I am not an expert of this area but I need help to answer this question: For a compact metric space $X$, let $C(X,X)$ be the space of all continuous functions $X\to X$, equipped with the uniform metric $$d(f,g)=\sup_x d(f(x),g(x)).$$ Can we say the space $C(X,X)$ is compact?
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3$\begingroup$ For $X=[0,1]$ it will be non-compact. $\endgroup$– erzCommented Mar 10, 2020 at 6:10
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$\begingroup$ @erz Thank you so much. $\endgroup$– Sh.M1972Commented Mar 10, 2020 at 6:16
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$\begingroup$ See the Arzelà–Ascoli theorem. $\endgroup$– abxCommented Mar 10, 2020 at 6:40
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2$\begingroup$ Note that a compact metric space is always complete and separable. $\endgroup$– YCorCommented Mar 10, 2020 at 7:27
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2$\begingroup$ @M.Shahryari it's quite trivial to find $X$ with $C(X,X)$ not compact, so maybe you should reformulate the question taking this into account (e.g., asking about a characterization of $X$ such that $C(X,X)$ is compact). Note that (for $X$ compact) this is equivalent to asking when $C(X,X)$ is closed in $X^X$, and hence a related question is this one. $\endgroup$– YCorCommented Mar 10, 2020 at 8:31
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Indeed this space is (almost, see the comment by YCor) never compact except for trivial cases (i.e. the compact set $X$ having only finitely many points). For more information on the topology (which is colloquially known as the compact-open topology) of the space you are asking about, see e.g. the book Engelking: General topology. (Chapters 2.6 and 3.4 will have most of the information you could possibly need).
Thanks for the clarification by @YCor, the never compact of the original answer was a bit hasty and comes from my use of these spaces where $X$ is always locally euclidean (i.e. the $X$ should be a manifold). Which is of course not necessary.
answered Mar 10, 2020 at 7:06
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3$\begingroup$ This seems to be in contradiction with this answer, which says that there exists an infinite connected compact metrizable space, in which every continuous self-map is identity or constant. $\endgroup$– YCorCommented Mar 10, 2020 at 7:37
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2$\begingroup$ Still for $X$ compact metrizable $C(X,X)$ is non-compact in many cases, notably (a) when $X$ has infinitely many connected components (b) when $X$ contains a non-trivial arc. Also, for $X$ compact metrizable $C(X,[0,1])$ is compact iff $X$ is finite. $\endgroup$– YCorCommented Mar 10, 2020 at 8:28
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$\begingroup$ @YCor so $C(A^G, A^G)$ is not compact, because $A^G$ is totally disconnected ($G$ is f.g). $\endgroup$– Sh.M1972Commented Mar 10, 2020 at 9:54
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$\begingroup$ @YCor: Thanks for the correction, I am always thinking nice metrisable spaces, i.e. manifolds... $\endgroup$ Commented Mar 10, 2020 at 12:08