The discussion centers on proving that a topological space X is Hausdorff if and only if the diagonal set Δ = {x × x | x ∈ X} is closed in the product space X × X. The proof is established in both directions: first, if X is Hausdorff, then the complement of Δ is open, confirming Δ's closedness; second, if Δ is closed, then distinct points in X can be separated by disjoint neighborhoods, confirming that X is Hausdorff. The use of product topology is essential in both proofs.
PREREQUISITESMathematics students, particularly those studying topology, as well as educators and anyone interested in understanding the foundational concepts of Hausdorff spaces and their implications in topology.
With so little work on your part shown to go by it's difficult to know where you're stuck.Symmetryholic said: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 said: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.
That's quite a leap from your previous post.Symmetryholic said:->
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 .
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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.
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!Symmetryholic said: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.