(connected and locally path-connected) => (path-connected)

[quasar987]

Homework Statement

I'm asked to show that for a metric space M, (connected and locally path connected) ==> (path connected).

The reason for my post is 3-fold.

1) Not 100% sure my solution is correct.
2) Want confirmation that my proof holds in a general topological space.
3) I'm interested in hearing about other methods of proof for this result.

The Attempt at a Solution

Let a be in M and let U be the biggest path-connected open nbhd of a. If U is all of M, then there is nothing to show. If U is not all of M, then first observe that bd(U), the boundary of U, is not empty. If it were, then U would be open and closed, which contradicts the fact that M is connected.

So let x be in bd(U). x must not be in U, otherwise, U would not be open. But since M is locally path-connected, there is an open nbhd V of x that is path-connected and that intersects U. This leads to a contradiction, either because it means x is in U or because U u V is a bigger path-connected open nbhd of a than U is.

 

[quasar987]

Well now I'm 100% sure that it's correct and I am even more convinced that it holds in a general topological space. My only doubt was related to this part:

If U is not all of M, then first observe that bd(U), the boundary of U, is not empty. If it were, then U would be open and closed, which contradicts the fact that M is connected.

that I had proven using 'cl(A)=A u bd(A)'. But I now see clearly that it also holds in a general topological space: If bd(U) is empty, then M\U is closed because U is open and is open also because let y be in M\U. there must be a nbhd W of y that is entirely contained in M\U because the contrary would mean that y is in bd(M\U)=bd(U) ==><==.