Thm: If X is Hausdorff, then the diagonal, D, is closed in X × X.Need to prove D is closed in X × X.Question: How do you prove a set is closed?a.) Show X × X − D is open.b.) Show D′⊂ D.c.) Show D =D.d.) Show D is a finite union of closed sets or an arbitrary intersection of closed sets.e.) etc.Question: What is given?X is Hausdorff. Hencea.) For all u, v ∈ X such that u 6= v, there exists sets U and V which are open in X such thatu ∈ U , v ∈ V , U ∩ V = ∅.b.) Every finite point set in X is closed.c.) 17.9 involving limit points, 1 7.10 i nvolving sequences, etc.Question: How can we relate what is given to what we need to prove. Which definitions/theoremsinvolving what we need to prove are most related to which definitions/theorems of what is given.The definition of Hausdorff gives open sets (in X). Hence we will try focusing on the definition Dclosed in X × X iff X × X − D is open in X × X.Concern: X Hausdorff gives open sets in X, but we need open set in X × X.We will ignore this concern for either of the following two reasons:1.) We have to try something. Might as well give this a try. Working blindly with litt le motivationas to why this mi ght work may be disconcerting, BUT it often results in t he correct answer. If youdon’t know what to do, try something (hopefully relating what is given to what you need to proveeven if weakly related). When successful, look back over your proof and see if you can figure outwhy it worked and how you could have come up with it with more motivation.2.) X × X has the product to po logy, thus we can create the needed o pen set in X × X by takingthe product of two open sets in X. Since X Hausdorff gives us two o pen sets in X, this is excell entmotivation.Thus we wil l try to show X × X − D is open in X × X. There are many ways to show a set is open.We’ll try the definition.Take (x, y) ∈ X ×X −D (not e since it often helps to be specific, I didn’t take a point p in X ×X −D,but instead took a n ordered pair (x, y)).1We need to find a set W open i n X × X such that (x, y) ∈ W ⊂ X × X − D. Since basis elementsare open, we could look for a basis element U × V where U and V are open in X such t hat(x, y) ∈ U × V ⊂ X × X − D (note usually it is easier to find a basis element because you knowwhat basis elements look like, but keep in mind that not all open sets are basis elements).Looking for a basis element is particularly useful for this problem since Hausdorff gives us open setsin X and we are now looking for U , V open in X such that x ∈ U , y ∈ V , and U × V ⊂ X × X − D.We will now try to apply the hypot hesis that X is Hausdorff. To apply Hausdorff, we need twodistinct points in X, preferably o nes related to what we are t rying to prove. Since (x, y) ∈ X ×X −D,x, y are points in X. Are they distinct? Since (x, y) 6∈ D = {(a, a) | a ∈ X}, x 6= y. Hence x, y aretwo distinct points in X. Hence there ex ists U, V open in X such that x ∈ U , y ∈ V , U ∩ V = ∅.Thus (x, y) ∈ U × V . Is U × V ⊂ X × X − D. Suppose (u, v) ∈ U × V . We have not yet usedU ∩ V = ∅ . u ∈ U, v ∈ V , U ∩ V = ∅ implies u 6= v. Hence (u, v) 6∈ D. Thus (u, v) ∈ X × X − D.Now you know how to relate X Hausdorff to D is closed in X × X to prove X Hausdorff implies Dis closed in X × X. Perhaps this relationship can also be used to prove D closed in X × X impliesX Hausdorff. Sometimes a different relationship is needed to prove the other direction.Note this pro of was actually an ”easy” proof. All it used was definitions. We used the definition ofclosed to start the proof and give us the two distinct points x, y ∈ X so we could apply Hausdorff.We used the definiti on of Hausdorff to find the U × V that we needed.Thus the proof for this direction is:Take (x, y) ∈ X × X − D. Since (x, y) 6∈ D, x 6= y. Thus there exists U, V open in X such thatx ∈ U, y ∈ V , U ∩ V = ∅. Thus (x, y) ⊂ U × V . If (u, v) ∈ U × V , then u ∈ U, v ∈ V . Thus,U ∩ V = ∅ implies u 6= v. Hence ( u, v) 6∈ D. Thus (u, v) ∈ X × X − D. Thus, U × V ⊂ X × X − D.Hence X × X − D is open in X × X. T hus D is closed in X ×