answer:The separation axiom class refers to the topological properties of a space. In the context of the Euclidean plane, we can consider the separation axioms T0, T1, T2, T3, and T4. Let's analyze each of them: 1. T0 (Kolmogorov) - For any two distinct points, there exists an open set containing one of the points but not the other. In the Euclidean plane, we can always find open balls around A and B that do not intersect, so the points A and B satisfy the T0 axiom. 2. T1 (Frechet) - For any two distinct points, there exist open sets containing each point such that the open sets do not intersect. Again, in the Euclidean plane, we can find open balls around A and B that do not intersect, so the points A and B satisfy the T1 axiom. 3. T2 (Hausdorff) - For any two distinct points, there exist disjoint open sets containing each point. In the Euclidean plane, we can find open balls around A and B that do not intersect, so the points A and B satisfy the T2 axiom. 4. T3 (Regular) - For any closed set and a point not in the set, there exist disjoint open sets containing the closed set and the point. In the Euclidean plane, we can find open balls around A and B that do not intersect, so the points A and B satisfy the T3 axiom. 5. T4 (Normal) - For any two disjoint closed sets, there exist disjoint open sets containing each closed set. In the Euclidean plane, we can find open balls around A and B that do not intersect, so the points A and B satisfy the T4 axiom. Thus, the points A(3,4) and B(5,6) in the Euclidean plane belong to all the separation axiom classes T0, T1, T2, T3, and T4.

answer:The T1-axiom (also known as the T1 separation axiom) states that for any two distinct points x and y in a topological space X, there exist open sets U and V such that x ∈ U, y ∉ U, and y ∈ V, x ∉ V. Let's check if the topological space X = {a, b, c, d} with the topology T = {{}, {a}, {a, b}, X} satisfies this axiom. 1. Take two distinct points x = a and y = b. We have the open sets {a} and {a, b} in T. We can see that a ∈ {a}, b ∉ {a}, and b ∈ {a, b}, a ∉ {a, b}. So, the T1-axiom holds for these points. 2. Now, let's consider x = a and y = c. We can't find open sets U and V in T such that a ∈ U, c ∉ U, and c ∈ V, a ∉ V. The only open set containing a is {a}, but it doesn't contain c. The only open set containing c is X, but it also contains a. Thus, the T1-axiom doesn't hold for these points. Since we found a pair of points for which the T1-axiom doesn't hold, the topological space X = {a, b, c, d} with the topology T = {{}, {a}, {a, b}, X} does not satisfy the T1-axiom.

answer:Proof that X is a Hausdorff space: Recall that a topological space X is a Hausdorff space (also called a T2 space) if for any two distinct points x and y in X, there exist disjoint open neighborhoods U and V containing x and y, respectively. Since X is a topological space in which any two distinct points can be separated by open neighborhoods, let x and y be any two distinct points in X. By the given property, there exist open neighborhoods U_x containing x and U_y containing y such that U_x and U_y are disjoint (i.e., U_x ∩ U_y = ∅). Thus, for any two distinct points x and y in X, we can find disjoint open neighborhoods containing them. Therefore, X is a Hausdorff space. Counterexample for a general topological space: Consider the topological space X = {a, b} with the topology τ = {∅, X}. This is a valid topological space, but it does not satisfy the separation axiom mentioned above. In this space, there are no open neighborhoods that can separate the points a and b, since the only non-empty open set is X itself, which contains both points. Therefore, this space is not Hausdorff. This counterexample shows that the given property (any two distinct points can be separated by open neighborhoods) is not true for a general topological space without the separation axiom.

answer:To show that X is a normal space, we need to prove that for any two disjoint closed subsets A and B of X, there exist disjoint open subsets U and V such that A ⊆ U and B ⊆ V. (⇒) Suppose X is a normal space. By definition, for any two disjoint closed subsets A and B of X, there exist disjoint open subsets U and V such that A ⊆ U and B ⊆ V. This is exactly what we need to prove, so the forward direction is done. (⇐) Now suppose that for any two disjoint closed subsets A and B of X, there exist disjoint open subsets U and V such that A ⊆ U and B ⊆ V. We need to show that X is a normal space. Let A and B be any two disjoint closed subsets of X. By our assumption, there exist disjoint open subsets U and V such that A ⊆ U and B ⊆ V. Since A and B are closed, their complements, X A and X B, are open. Note that U = X A and V = X B, as A ⊆ U and B ⊆ V. Now, we have that A ∩ V = ∅ and B ∩ U = ∅, since A and B are disjoint and U and V are disjoint. This means that A ⊆ X V and B ⊆ X U. Since X V and X U are open, we have shown that any two disjoint closed subsets of X can be separated by open sets, which is the definition of a normal space. Therefore, we have proved that a topological space X is normal if and only if for any two disjoint closed subsets A and B of X, there exist disjoint open subsets U and V such that A is a subset of U and B is a subset of V.