2
$\begingroup$

I have difficulty coming up with a proof that if two points in a topological space $X$ are connected by a path, then there is an injective continuous map of $[0,1]$ to $X$ that sends $0$ and $1$ to those two points. Is this true? How is this proved?


Edit: As answered by Zev Chonoles, this is obviously false for general spaces. I would also like to have an answer for Hausdorff spaces.


Update: it turns out the hard part of this question is a duplicate of A question about path-connected and arcwise-connected spaces

$\endgroup$
  • $\begingroup$ General spaces? Hausdorff spaces? $T_1$ spaces? $\endgroup$ – Asaf Karagila Mar 6 '13 at 8:20
  • $\begingroup$ As stated, i am asking about general spaces. In fact, i am more interested in Hausdorff spaces. $\endgroup$ – Alexey Mar 6 '13 at 8:23
2
$\begingroup$

Let $X=\{a,b\}$ with the topology $T=\{\varnothing,\{a\},X\}$ (which is non-Hausdorff). The map $f:[0,1]\to X$ defined by $$f(t)=\begin{cases} a & \text{ if }t\in [0,1),\\ b & \text{ if }t=1 \end{cases}$$ is continuous, but there is clearly no injective map from $[0,1]$ to $X$.


Your conjecture is true if we require $X$ to be Hausdorff. According to Wikipedia, a path-connected Hausdorff space is necessarily arc-connected. Thus, if $f:[0,1]\to X$ is the path from $a$ to $b$, the subspace $f([0,1])$ of $X$ is path-connected and Hausdorff, hence arc-connected, hence there is an arc connecting $a$ and $b$, which is in particular an injective continuous path $[0,1]\to X$.

$\endgroup$
  • $\begingroup$ Thanks, it was obvious! Excuse me for not accepting your answer because i am more interested in Hausdorff spaces, i am going to edit the question. $\endgroup$ – Alexey Mar 6 '13 at 8:27
  • $\begingroup$ Thanks for pointing me to arc connected spaces, i forgot that those are different terms. I have also found that this question is a duplicate. If i find somewhere a link to a proof, i will accept your answer, otherwise i am interested in a link to a proof. $\endgroup$ – Alexey Mar 6 '13 at 9:03

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.