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Let $ A=\left\{ f\in C\left[ 0,1\right] |\text{ }f\text{ is strictly increasing and }f\left( 0\right) =0\text{ and }f\left( 1\right) =1\right\} $, where $ C\left[ 0,1\right]$ is the set of continuous functions over $\left[ 0,1\right]$. Is $A$ closed with respect to sup metric (or, some other nontrivial metric)?

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    $\begingroup$ What did you try? What is a nontrivial metric? $\endgroup$ – José Carlos Santos Apr 17 at 13:24
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No, it isn't. Think of a continuous (non-strictly) increasing function $f$ with $f(0) = 0$ and $f(1) = 1$ that has a "plateau" around $x = 1/2$. That means, $f(x) = 1/2$ for all $x$ in a neighborhood of $x=1/2$. It oculd even by piecewise linear. You can now easily think of a sequence of functions in $A$ that converge uniformly to $f$, can you?

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