Theorem 4.19 in Kechris' Classical Descriptive Set Theory says that the space of continuous functions from a compact metric space to a Polish space is Polish. It is therefore obvious that the space of continuous functions from a compact Polish space to a Polish space is Polish.

The space of continuous function is equipped with a product metric, for example, the Chebyshev metric.

My question is, can we drop the "compactness"?

What anomaly would arise?

If we drop the "continuity", then some problems would arise for cardinality. However, with continuity, we don't have this problem.

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    $\begingroup$ Consider $C(\mathbb R,\mathbb R)$. $\endgroup$ Jan 26 '20 at 19:45
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    $\begingroup$ $K$ compact, $X$ metric: For space $C(K,X)$ use uniform convergence. (I assume Kechris says this somewhere before 4.19.) With other domains, for $C(Y,X)$ try the compact-open topology en.wikipedia.org/wiki/Compact-open_topology When $Y$ is locally compact Hausdorff, this still has some good properties. $\endgroup$ Jan 26 '20 at 20:25
  • $\begingroup$ I suppose that if this is the question you wanted to ask over there (mathoverflow.net/questions/351160/…), you should edit the other question and delete this one. $\endgroup$
    – Arno
    Jan 26 '20 at 22:01
  • $\begingroup$ @Arno. Re: "you should edit the other question and delete this one"... Please keep in mind that High GPA is a new MathOverflow poster. Let's try to be gentle and welcoming. $\endgroup$ Jan 27 '20 at 0:23
  • $\begingroup$ Chebyshev? -- r'u sure? $\endgroup$
    – Wlod AA
    Jan 27 '20 at 2:02

As already answered here ($X$ is Polish and $N$ is countable. Is $N^X$ Polish?):

Let us work in a convenient category of topological spaces, ie assume that we have function spaces and sufficiently many nice spaces around. My favourite such category is the category of $\mathrm{QCB}_0$-spaces, which are $T_0$ quotients of countably-based spaces [1].

Inside the category of $\mathrm{QCB}_0$-spaces, Schröder has obtained a characterization of the Hausdorff spaces $\mathbf{X}$ such that $\mathbb{R}^\mathbf{X}$ is Polish [2]. These spaces are now called coPolish spaces. A Polish space is coPolish iff it is locally compact. So we can weaken compactness to local compactness, but not further.

A space such as $2^{(\mathbb{N}^\mathbb{N})}$ even fails to be countably based.

[1] https://www.sciencedirect.com/science/article/pii/S0304397501001098

[2] https://onlinelibrary.wiley.com/doi/abs/10.1002/malq.200310111

  • $\begingroup$ The other question does not involve continuity, though, so in my humble opinion, they are quite different $\endgroup$
    – High GPA
    Jan 27 '20 at 5:03

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