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This question already has an answer here:

Let $M$ be a manifold. Prove that $M$ is path connected if and only if $M$ is connected.

My attempt:

($\Rightarrow$) Let $M \subset \mathbb{R}^n$ be path connected. Every path connected subset of $\mathbb{R}^n$ is connected.

($ \Leftarrow$) Let $M \subset \mathbb{R}^n$ be connected. I think that the goal here is to show that for any $x,y \in M$, there exists $a,b \in \mathbb{R}$ and a path $\Phi:[a,b] \rightarrow M$ such that $\Phi(a)=x$ and $\Phi(b)=y$.

I know that since $M$ is a manifold, there exists an open neighborhood $U$ around $x,y \in M$. I also know that $M$ contains a diffeomorphism.

Not sure how to put this all together in my proof..

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marked as duplicate by Moishe Kohan, Misha Lavrov, C. Falcon, Daniel W. Farlow, Zain Patel Apr 23 '17 at 5:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Fix $p\in M$, and define $\Sigma_p$ to be the set of all $q\in M$ for which there is a path in $M$ from $p$ to $q$. Show that $\Sigma_p$ is nonempty, open, and closed in $M$. Then use connectedness of $M$. $\endgroup$ – kobe Apr 22 '17 at 20:31
  • $\begingroup$ @MoisheCohen My question is specifically about my own proof $\endgroup$ – combo student Apr 22 '17 at 22:40
  • $\begingroup$ @MoisheCohen but your link made me think... are we trying to prove here connected spaces or connected components? $\endgroup$ – combo student Apr 22 '17 at 22:49