0
$\begingroup$

Suppose $(X, \tau_X )$ is path-connected. Then is it true that there is a continuous function from $(X, \tau_X )$ to $([0,1],τ_E)$ with $f(x) = 0$ and $f(y) = 1$ ?

My answer is no, the counterexample I give is $X=[0,1) \cup (1,2]$, $x=0$ and $y=\frac\pi2$, so that $f(z)=\sin(z)$ satisfies $f(0)=0$ and $f(\frac\pi2)=1$ and $f$ is clearly continuous.

I guess there is a much easier counterexample, any ideas ?

$\endgroup$
3
  • 1
    $\begingroup$ Are you claiming that $X$ is path-connected? $\endgroup$ Jan 25 at 12:33
  • $\begingroup$ @JoséCarlosSantos What a misunderstanding from me, I'll fix that asap $\endgroup$
    – Kilkik
    Jan 25 at 12:36
  • $\begingroup$ The $X$ you give is not path-connected. $\endgroup$
    – Dan Rust
    Jan 25 at 12:49

2 Answers 2

1
$\begingroup$

Let $X$ be a topological space which consists of a single point. Then $X$ is path-connected, but the range of any map from $X$ to $[0,1]$ consists of a single point.

$\endgroup$
3
  • $\begingroup$ Sure ! But what about a space with at least two distinct points ? $\endgroup$
    – Kilkik
    Jan 25 at 12:43
  • $\begingroup$ That was not part of the problem when I posted my answer. $\endgroup$ Jan 25 at 12:44
  • $\begingroup$ Yes sorry, should I bring back the old version, accept the answer and post a new question ? $\endgroup$
    – Kilkik
    Jan 25 at 12:45
0
$\begingroup$

If $X$ is $\Bbb R$ in the indiscrete (trivial) topology, then $X$ is path-connected (any map into an indiscrete space is continuous) but there is no non-constant continuous map to $([0,1], \tau_E)$.

$\endgroup$

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.