset B$ is a sequence such that $g_n \to g$, and let $a\in A$. That $g_n \to g$ just means that $g_n$ converges to $g$ uniformly. So if $g_n(a) = 0$, then what can you say about $g(a)$?

dc2814
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First proof it taking $A$ as a singleton which is shown by someone else above then it is obvious because intersection of closed sets is closed.

user577215664
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Manoranjan Singha
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