Locally path connected implies path connected

[Mosis]

Homework Statement

We say the metric space X is locally path connected (lpc) if all balls are path connected sets.

Suppose X is lpc and that E is an open and connected subset of X. Prove that E is path connected.

Homework Equations

A set is open if every point is an interior point. A set is connected if it cannot be written as the union of two open, disjoint, non-empty sets. A set is path connected if given any two points x,y in the set, there exists a path between them.

The Attempt at a Solution

None really. My professor suggested that for a fixed x, I consider the set Y of all y that are path connected to x and show that this set is all of E by somehow showing that if it were not, I could disconnect the space using Y and Y complement.

I just don't see the direction that this problem could take. Is it long? Is it short but requires ingenuity? What should I try to do? Where am I trying to end up?

Help, it's hard!

 

[Dick]

It's not hard and doesn't require any stunning ingenuity. You just have to think clearly. Can you show Y is open? The main ingredient you need is that if a,b and c are three points, and a is path connected to b and b is path connected to c then a is path connected to c. Now can you show E-Y (the set of all points in E that aren't in Y) is also open? That's your disconnection.