Proving Connectedness of a Cartesian Product with Proper Subsets

[radou]

Homework Statement

Let X and Y be connected spaces, and A and B proper subsets of X and Y, respectively. One needs to show that (X x Y)\(A x B) is connected.

The Attempt at a Solution

(X x Y)\(A x B) can be written as ((X\A) x Y) U (X x (Y\B)). All one needs to show is that ((X\A) x Y) and (X x (Y\B)) are connected. Since these sets have a point in common, (choose (x, y) so that x is any point in X\A and y any point in Y\B), their union is connected. So, let's go for it.

I know that a finite cartesian product of connected spaces is connected, so I only need to show that X\A and Y\B are connected. Assume X\A is not connected. Then there exist two non-empty, disjoint open sets U and V whose union is X\A. Hence X\A is open. It follows that A is closed. Since a can be any set, without loss of generality, one can assume that A is open. But then we arrive at a contradiction: A is both open and closed, and the only sets which are both open and closed in X are X and the empty set, assuming X is connected. Hence, X\A is connected, which proves the problem.

I hope this works, thanks for checking.

 

[Office_Shredder]

If X=[0,1] and A=(0,1) X\A={0,1} is not connected.

U and V are open in the relative topology of X\A, that doesn't mean that they are open in X. So you can't conclude that X\A has to be open.

Also, I would be very wary of assuming that WLOG an arbitrary set can be assumed to be open

 

[radou]

U and V are open in the relative topology of X\A, that doesn't mean that they are open in X. So you can't conclude that X\A has to be open.

Right, that was a stupid mistake. I'll have to think this over.

Also, I would be very wary of assuming that WLOG an arbitrary set can be assumed to be open

That's an interesting point. But if we make a statement about an arbitrary set, it must hold for any set, right? i.e. for an open set, too?