# Showing that a space is Hausdorff

Suppose $X$ is compact and Hausdorff and that $f:X \to Y$ is continuous, closed, and surjective. How can I show that $Y$ is Hausdorff?

• Do you know what you need to prove in order to show that? Any ideas or tries? – Alejandro Nasif Salum Jan 8 '18 at 2:04
• Try reading here. – Michael Lee Jan 8 '18 at 4:01

Since $X$ is compact Hausdorff, $X$ it is normal. Next, since $f$ is closed, the single-element subsets of $Y$ are closed (as $\{y\} = f''\{x\}$ for some $x \in X$.) It follows that for each pair of distinct $y_0, y_1 \in Y$, the sets $X_0 = f^{-1}(y_0)$ and $X_1 = f^{-1}(y_1)$ are disjoint closed subsets of $X$. Noting that $X$ is normal, you can separate $X_0$ and $X_1$ using open-sets with disjoint closures. The remaining bit is to turn the open-sets separating $X_0$ and $X_1$ into open subsets of $Y$ separating $y_0$ and $y_1$.
• "Turning $X_0$ and $X_1$ into open sets.... " seems problematic. Can you finish this, or give a hint? – DanielWainfleet Jan 9 '18 at 15:33
• Given any separation of $X_0$ and $X_1$, by disjoint open sets, $U_0$ and $U_1$. The image under $f$ (a closed map) of the complement of $U_k$, namely $C_k = f''(X\backslash U_k)$ is a closed subset of Y with $y_{1-k}$ \in C_k$. and$y_k\not\in C_k$. Hence – Not Mike Jan 9 '18 at 15:53 • .. letting$V_k = Y \backslash C_k$be the open complement of$C_k$, you have$y_k \in V_k$and$y_{1-k} \in V_{1-k}$. Moreover,$V_{k} \cap V_{1-k} = Y \backslash (C_k \cup C_{1-k}) =\emptyset$– Not Mike Jan 9 '18 at 16:04 Let$y_1 \neq y_2$be distinct points of$Y$. By surjectivity we have$x_1 ,x_2 \in X$such that$f(x_1) = y_1 , f(x_2) = y_2$. In particular,$f[\{x_i\}] = y_i$for both$i$and as in$X$singletons are closed (Hausdorff implies$T_1$) and$f$is a closed map (first time we use it), we have that both$\{y_i\}$are closed in$Y$. Now, define$F_i = f^{-1}[\{y_i\}] \subseteq X$,$i=1,2$.$F_1$and$F_2$are disjoint closed (by continuity of$f$) subsets of$X$, which is compact and Hausdorff and thus$T_4$or normal. So in$X$we can find open disjoint subsets$U_1, U_2$such that$F_1 \subseteq U_1$and$F_2 \subseteq U_2$. Now define$O_i = Y\setminus f[X\setminus U_i]$for$i=1,2$. These sets are open as$f$is a closed map and complements of open sets are closed and complements of closed sets are open. For each$iy_i \in O_i$: suppose$y_i \notin O_i$then by definition$y_i \in f[X\setminus U_i]$so$y_i = f(p)$for some$p \in X \setminus U_i$. But by definition$p \in F_i$as it maps to$y_i$and so should be in$U_i$by$F_i \subseteq U_i$, a contradiction with$p \in X \setminus U_i$. This contradiction shows that$y_i \in O_i$.$O_1 \cap O_2 = \emptyset$. Suppose that$y$lies in both. Then for some$x \in X$,$f(x) = y$. As$U_1$and$U_2$are disjoint,$x$must lie in$X\setminus U_1$or in$X \setminus U_2$(or both); suppose it lies in$X \setminus U_j$for some$j \in \{1,2\}$. But then$y = f(x) \in f[X\setminus U_j]$which means$y \notin O_j$, which is our contradiction. So$O_1$and$O_2$are the required disjoint open neighbourhoods of$y_1$resp.$y_2$. Sanity check: we used the closedness of$f$and its Hausdorffness too, (also to get normality from the combination with compactness), and the continuity of$f$and its surjectivity. The proof would also have worked if$X$would have been merely$T_4$(normal and$T_1$) and$f$the same (continuous closed surjection). Let$y_1 \neq y_2$in Y. Pick$x_1 ,x_2$such that$f(x_1)=y_1$and$f(x_2)=y_2$Claim: there exist neighborhoods U and V of$x_1$and$x_2$such that$f(U) \cap f(V)$is empty. [ This is where continuity is used]. If not, for any neighborhoods U and V there exist points$t_{UV}$,$s_{UV}$in U and V such that$f(t_{UV})=f(s_{UV})$Order the pairs (U,V) by$(U,V) \geq (S,T)$if U is a subset of S and V is a subset of T. We get nets$\{t_{UV}\}$and$\{t_{UV}\}$converging to$x_1$and$x_2$respectively. Since f is continuous and$f(t_{UV})=f(s_{UV})$we get$y_1=f(x_1)=f(x_2)=y_2$which is a contradiction. This proves the claim. Next pick open sets A, B containing$x_1$and$x_2$such that their closures are contained in U and V respectively. Now the complements of the images of the closures of A and B are disjoint neighborhoods of$y_1$and$y_2\$.