1
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ Related/duplicate? – Existence of non-constant continuous functions $\endgroup$ – Martin R May 29 at 12:05
  • $\begingroup$ By "the indescrete topology", do you mean the trivial topology (i.e. $\{X, \emptyset\}$)? $\endgroup$ – 5xum May 29 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 at 12:14
  • $\begingroup$ Yes, that is what I mean by the indiscrete topology. $\endgroup$ – Daniel W. May 29 at 12:17
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.