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Let $$X=\{f:[0,1]\rightarrow[0,1], f\text{ is Lipschitz continuous}\} $$with the supremum metric .

What can we say about the compactness of$ (X,d).$

I think the result that ''A space is compact iff every continuous real valued function on X is bounded" might be useful.

I can't think of anything else. Kindly help !!

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  • $\begingroup$ "A space is compact iff every continuous real function is bounded." That is not true. $\endgroup$ – Mars Plastic Mar 12 '19 at 12:59
  • $\begingroup$ Is there a different version of result ?@MarsPlastic $\endgroup$ – Devendra Singh Rana Mar 12 '19 at 13:04
  • $\begingroup$ A topological space $X$ with the property that every continuous $f\colon X\to \Bbb R$ is bounded is called pseudo-compact. For the relation other types of compactness, this is a good overview. $\endgroup$ – Mars Plastic Mar 12 '19 at 13:06
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If $f_n(x)=x^n$ ($x\in[0,1]$), then $f_n$ is Lipschitz continuous. However, the sequence $(f_n)_{n\in\mathbb N}$ has no convergente subsequence. Therefore, your space is not compact.

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Posting just for completeness since the answer was already provided, and maybe this comment will be useful for those stumbling upon this question in the future.

A more insightful topology defined on $X$ is the one which present $X$ as a union of compact spaces.

If we call $$ X_n=\{f:[0,1]\to[0,1]\ |\ \mathrm{Lip}(f)\leq n,\ \|f\|_\infty\leq n\} $$ which we know are compact from the Ascoli Arzelà theorem.

The most natural topology on $X=\bigcup_n X_n$ is given by the finest topology which makes all the embeddings $i_n: X_n\to X$ continuous.

The compact subsets of $X$ with this topology are those closed sets which are contained in some $X_n$.

Your metric space $(X,d)$, which we will call $Y$ for clarity, "contains" our space $X$, i.e. we have a continuous mapping $$ X\to Y. $$ This tells us that all the compacts subspaces of $X$ are compacts in $Y$ but a priori $Y$ should have more compacts than just those defined by $X$.

An example of such a compact set given by $$ C=\left\{f_n\in Y\ |\ f_n(t)=\frac{\sin(n^2t)}{n}\right\}\cup\{0\} $$ which converges present a sequence converging uniformly to $0$ but not in any $X_n$.

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