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I've seen that are one or two questions like this, but I'm not fully convinced that are right. Both pretty much say the same thing, so I leave the easier to understand:

path connectedness implies connectedness

The thing I'm not quite sure, is that if $f$ is a path, then I have that $f(0)=a$ and $f(1)=b$ for some $a,b \in X$. But the demonstration, uses that $\DeclareMathOperator{\Im}{Im}\Im(f)=X$ which, as far as I know, it's not necessarily true.

One thing I thought, but I'm not sure if is right as a proof, is that since every point in $X$ can be joined with a path, and I can join two different paths, I can make an uncountably union of paths to join every point of $X$ and then I could apply the link before, but, for example, I might have a problem if $X$ has more elements than $\mathbb{R}$.

What I was thinking is mostly the same, but when I was about to take $$f^{-1}(\Im(f)) = f^{-1}(\Im(f) \cap M) = f^{-1}(\Im(f) \cap (U \cup V)) = f^{-1}((\Im(f)\cap U) \cup (\Im(f) \cap V))$$ I realized that $\Im(f) \cap U$ might not be open, so this wouldn't imply that $[0,1]$ is disjoint, so I don't have a contradiction.

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    $\begingroup$ I fail to see where the proof you linked to uses Im$(f) = X$. Could you explain where you think this was used? In case you are referring to the statement $[0,1] = f^{-1}(X)$, note that this is not equivalent to $f([0,1]) = X$ but only to $f([0,1]) \subseteq X$. $\endgroup$ – Matthias Klupsch May 15 at 13:23
  • $\begingroup$ You're right! I thought it all wrong. Thanks $\endgroup$ – Dani Seidler May 15 at 13:36

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