Proving the Hausdorff Property: The Diagonal in X x X

[Symmetryholic]

Homework Statement

Show that X is Hausdorff if and only if the diagonal \Delta = \{x \times x | x \in X \} is closed in X \times X.

Homework Equations

Definition of Hausdorff Space (T2) : A topological space in which distinct points have disjoint neighborhoods.

The Attempt at a Solution

 

[Unco]

Homework Statement

Show that X is Hausdorff if and only if the diagonal \Delta = \{x \times x | x \in X \} is closed in X \times X.

Homework Equations

Definition of Hausdorff Space (T2) : A topological space in which distinct points have disjoint neighborhoods.

The Attempt at a Solution

With so little work on your part shown to go by it's difficult to know where you're stuck.

If you haven't already, express the closedness of \Delta in X\times X (which I assume has the product topology) in terms of the openness of its complement.

Now have a go at proving each direction (neither is more difficult than the other) if you haven't already, and please show us your efforts.

 

[Symmetryholic]

If you haven't already, express the closedness of \Delta in X\times X (which I assume has the product topology) in terms of the openness of its complement.

->
If X is Hausdorff, the diagonal \Delta is closed in X\times X.

Assume X is Hausdorff. Now, we have two distinct points x, y and disjoint open sets U, V containing x, y, respectively. The basis element U \times V containing (x,y) \in X \times X should not intersect \Delta by the assumption given to the Hausdorff property.
For every (x,y) \notin \Delta, we have a basis element in X \times X containing (x,y), which does not intersect \Delta.
Thus, X \times X \setminus \Delta is open and we conclude \Delta is a closed set in X \times X .

<-
If the diagonal \Delta is closed in X\times X, X is Haudorff.

Supppose \Delta is closed in X\times X. Then, X \times X \setminus \Delta is open. Let (x,y) \in X \times X and x \neq y. For (x,y) \notin \Delta, we have a basis element U \times V in X \times X containing (x, y).
We remain to show U and V are disjoint. Suppose on the contrary that U and V are not disjoint. Then, there is an element (z,z) \in X times X which belongs to both U and V. Contradicting the fact that x and y are distinct.
Thus, X is Hausdorff.

 

[Unco]

->
If X is Hausdorff, the diagonal \Delta is closed in X\times X.

Assume X is Hausdorff. Now, we have two distinct points x, y and disjoint open sets U, V containing x, y, respectively. The basis element U \times V containing (x,y) \in X \times X should not intersect \Delta by the assumption given to the Hausdorff property.
For every (x,y) \notin \Delta, we have a basis element in X \times X containing (x,y), which does not intersect \Delta.
Thus, X \times X \setminus \Delta is open and we conclude \Delta is a closed set in X \times X .

<-
If the diagonal \Delta is closed in X\times X, X is Haudorff.

Supppose \Delta is closed in X\times X. Then, X \times X \setminus \Delta is open. Let (x,y) \in X \times X and x \neq y. For (x,y) \notin \Delta, we have a basis element U \times V in X \times X containing (x, y).
We remain to show U and V are disjoint. Suppose on the contrary that U and V are not disjoint. Then, there is an element (z,z) \in X times X which belongs to both U and V. Contradicting the fact that x and y are distinct.
Thus, X is Hausdorff.

That's quite a leap from your previous post.

Now how 'bout you tackle your https://www.physicsforums.com/showthread.php?t=287038" without copying down the solution from an external source. It's the only way you'll learn topology.

 

[Symmetryholic]

Can you please give me a link of an external source you mentioned?
It's my self study of topology (I am not even majoring in math) and I don't need to copy the external source to show you something to impress you. Rather, I just did for my self study purpose and asked an advice if someone finds an error in my attempt to the solution.

 

[Unco]

Can you please give me a link of an external source you mentioned?
It's my self study of topology (I am not even majoring in math) and I don't need to copy the external source to show you something to impress you. Rather, I just did for my self study purpose and asked an advice if someone finds an error in my attempt to the solution.

Well, in that case, your work is quite error-free indeed! I would only suggest rephrasing picking an element (x,y)\not\in \Delta as picking an element (x,y)\in (X\times X)\backslash \Delta. Now, on to your other problem!