Let $X, Y$ be topological spaces.

I want to understand better the structure of the space $C(X,Y)$ of all continuous functions from $X$ to $Y$. Clearly, if $X$ has the indiscrete topology and $Y$ has the discrete topology, then the only continuous functions are the constants.

Now come my questions:

1.) If $X$ is not indiscrete, is there always a non-constant continuous function?

2.) If $Y$ is not discrete, is there always a non-constant continuous function?

3.) If $X$ is not indiscrete and $Y$ is not discrete, is there always a non-constant continuous function?

It seems to be very hard to construct a non-constant continuous map just by knowing that there is one non-trivial open set in $X$ and/or one set in $Y$ not being open. But on the other hand I am not able to construct a counterexample.

Thanks in advance for all help!

  • 2
    $\begingroup$ Related/duplicate? – Existence of non-constant continuous functions $\endgroup$ – Martin R May 29 '19 at 12:05
  • $\begingroup$ By "the indescrete topology", do you mean the trivial topology (i.e. $\{X, \emptyset\}$)? $\endgroup$ – 5xum May 29 '19 at 12:05
  • 1
    $\begingroup$ There is no nonconstant continuous function from $\Bbb R$ to $\Bbb R_{\text{cocountable}}$, where $\Bbb R_{\text{cocountable}}$ is the cocountable topology (countable complement topology). [proof: $f(\Bbb Q)$ is countable, hence closed in $\Bbb R_{\text{cocountable}}$. $f^{-1}(f(\Bbb Q))$ is a closed set containing $\Bbb Q$, hence equal to $\Bbb R$. $f^{-1}(f(\Bbb Q))=\Bbb R$ implies $f(\Bbb R)=f(f^{-1}(f(\Bbb Q)))\subseteq f(\Bbb Q) $, hence $f(\Bbb R)$ is countable. A countable, connected subset of $\Bbb R_{\text{cocountable}}$ must be a one-point set. Hence $f$ is constant.] $\endgroup$ – YuiTo Cheng May 29 '19 at 12:14
  • $\begingroup$ Yes, that is what I mean by the indiscrete topology. $\endgroup$ – Daniel W. May 29 '19 at 12:17

Let $X$ be topology on $\{1, 2, 3\}$ with base of $\{1\}, \{1, 2\}, \{1, 3\}$ and $Y$ be any Hausdorff space.

Assume $f(1) \neq f(2)$. Take open $U \subset Y$ s.t. $f(2) \in U$, $f(1) \notin U$. Then $2 \in f^{-1}(U)$, $1 \notin f^{-1}(U)$, so $f^{-1}(U)$ is not open. So $f(1) = f(2)$.

Analogously, $f(1) = f(3)$, and so $f$ is constant.


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