# connected manifolds are path connected

prove every connected manifold is path connected manifold .

my thought:

connected space : Let $X$ be a topological space. A separation of $X$ is a pair $U, V$ of disjoint
nonempty open subsets of $X$ whose union is $X$. The space $X$ is said to be connected
if there does not exist a separation of $X$ .

Components: Given $X$ , define an equivalence relation on $X$ by setting x~y if there
is a connected subspace of $X$ containing both $x$ and $y$ . The equivalence classes are
called the components (or the “connected components”) of $X$.

Path Component: I define another equivalence relation on the space $X$ by defining x ~ у
if there is a path in $X$ from $x$ to $y$ . The equivalence classes are called the path components of $X$ .

Theorem : The path components of $X$ are path connected disjoint subspaces
of $X$ whose union is $X$ , such that each nonempty path connected subspace of $X$ intersects only one of them.

thank you so much

#### Solutions Collecting From Web of "connected manifolds are path connected"

Let $x\in X$. Consider $U:=\{y\in X\mathop{|}\text{there is a path from$x$to$y$}\}$. So $U$ is nonempty: $x\in U$. Claim: $U$ and $U^c:=X\smallsetminus U$ both are open. To prove this use the fact that given any point $z\in X$ there is a neighbourhood of $z$ which is homeomorphic to an open ball of $\mathbb{R}^n$ for some $n$. The homeomorphic image of a path connected set is path connected, so $U$ and $U^c$ both are open but $U \cup U^c = X$ which implies that $U=X$ since $U$ is nonempty.

path connected sets are also connected, conversely is not always true (there is a famous counterexample which shows this fact). However a necessary condition To make the converse of the previous statement possible is that the connected sets are required to be locally Euclidean. So a topological manifold, by definition is also a locally Euclidean space, if you can prove the previous statement you can get the answer. Hope it helps