# $M$ is path connected if and only if $M$ is connected [duplicate]

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..

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

• 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 $M$. – kobe Apr 22 '17 at 20:31