Question

Consider a topological space (X,T). Show that the following four properties are equivalent. (Try to show a cycle of implications rather than pairwise equivalence.) (a) X is Hausdorff. (b) The diagonal =Δ {(x,x) ∣x∈ X} is closed in X×X with the product topology. (c) For any point x in X the singleton set {x} is a closed subset of X. (d) The convergence of nets in X is unique. You can proceed with proving the equivalence of these properties as stated in the scenario.

Answer 1

In a topological space (X,T), the four properties of Hausdorffness, the diagonal being closed, singletons being closed sets, and unique convergence of nets are equivalent, demonstrated through a cyclical series of implications from a to d.

Step-by-step explanation:

To prove that the four properties in a topological space (X,T) are equivalent, we need to show that each one implies the next in a cyclical fashion.

1. a → b: Assume that X is Hausdorff. For any two distinct points x,y ∈ X, there exist disjoint open sets U and V such that x ∈ U and y ∈ V. The complements of these sets are closed in X, and thus the product (X \ U) × (X \ V) is closed in X × X. The diagonal Δ is the intersection of all such products, which is also a closed set.
2. b → c: If the diagonal Δ is closed in X × X, then for any point x ∈ X, {x} × X intersects Δ only at (x,x). The projection onto the first factor, which is a closed map, will send this set to {x}, showing that singletons are closed.
3. c → d: Given that singletons are closed, assume that a net in X converges to some point x. If there were another limit point y
   eq x, the net would also eventually be in X \ {y}, which contradicts the convergence to y, ensuring unique convergence of nets.
4. d → a: If the convergence of nets in X is unique, then for any two distinct points x and y, no net can converge to both, implying the existence of disjoint neighborhoods which is the definition of a Hausdorff space. Thus, we have come full circle from a to d, proving their equivalence.