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I have a Metric Spaces exam on Tuesday and the following question comes up a lot. I have attempted it multiple times but i cant seem to come up with a valid solution. I could really do with a solution as soon as possible. Thanks in advance.

consider the metric space $(X,d)$ where $X$ is the set of functions $f:[0,1]\to[0,1]$, and the distance $d$ is given by

$$d(f,g)=\Vert f−g\Vert_{\infty}=\sup\{|f(t)−g(t)|:t\in[0,1]\}$$ let

$J=\{f\in X|\forall x,y\in [0,1],x\le y:f(x)\le f(y)\}$ be the set of non-decreasing functions in $X$

(a) is $J$ closed in $(X,d)$ (12 marks)

(b)is $J$ open in $(X,d)$ (12 marks)

(c) is $J$ compact (12 marks)

Norse
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    Please write out explicitly your attempts in the question. What did you do when attempting to solve this problem? – Clement Yung Jan 12 '20 at 14:34
  • Also, for future reference, when asking question, you should format the math in mathJax as described here – Norse Jan 12 '20 at 14:35
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    Apologies, I am new to this website. –  Jan 12 '20 at 14:37
  • @ClementYung My understanding of the topic is quite poor and despite research I was unable to find anything useful relating to this question. My attempts are non-sensical. I attempted to show that if J was closed then a a sequence in J would be cauchy and then the sequence would also be cauchy in X. –  Jan 12 '20 at 14:39
  • $J$ is closed if it contains all its limit points, i.e. if for all sequences of function ${f_i}_{i\in \Bbb N} \subseteq X$ which converges to some function $f$, we have that $f \in J$. $J$ is compact if it is closed and bounded. – Norse Jan 12 '20 at 14:43
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    @Norse: There is a bit of disorderliness in the applying of universal quantifiers in the above characterization of closedness: one says that a subset $M \subseteq X$ is closed in the (unmentioned) topology of space $X$ if it contains all its adherent points (it is true that this is equivalent to requiring that it contains all its accumulation points, however what you describe in your comment above seems to be a general adherent point, so let us be precise). – ΑΘΩ Jan 20 '20 at 08:21
  • @Norse: Next, in a first countable space, a point $x \in X$ will be adherent to subset $M$ if and only if there exists a sequence $t \in M^{\mathbb{N}}$ of points in $M$ converging to $x$. Therefore, the correct quantification in your above description would be: ''$M$ is a closed subset if it contains any point $x \in X$ which occurs as a limit of some sequence $t$ of points in $M$''. – ΑΘΩ Jan 20 '20 at 08:27

1 Answers1

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Let $\{f_n\}_{n \in \Bbb N}\subseteq J$ be a convergent sequence of functions in $J$. Let $f$ denote the limit of the sequence, i.e. $f(x) = \lim_{n \to \infty} f_n(x)$.

We prove that $f\in J$ proving that $J$ is closed: Let $x, y \in [0,1]$ such that $x \le y$, then $$ f(x)=\lim_{n\to\infty}f_n(x) \le \lim_{n\to\infty}f_n(y)=f(y) $$ since $f_n\in J$ for all $n$. Hence $f$ is non-decreasing, so $f \in J$, so $J$ contains all its limit points.

Norse
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  • Thanks so much Norse. –  Jan 12 '20 at 14:56
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    @Gavink: as to the remaining two questions, since your given metric space is arc-connected (given $f, g \in X$ the map $[0,1] \to X, \lambda \mapsto \lambda f+(1-\lambda)g$ is continous) hence connected, so its only closed-open sets are the extreme ones (the full space itself and the empty subset); therefore, as $J$ is closed non-empty and proper, it cannot be also open. For c), $(X, d)$ is compact and therefore any closed subset is also compact. – ΑΘΩ Jan 12 '20 at 15:11
  • @ΑΘΩ Thank you so much. I really appreciate the help. –  Jan 12 '20 at 15:16
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    @GavinK14: I actually made a big and ugly blunder in hastily replying without properly checking my background reasoning. What I said about b) is valid indeed, however what I said about c) is wrong: the topology given on $X$ is that of uniform convergence, and in order to discern the compactness of subset $M$ (I dislike the choice of $J$ for that set) one resorts to the Arzela-Ascoli theorem. However, $M$ will not be equicontinuous in points $x \in [0 ,\frac{1}{2})$, since for $\epsilon=\frac{1}{3}$ and for any $\delta>0$ there will exist increasing functions – ΑΘΩ Jan 14 '20 at 02:13
  • @GavinK14 (continuation of the above): increasing functions such as the one (call it $f$) acting identically on $[0, x]$, affine and strictly increasing in mapping $[x, x+\frac{\delta}{2}]$ to $[x, 1]$ and then constantly equal to $1$ on $[x+\frac{\delta}{2}, 1]$. It will then be the case that $\left|f(x+\frac{\delta}{2})-f(x)\right|=1-x>\frac{1}{2}>\epsilon$, prohibiting equicontinuity. – ΑΘΩ Jan 14 '20 at 02:18