# A local path connectedness problem

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## Homework Statement

Let X be locally path connected. Show that every connected open set in X is path connected.

## The Attempt at a Solution

Let U be a connected open subset of X. Since, X is path connected, for any x in X and any neighborhood N of X, there exists a path connected neighborhood Vx of x contained in N. For any x in U, pick a basis element containing x, contained in U, and for that basis element, pick the path connected neighbourhood Vx contained in this basis element. Then U can be written as the union of the Vx's, where x is in U.

Now, since U is connected, this union must have a nonempty intersection. Since if this would not hold, we could pick any Vx, and then this Vx and its complement in U would be two nonempty disjoint and open sets whose union is U, contradicting the fact that U is connected.

Let x, y be any elements of U. Then x is in Vx and y in Vy. Take an element z from the intersection of Vx and Vy. Let f1: [a, b] --> Vx be the path from x to z, and f2 : [c, d] --> Vy the path from z to y. Let [b, d] be some segment homeomorphic to [c, d]. Then, by the pasting lemma, the function f : [a, d] --> Vx U Vy (given with f = f1 is t is in [a, b] and f = f2 if t is in [b, d]) is continuous, and hence a path from x to y.

So, U is path connected.

Sorry if this is a bit long, but I tried to be precise.

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I am not able to follow your argument, but what about this idea: choose x. Consider the set of all points path connected to x. Can you prove that it is an open set? On the other hand consider the set of all points that cannot be connected by a path with x. Can you prove that it is an open set?

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My argument followed the definition of local path connectedness and the fact that any two sets with a point in common are path connected (if this is true). I'd like to see if this works before considering any other options.

You wrote: "Since, X is path connected". But the assumption was that X is "locally path connected".

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You wrote: "Since, X is path connected". But the assumption was that X is "locally path connected".
Sorry, my mistake. X is locally path connected.

Office_Shredder
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Then U can be written as the union of the Vx's, where x is in U.

Now, since U is connected, this union must have a nonempty intersection. Since if this would not hold, we could pick any Vx, and then this Vx and its complement in U would be two nonempty disjoint and open sets whose union is U, contradicting the fact that U is connected.
I assume by "this union must have a nonempty intersection" you mean every pair must have non-empty intersection (since this is what you assume next). This isn't true. You could have thousands of basis elements Vx that you're unioning together, and no two of them have to touch. For example, (0,1) is the union of all elements of the form (i/1000,(i+2)/1000) but most of them have empty intersection.

Since U is connected, it contains no non-trivial open and closed sets. arkajad's post points in the right direction

Then, what do you mean by this: "for any x in X and any neighborhood N of X"? Another mistake? Or what?

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I assume by "this union must have a nonempty intersection" you mean every pair must have non-empty intersection (since this is what you assume next). This isn't true. You could have thousands of basis elements Vx that you're unioning together, and no two of them have to touch. For example, (0,1) is the union of all elements of the form (i/1000,(i+2)/1000) but most of them have empty intersection.

Since U is connected, it contains no non-trivial open and closed sets. arkajad's post points in the right direction
I always make mistakes when considering "such unions", I'll have to think about it more thoroughly.

I am not able to follow your argument, but what about this idea: choose x. Consider the set of all points path connected to x. Can you prove that it is an open set? On the other hand consider the set of all points that cannot be connected by a path with x. Can you prove that it is an open set?
OK, let x be an element of U. Let S be the set of all points of U which can be path connected to x. If y is in U, then for any neighborhood of y, we have a path connected neighborhood V of y contained in this neighborhood (since X is locally path connected). Any point of this neighborhood is path connected to x (because path connectedness is an equivalence relation). So the set of points path connected to x is open.

Now, consider its complement, i.e. the set of all points of U which can't be path connected to x. Again, for any such point y, and for any neighborhood of y, we have a path connected neighborhood contained in the primary neighborhood, which contains y. But none of the points of this neighborhood are connected to x. Hence, the set of all points not connected to x is open, i.e. S is closed, and hence clopen, so, because of connectedness of U, it has to equal U. So U is path connected.

Actually, I just realized this is a very similar problem to the one I've done yesterday:

"If U is an open connected subspace of R^2, then U is path connected."