Another path connectedness issue

[radou]

Homework Statement

Here's an exercise which seems a bit bizzare.

Show that if U is an open connected subspace of R^2, the U is path connected.
(Hint: show that, given any x0 in u, the set of points that can be connected to x0 by a path in U is clopen in U.)

Now, given the fact that I know that R^2 is path connected, this seems incredibly trivial.

Obviously there's another way to solve it, ignoring the fact above. In this case, I don't even have a clue where to start. The hint seems helpful, since the only clopen sets in a connected space are the empty set and U (in this case) itself, so this proves the fact, but nevertheless I don't have a clue where to start.

 

[Dick]

Knowing R^2 is path connected does not prove that a subset is path connected. So it's not incredibly trivial. Now why don't you think a little more seriously about the other way? Open balls are path connected, right?

 

[radou]

Knowing R^2 is path connected does not prove that a subset is path connected. So it's not incredibly trivial. Now why don't you think a little more seriously about the other way? Open balls are path connected, right?

Of course, I completely ignored that the path needs to be in the subset of interest. Thanks.

OK, let U be an open connected subspace of R^2.

U can be written as a union of open balls, i.e. [itex]U = \cup_{x \in U} B_{x}[/itex], every of which is path connected. The intersection of these open balls is nonempty, since if it were empty, we could find a separation of U. So, all of these open balls have at least one element in common.

Now, take any x and y from U. There exist balls Bx and By in U which contains x and y. Their intersection is nonempty, and let z be an element contained in it. Now, since Bx is path connected, we can find a function f1 : [0, 1] --> Bx such that f1(0) = x and f1(1) = z, and the same for By, f2 : [0, 1] --> By, with f2(0) = z and f2(1) = y.

Now, if we define g(t) with g(t) = (1 - t)*f1(t) + t* f2(t), g is continuous, and g(0) = f1(0) = x, g(1) = f2(1) = y.

Does this work?

 

[Dick]

Of course, I completely ignored that the path needs to be in the subset of interest. Thanks.

OK, let U be an open connected subspace of R^2.

U can be written as a union of open balls, i.e. [itex]U = \cup_{x \in U} B_{x}[/itex], every of which is path connected. The intersection of these open balls is nonempty, since if it were empty, we could find a separation of U. So, all of these open balls have at least one element in common.

Now, take any x and y from U. There exist balls Bx and By in U which contains x and y. Their intersection is nonempty, and let z be an element contained in it. Now, since Bx is path connected, we can find a function f1 : [0, 1] --> Bx such that f1(0) = x and f1(1) = z, and the same for By, f2 : [0, 1] --> By, with f2(0) = z and f2(1) = y.

Now, if we define g(t) with g(t) = (1 - t)*f1(t) + t* f2(t), g is continuous, and g(0) = f1(0) = x, g(1) = f2(1) = y.

Does this work?

Is it really so obvious that the image of your g(t) is in BxUBy? I think it is because the balls are convex, but there's a much better way to join the two paths together. But once you make it work, you've basically just shown that the union of two path connected sets that have a nonempty intersection is path connected. That's fine. But that's not what you are being asked to prove. Think about using the clopen hint.

 

[radou]

For some reason, every time I totally ignore the fact that a path of interest must lie in the set I'm looking at.

OK, but the clopen argument leaves me a bit confused; given some x in U, I need to show that the set of points that can be joined to x by a path in U is clopen in U. Let's denote this set with S, then if y is in S, there exists a continuous function f : [0, 1] --> U such that f(0) = x and f(1) = y. Now I have no idea of what to do..?

 

[radou]

Is it really so obvious that the image of your g(t) is in BxUBy? I think it is because the balls are convex,

Btw, just to make this clear, if a set in R^2 is convex, then every line joining some two points of this set lies in the set, right? And the function g is linear function with boundary values "at these points" right? Clearly an open ball is a convex set, but the union of open balls isn't, in general, right?

 

[Dick]

For some reason, every time I totally ignore the fact that a path of interest must lie in the set I'm looking at.

OK, but the clopen argument leaves me a bit confused; given some x in U, I need to show that the set of points that can be joined to x by a path in U is clopen in U. Let's denote this set with S, then if y is in S, there exists a continuous function f : [0, 1] --> U such that f(0) = x and f(1) = y. Now I have no idea of what to do..?

Start by showing S is open. If y is path connected to x then there is an open ball around y that's path connected to x (and therefore in S). Then show U-S is also open.

 

[Dick]

Btw, just to make this clear, if a set in R^2 is convex, then every line joining some two points of this set lies in the set, right? And the function g is linear function with boundary values "at these points" right? Clearly an open ball is a convex set, but the union of open balls isn't, in general, right?

Right. The union of the balls may not be convex. There's much easier way to make your path g. Just let g on [0,1/2] follow f1 and g on [1/2,1] follow f2.

 

[radou]

Start by showing S is open. If y is path connected to x then there is an open ball around y that's path connected to x (and therefore in S)

This may seem like a dumb question, but how do I know that there's an open ball around y that's path connected to x?

Btw, just a small digression. I just tried to prove that open balls are path connected in R^n. If I define an open ball at the origin B(0, ε) = {x in R^n : ||x|| < ε}, then, given any two points x and y in B, the function f(t) = (1 - t) x + t y lies in B, for ||f(t)|| <= (1 - t)||x|| + t||y|| < ε. Now, can I conclude that any open ball of radius ε in R^n is path connected, since it's homeomorphic to the open ball B(0, ε)? I assume homeomorphisms preserve path connectedness, right?

 

[Office_Shredder]

This may seem like a dumb question, but how do I know that there's an open ball around y that's path connected to x?

Start with any open ball around y (preferably one inside of U). Find a path from x to y. Find a path from y to a point in the open ball. Glue them together

Btw, just a small digression. I just tried to prove that open balls are path connected in R^n. If I define an open ball at the origin B(0, ε) = {x in R^n : ||x|| < ε}, then, given any two points x and y in B, the function f(t) = (1 - t) x + t y lies in B, for ||f(t)|| <= (1 - t)||x|| + t||y|| < ε. Now, can I conclude that any open ball of radius ε in R^n is path connected, since it's homeomorphic to the open ball B(0, ε)? I assume homeomorphisms preserve path connectedness, right?

This is right. If you have a homeomorphism, you can just compose it with a path in the domain to get a path in the codomain. If you're interested, there's an even easier proof that all balls are path connected, which doesn't require convexity or homeomorphisms: to get from x to y, just take the path from x to the origin, then the origin to y that goes on straight lines.

 

[radou]

If you're interested, there's an even easier proof that all balls are path connected, which doesn't require convexity or homeomorphisms: to get from x to y, just take the path from x to the origin, then the origin to y that goes on straight lines.

I'm not sure I understood this? You take any open ball of R^n, right? Take any points x and y lying inside it. Then..? How can a path go to the origin, it goes outside the ball then, no? I guess there's a misunderstanding here..

 

[radou]

Start by showing S is open. If y is path connected to x then there is an open ball around y that's path connected to x (and therefore in S). Then show U-S is also open.

OK, if y is path connected to x, take an open ball around y in U (we can do this, since U is open). Then every point of this open ball is path connected to y, and hence path connected to x. So, the set of all points path connected to x is a union of open balls, which is again open. The complement of this set in U is the set of all points not path connected to x. But regardless of this, since these points are in U, for all of them there exist open balls in U around them, and hence the set equals a union of such open balls, so it's open. Hence, the set of all points path connected to x is closed. So, it must equal U (it cannot be empty, since y it contains y). Since this holds for any x, it follows that U is path connected.

Now, if this is correct, there's only one step I don't quite understand, and it's the first one. We only assumed that y is path connected to x. But this might not be true..?

 

[Dick]

OK, if y is path connected to x, take an open ball around y in U (we can do this, since U is open). Then every point of this open ball is path connected to y, and hence path connected to x. So, the set of all points path connected to x is a union of open balls, which is again open. The complement of this set in U is the set of all points not path connected to x. But regardless of this, since these points are in U, for all of them there exist open balls in U around them, and hence the set equals a union of such open balls, so it's open. Hence, the set of all points path connected to x is closed. So, it must equal U (it cannot be empty, since y it contains y). Since this holds for any x, it follows that U is path connected.

Now, if this is correct, there's only one step I don't quite understand, and it's the first one. We only assumed that y is path connected to x. But this might not be true..?

Ok, so if y is in S then you agree there is an open ball around y that is path connected to x. Right? Now if z is in U-S, yes, there is a ball B around z in U. Isn't it pretty clear that if z is NOT path connected to x then no point in B can be path connected to x? Hence the ball is in U-S.

 

[radou]

Ok, so if y is in S then you agree there is an open ball around y that is path connected to x. Right? Now if z is in U-S, yes, there is a ball B around z in U. Isn't it pretty clear that if z is NOT path connected to x then no point in B can be path connected to x? Hence the ball is in U-S.

OK, I got it all now. Thanks for your patience.