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Let $N$ be a normed vector space, let $M$ be a metric space, and let $A$ be a subset of $M$. Let $V$ denote the space of all bounded functions from $A$ to $N$, with the sup norm. Let $C_b$ denote the set of all bounded continuous functions from $A$ to $N$. Show that $C_b$ is closed in $V$.

I think I'm suppose to use Arzela-Ascoli Theorem but do not know how to go about it

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  • $\begingroup$ What is $C_b{}$? What topology are you putting on the space of all bounded functions on $A$? $\endgroup$ Oct 30, 2015 at 3:36
  • $\begingroup$ C={f€V| f is continuous} where V is the set of all functions f: A to N. N is a normed vector space and A € M and M is a metric space. $C_b$ is a subset of C consisting of bounded functions. $\endgroup$
    – 3141
    Oct 30, 2015 at 3:47

1 Answer 1

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Hint: Arzela-Ascoli won't be helpful here. To show that $C_b$ is closed, suppose you have a sequence $f_n$ of elements of $C_b$ which converge to some function $f\in V$. You want to show that $f$ is in $C_b$. To show this, you can use the fact that a uniform limit of continuous functions is continuous.

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  • $\begingroup$ That hint was given to me in the back of the book but that made me more confused. $\endgroup$
    – 3141
    Oct 30, 2015 at 4:04
  • $\begingroup$ I have added a bit more detail. Are you still confused? $\endgroup$ Oct 30, 2015 at 4:06
  • $\begingroup$ So since fn converges to a function f in Cb, fn is closed and therefore Cb is closed in V? $\endgroup$
    – 3141
    Oct 30, 2015 at 4:12
  • $\begingroup$ What do you mean by "$f_n$ is closed"? $\endgroup$ Oct 30, 2015 at 4:21
  • $\begingroup$ Fn is closed in Cb because it converges to f and f is in Cb right? $\endgroup$
    – 3141
    Oct 30, 2015 at 12:55

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