By definition, I need to show that every point in $A$ is an interior point in $A$. Do I go about this by showing $A'$ is closed? Please help.
Correction in the title: For all $t\in{[0,1]}$.
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You mean $t\in[0,1]$ in the title. – Mhenni Benghorbal Feb 20 '14 at 05:23
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1Assuming $;A;$ is closed won't give you a contradiction proof. For that you need to assume $;A;$ isn't open (which is not the same as being closed) – DonAntonio Feb 20 '14 at 05:24
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Hint: $[0,1]$ is compact; what does this imply about the maximum / minimum values of $f$?
Given this, convince yourself that
$$\|f - g\| < \epsilon \implies \max_{t \in [0,1]} |g(t)| < \max_{t \in [0,1]} |f(t)| + \epsilon$$
Can you finish it from here?
Remark: Showing that $A$ is closed doesn't help at all, since open and closed are not opposites of each other.
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We haven't talked about compactness but I'm guessing the minimum and maximum values of $f$ are $0$ and $1$ respectively. – mmh0015 Feb 20 '14 at 05:43
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@user121087 No information is given about the maximum and minimum values of $f$; for example, $f$ could be the constant function $1/2$. The relevance of compactness is the extreme value theorem which implies that $f$ actually attains its maximum in $[0,1]$. – Feb 20 '14 at 05:45
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Show its complement is closed. Let $f_n$ be a convergent sequence such that $f_n(t_n)\geqslant 1$ for a sequence of points $t_n\in [0,1]$. Let $t_{n_k}$ be a convergent subsequence, guaranteed by Bolzano Weiertrass. I claim that if $t_0=\lim t_{n_k}$, then $f(t_0)\geqslant 1$. Since $f_n\to f$ uniformly, in particular $f_{n_k}\to f$ uniformly, so $1\leqslant f_{n_k}(t_{n_k})\to f(t_0)$.
Pedro
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Thank you. I'm not sure how this way involves the supremum metric though. – mmh0015 Feb 20 '14 at 05:59
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