In $C\left( {\left[ {0,1} \right]} \right)$, we define metric $$d\left( {f,g} \right) = \mathop {\max }\limits_{t \in \left[ {0,1} \right]} t\left| {f\left( t \right) - g\left( t \right)} \right|.$$ We define $A = \left\{ {f \in C\left( {\left[ {0,1} \right]} \right):f\left( 0 \right) = f\left( 1 \right)} \right\}$. Is $A$ an open or closed set in $C\left( {\left[ {0,1} \right]} \right)$ with metric $d$??
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1Yes, it is definitely either open or closed. – Jun 13 '18 at 03:55
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2@SaucyO'Path: it's certainly not the case that every set must be either open or closed. – Greg Martin Jun 13 '18 at 05:31
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Hint: considering $f(t)=0$ and $g(x) = \epsilon t$ should enable you to answer one of your questions. – Greg Martin Jun 13 '18 at 05:36
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@GregMartin I agree, and it's all the more the case for vector subspaces, since vector subspaces are open only if they are closed. – Jun 13 '18 at 05:38
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1@SaucyO'Path: This claim is more than wrong. Subspaces are open if and only if they equal the entire space! – gerw Jun 13 '18 at 06:28
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@gerw Where are you negating the fact that they are open only if they are closed? – Jun 13 '18 at 06:56
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@SaucyO'Path: Because this "fact" is plainly wrong! Do you have a good reason to believe that subspaces are open iff they are closed? – gerw Jun 13 '18 at 09:31
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"only if" is not the same as "if and only if". @gerw – Jun 13 '18 at 09:48
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If a subspace is open, then it is closed. Some people were confused by "only if". – GEdgar Jun 13 '18 at 11:43
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@SaucyO'Path: Oh, I always read "if and only if". Sorry for the noise. – gerw Jun 13 '18 at 14:16
1 Answers
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$A$ is not open
Consider the sequence $(f_n)$ defined by $$f_n(x)=\begin{cases} 1 -nx & 0 \le x \le 1/n\\ 0 & 1/n \le x \le 1 \end{cases}$$
It is a sequence of continuous functions. You'll verify that $d(f_n,0) = 1/4n \to 0$, $0 \in A$ and $f_n \notin A$ for all $n \in \mathbb N$. Therefore $A$ can't be open.
$A$ is not closed in $\mathcal C([0,1])$
Consider the sequence:
$$g_n(x)=\begin{cases} 1 -nx & 0 \le x \le 1/n\\ \frac{n}{n-1}\left(x-\frac{1}{n}\right) & 1/n \le x \le 1 \end{cases}$$
of elements of $A$. It converges in $\mathcal C([0,1])$ to the identity map $i$ which does not belong to $A$. Hence $A$ is not close.
mathcounterexamples.net
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I would appreciate explaining one aspect connected with the reason why the set $A$ is not open. I understand every step, but the idea isn't clear for me. We have shown that $f_n$ (which are not in the set $A$) converges to $0$ in our metric and its limit belongs to $A$. Intuitively it shouldn't be possible of course, but my main question is: how is this associated with checking if $A$ is an open set? I can't see here relationship with a definition of open set. Thanks in advance. – Novice Jul 13 '20 at 15:54
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1If $A$ would be open, as $0 \in A$, there would be an open ball centered on $0$ and included in $A$. This can't be as $f_n \notin A$ for all $n \in \mathbb N$. – mathcounterexamples.net Jul 13 '20 at 16:03
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The part on the closedness of $A$ is not correct. The question asks if $A$ is a closed subset of the continuous functions, so giving a sequence that converges to a non-continuous function cannot answer that. – MaoWao Jul 14 '20 at 08:20
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@MaoWao You're right and $A$ is in fact closed in $\mathcal C([0,1])$. Response updated. – mathcounterexamples.net Jul 14 '20 at 09:35
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But does convergence with respect to $d$ imply convergence at $0$? With the factor $t$ in the definition of $d$, this seems at least not clear. – MaoWao Jul 14 '20 at 10:16
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Not good to come back after two years and two quickly on an answer. My initial answer is correct as the identity map is continuous. Your initial comment is wrong. – mathcounterexamples.net Jul 14 '20 at 10:26
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The initial version was correct, but then you somehow changed it to "converging to a non-continuous map". Anyway, everything is back to the correct version now, so that's fine. – MaoWao Jul 14 '20 at 11:38