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hina

Guest

topology

Let X,Y be topological spaces. Consider Z = X x Y and the product topology genearted by the projections p_x, p_y. Let A be a subset of Z.
Suppose that X,Y are locally path connected. Show that A is connected if and only if A is path connected.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: topology

You have a nifty little theorem that says:

A locally path connected space is path connected if and only if it is connected.

I really hope you can use that.  Otherwise, you'd have to prove it as well.  But under the assumption that you can, all you need to do is prove that A is locally path connected.  I think that should be straightforward, but it's been a while since I studied topology.  If it isn't, just come back and I'll see what I can do.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

hina

Guest

Re: topology

i have this thm : every path connected space is connected
is this same as yours

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: topology

That's half of it.  Now using that, all you need to prove is that if a space is connected and locally path connected, then it is also path connected as well.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

hina

Guest

Re: topology

how do i start writing this proof using the above thm

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: topology

Start out with the implication going one way:

Assume it's path connected.  The by that theorem, it is certainly connected.

Now you must do the converse:

Assume it's connected.  We will show that a connected space which is locally connected is path connected.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

hina

Guest

Re: topology

Please if you can check this proof and tell me if it is correct

<=  by thm every path connected space is connected

=> let see an example where a subset of R x Rwhich is connected but not path- connected
Let B = {(x, sin(1/x)): x in R-{0}}
Theorem: Putting a limit point of a connected set into a connected set does not disconnect the set (i.e., the set remains connected).
Thus, since B is connected and (0,0) is a limit point of B, it follows that B U {(0,0)} = A is connected.
To show that A is not path connected, suppose to the contrary that there is a path f : [0,1]->A with f(0) = ((1/pi),0)and f(1)=(0,0) Clearly (0,1) not in f([0,1])but (0,1)in cl(f([0,1])) Hence f([0,1]) is not closed and therefore f([0,1])is not compact.

On the other hand, f([0,1])is supposed to be compact since it is the continuous image of a compact set.

This is a contradiction and we are forced to conclude that A is not path connected.

simron

Real Member

Registered: 2006-10-07

Posts: 237

Re: topology

That's great, but you might want to turn smilies off.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: topology

=> let see an example where a subset of R x Rwhich is connected but not path- connected

No.  You want to show that if a connected space is also locally path connected, then it is path connected.  This needs to be a proof, not a counter example.

The idea behind the proof is that the open sets for the locally path connected property coincide to connected subspaces.  Then, once you have this, you can take such open sets to be the entire space because the entire space is connected.  This means that the entire space is path connected, which would end the proof.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

hina

Guest

Re: topology

is this proof right if not please can you correct it

Let C be a connected set that is also locally path connected.  Pick any point x in C, and let U be the set of points in C that are path connected to x.  Thus U is a subset of C.
Let y be a point in U.  Enclose y in an open set H in C, such that y is path connected to all of H.  Since an arc can run from x to y to anything in H, H is in U.  Therefore U is the union of open sets and is open, relative to C.

Let y be a point in C that is a limit point of U.  Put an open set H around y such that H is path connected.  Let z be common to H and U.  Now x connects to z connects to y, and y is in U.

Since U contains its limit points it is closed.  thus U is both open and closed in C.  If U is not all of C, separate U and the rest of C in open sets.  This contradicts the fact that C is connected.  Therefore U is all of C, and C is path connected.

In n dimensional space, every open ball is path connected, and every open set is locally path connected, hence every open connected set is path connected.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: topology

In a lot of places, your terminology is poor.  I know what you mean, but it's completely different from what you say.  The differences may look small, but it is the difference between a sloppy proof and a good proof.

Pick any point x in C, and let U be the set of points in C that are path connected to x.

Let U be the set of all points in C that are path connected to x.

Enclose y in an open set H in C, such that y is path connected to all of H.

This statement needs justification.  Why does such a set H exist?

Since U contains its limit points it is closed.

You haven't shown this.  You let y be a point in C that is a limit point of U.  But you need to show it for any point that is just a limit point of U.   For all you know, U could be equal to C.  Of course, weirder things can also occur.

If U is not all of C, separate U and the rest of C in open sets.  This contradicts the fact that C is connected.  Therefore U is all of C, and C is path connected.

This fails if U is the null set.  There is a very simple reason why U can't be the null set, and it is almost too obvious to write.  Almost.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."