Could someone please help me with this question?

Given pix- y is a quotient map and P" ( Ey3) is closed for all yEy, prove or disprove that I must be Hausdorff. Proof: We will disprove that Y must be Hausdorff by providing a counter-example. Let p(x) = {x + z : z EZ} be the quotient map from R - R/~ defined by x ~ y if and only if x - y E Z. We note that by Definition (6.4.6), this is a natural quotient map. This is the set of all real numbers differing from x by an integer. Then by construction, p-'({y}) = [x], which are all closed subsets of R since they are all discrete in R. Discrete sets in R are closed since by Corollary (3.3.7), they contain all their limit points. The quotient space, R/ ~ is not Hausdorff by Definition (11.1.3) because the open neighborhoods of different equivalence classes over- lap. For example, the open neighborhoods of Is this (0] = {..., -1, 0, 1, ..} and [0.5] = {... - 0.5, 0.5, 1.5, ..} overlap, so Correct ? they are not disjoint. Are the open neighborhoods of these equivalence Finally, we verify that a ~ y. sheode vida classes disjoint Reflexivity holds: For any T ER, T - T = 0 E Z. Hence, I ~ x. because the open Symmetry holds: If x ~y, then x - y E Z. neighborhoods contain Then y - x = -(x - y) E Z. Hence, y ~ x. S molder More points than Transitivity holds: If r ~ y and y ~ z, then x - y E Z and y - z E Z. the points in these Hence, x - z = (x - y) + (y - z) EZ. equivalence Thus, ~ is an equivalence relation. classes ? For example, Therefore, if p : X - Y is a quotient map and p-({y}) is closed do they both for all y E Y, then Y may not be Hausdorff. Contain 0. 3 ? Definition ( 11. 1. 3) A topological Space X is Hausdorff if for each X y there exists neighborhoods UDX and Vay such that UnV = D