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Let $(X,d)$ be a path connected metric space. Prove that $X$ is a connected space. Towards contradiction assume that $X$ is not connected: $$\exists U,V \subset X \text{ open, unempty} : U \cap V = \emptyset, U \cup V = X $$ Let $x \in U, y \in V$. Since $X$ is path connected, we find a continuous map $f: [0,1] \rightarrow X$ such that $f(0) = x \text{ and } f(1)=y$.

since $U,V$ are open with $U \cap V = \emptyset$ we get $S :=f([0,1])\cap(\delta U \cup\delta V) \neq \emptyset $

Now let $p \in S \Rightarrow p \notin U \cup V = X$ which is a contradiction. So X must be connected.

Is there a problem with that proof?

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    $\begingroup$ What are $\delta U$ and $\delta V$? $\endgroup$ – Matthias Klupsch Sep 5 '17 at 11:03
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    $\begingroup$ Perhaps you should argument why $f([0,1])\cap(\delta U\cup\delta V)\neq\emptyset$, unless it is already done a previous exercise or theorem. $\endgroup$ – ajotatxe Sep 5 '17 at 11:04
  • $\begingroup$ The line where you introduce $\delta U$ etc. is inconclusive. Where did you use that $f$ is continuous? What do you get when you use that $f$ is continuoous and $U,V$ are open? $\endgroup$ – Hagen von Eitzen Sep 5 '17 at 11:05
  • $\begingroup$ I agree with @HagenvonEitzen that I dislike the line where you introduce $\delta U$ (and the one after it). It doesn't feel like the right way to go but it also doesn't feel very much related to paths or connectedness. I feel like the proof should morally say "if the space is not connected then we can split a path in two". My favourite definition for connectedness is that a space is disconnected if it subjects (continuously) into $\{0,1\}.$ I find it is much more natural for proving things like this; perhaps you will find the same. $\endgroup$ – Dan Robertson Sep 5 '17 at 11:38
  • $\begingroup$ Since $f$ is continuous the path has to cross the boundaries $\delta U$ and $\delta V$. $\endgroup$ – dba Sep 5 '17 at 15:06
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If you do not use the continuity of $f$, your proof will go wrong! Let $U,V$ and $f$ as in your proof.

Then we have

  1. $[0,1]=f^{-1}(X)=f^{-1}(U \cup V)=f^{-1}(U) \cup f^{-1}(V)$

and

  1. $\emptyset=f^{-1}(U \cap V)=f^{-1}(U) \cap f^{-1}(V)$

and, since $f$ is continuous,

  1. $f^{-1}(U)$ and $f^{-1}(V)$ are open in $[0,1]$.

Hence, it follows from 1. and 2. ,that $[0,1]$ is not connected, a contradiction.

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