Source of most content: Gunhee Cho and Lecture note by Matsumura
Hausdorff Space
Def.
Given T.S. $(X, T_X)$, $\{ a_n\}_{n=1}^\infty \subseteq X$ and $a \in X$. We say $\{ a_n \}$ converges to $a$, $a_n \stackrel{n \rightarrow \infty}{\rightarrow} a$ if $\forall$ nbhd $V$ of $a$, $\exists N \in \mathbb{N}$ s.t. $\forall n \geq N$, $a_n \in V$.
NB1; Give a toplogy $T = \{ \{ b \}, \{ a,b \}, \{ b,c \}, X, \emptyset \}$ on $X$. Define seq. $\{ x_n \}$ s.t. $\forall n \in \mathbb{N}, x_n = b$. Then, $x_n \stackrel{n \rightarrow \infty}{\rightarrow} b$ and $x_n \stackrel{n \rightarrow \infty}{\rightarrow} a$ and $x_n \stackrel{n \rightarrow \infty}{\rightarrow} c$.
- $x_n \stackrel{n \rightarrow \infty}{\rightarrow} b$ : nbhd of $b$ can be $\{ b \}, \{ a,b \}, \{ b,c \}$, or $X$. $\forall n \geq 1$, $b$ belongs to all nbhds.
- $x_n \stackrel{n \rightarrow \infty}{\rightarrow} a$ : nbhd of $a$ can be $\{ a,b \}$ or $X$. $\forall n \geq 1$, $b$ belongs to all nbhds.
- $x_n \stackrel{n \rightarrow \infty}{\rightarrow} c$ : nbhd of $c$ can be $\{ b,c \}$ or $X$. $\forall n \geq 1$, $b$ belongs to all nbhds.
NB2; Given $(\mathbb{N}, T_{cof})$, where $T_{cof}:= \{ A \subseteq \mathbb{N} : A = \emptyset \text{ or } \mathbb{N}\setminus A\text{ is finite set}\}$. Let $x_n = n.$ Then, $\forall m \in \mathbb{N}, x_n \rightarrow m$.
Proof
Take any $m \in \mathbb{N}$ and its any nbhd $U_m$. Then, by definition of open set, $| U_m^c | < \infty$. Thus, there exists $k \in \mathbb{N}$ s.t. $\forall a \in U_m^c, a \leq k$. This implies that $\{ m, k+1, k+2, k+3, \ldots \} \subseteq U_m$. Thus, $\forall n \geq k+1, x_n \in U_m$.
Thus, we should put an additional axiom in order to distinguish $x,y \in X$ s.t. $x\ne y$. More specifically, we want if $x\ne y$, $\exists$ nbhd $V_x, V_y$ s.t. $V_x \cap V_y = \emptyset$, which is called separation axiom.
Def.
Given T.S. $(X, T_X)$, $X$ is a Hausdorff space if $\forall x,y \in X$ s.t. $x\ne y$, $\exists$ disjoint nbhd $U_x, U_y$, i.e., $U_x \cap U_y = \emptyset$.
Def.
Given T.S. $(X, T_X)$, $X$ is a T1 space if every finite set in $X$ is closed.
NB3; Hausdorff space is T1 space but not conversely.
Properties
\[\text{Every finite set in a Hausdorff space is closed.}\nonumber\]Proof
It is equivalent to show every singleton set is closed, i.e., for $x_0 \in X$, show $ \overline{\{x_0\}}= \{ x_0 \}$.
Since $\overline{\{x_0\}} \supseteq \{ x_0 \}$ is trivial by definition, it remains to show $\overline{\{x_0\}} \subseteq \{ x_0 \}$.
Let $x \in \overline{\{x_0\}}$. If $x= x_0$ then, it holds.
Let $x \ne x_0$. Since $X$ is Hausdorff space, $\exists$ nbhd $U_x$ and $U_{x_0}$ s.t. $U_x \cap U_{x_0} = \emptyset$.
Since $\{ x_0 \} \subseteq U_{x_0}$, $U_x \cap \{x_0\} = \emptyset \Leftrightarrow U_x \cap (\{x_0\} - (\{x\})$, i.e., $x$ is not a limit point of $\{x_0 \}$.
Thus, if $x \in \overline{\{x_0\}}$ and $x \ne x_0$, then $x \not\in \overline{\{x_0\}}$.
Hence, $\overline{\{x_0\}} = \{ x_0 \}$, i.e., every singleton set is closed.
\[\text{Given T1 space } X \text{ and } A \subseteq X, x \in A' \text{(lim. pt.) iff } \forall \text{nbhd } V \text{ of } x, | V \cap A | = \infty. \nonumber\]
NB4; This means every nbhd contains infinitely many points in $A$.
Proof
- $(\Leftarrow)$ $\forall$ nbhd $V$ of $x$, $V \cap (A - \{ x\} ) \ne \emptyset$, which is the definition of limit point, i.e., $x \in A’$.
- $(\Rightarrow)$ Suppose $\exists$ nbhd $V$ of $x$ s.t. $| V \cap A | < \infty$. Set $A \cap (A - \{ x\} ) = \{x_1, \ldots, x_m \}$. Since $X$ is a T1 space, $ \{x_1, \ldots, x_m \}$ is closed. Then, $x \not\in \{x_1, \ldots, x_m \} \Leftrightarrow x \in \{x_1, \ldots, x_m \}^c =U$, where $U$ is open. Then, $V \cap U$ is a nbhd of $x$. Since $x \in A’$ by Asm, $(V \cap U) \cap (A - \{ x \}) \ne \emptyset$. $\Leftrightarrow [ V \cap (U - \{ x \}) ]\cap A \ne \emptyset $ $ \Leftrightarrow A \cap [V \cap \{ x \}^c ] \cap \{x_1, \ldots, x_m \}^c \ne \emptyset$. However, LHS is $\emptyset$, i.e., contradiction. Thus, $\forall \text{nbhd } V \text{ of } x, | V \cap A | = \infty$.
\[\text{Givne Hausdorff sp. } (X, T_X). \text{ Let } \{ x_n \} \text{ be a seq. in } X \text{ satisfying } x_n \rightarrow a \text{ and } x_n \rightarrow b. \text{ Then, } a= b. \nonumber\]
NB5; This means the uniqueness of a limit point in Hausdorff space.
Proof
If $a \ne b$, there exists nbhd $U_a$ and $U_b$ s.t. $U_a \cap U_b = \emptyset$. Since $x_n \rightarrow a$, $\exists N_1 \in \mathbb{N}$ s.t. $\forall n \geq N_1$, $x_n \in U_a$. Thus, $U_b$ contains at most $x_1, \cdots, x_{N_1-1}$. On the other hand, $b$ is a limit point of $\{x_n\}$. However, $U_b$ contains only finitely many $x_i$’s, which is a contradiction.
Continuity
Def.
Given T.S. $(X, T_X)$ and $(Y, T_Y)$, we say $f: X \rightarrow Y$ is continuous if $\forall V \in T_Y$, $f^{-1}(V) \in T_X$.
Properties
\[\text{Given T.S. } (X, T_X) \text{ and } (Y, T_Y), \text{ TFAE.} \nonumber\]
- $f: X \rightarrow Y$ is cts, i.e., $\forall V \in T_Y$, $f^{-1}(V) \in T_X$.
- $\forall p \in X$, $\forall$ nbhd $V$ of $f(p)$, $\exists U$ of $p$ s.t. $f(U) \subseteq V$.
- $\forall A \subseteq X$, $f(\bar{A}) \subseteq \overline{f(A)}$.
- $\forall$ closed $F \subseteq Y$, $f^{-1}(F)$ is closed in $X$.
Proof
- ($1 \Leftrightarrow 4$) Note that $\forall B \subseteq Y, f^{-1}(B^c)= X - f^{-1}(B)$.
a. $(\Rightarrow)$ Take any closed $F$ in $Y$. Then, $f^{-1}(F^c) = X - f^{-1}(F)$. Since $f$ is cts, $f^{-1}(F^c)$ is open, i.e., $f^{-1}(F)$ is closed in $X$.
b. $(\Leftarrow)$ For any closed $F$ in $Y$, $(f^{-1}(F))^c \in T_X$. Then, $F^c \in T_Y$, $f^{-1}(F^c) = X - f^{-1}(F)= X \cap (f^{-1}(F))^c \in T_X$. Thus, $f$ is cts. - ($1 \Leftrightarrow 2$) $\forall p \in X$, let $V$ be a nbhd of $f(p)$. Then, $f(p) \in V \Leftrightarrow p \in f^{-1}(V)$.
a. $(\Rightarrow)$ Let $U=f^{-1}(V)$. Then, $f(U) = f(f^{-1}(V)) \subseteq V$ by a general fact.
b. $(\Leftarrow)$ Take any open $V\in T_Y$. By Asm (2), $\forall p \in f^{-1}(V), \exists$ nbhd $U_p$ of $p$ s.t. $f(U_p) \subseteq V$, i.e., $U_p \subseteq f^{-1}(V)$. Then, $f^{-1}(V) = \cup_{p \in f^{-1}(V)}U_p$. This means that $f^{-1}(V)$ is a union of open sets, i.e., open in $X$. -
($2 \Rightarrow 3$) Take any $A \subseteq X$. Let $p \in \bar{A}$, then $f(p) \in f(\bar{A})$ holds. Take any nbhd $V$ of $f(p)$. By Asm (2), $\exists$ nbhd $U$ of $p$ s.t. $f(U) \subseteq V$. Since $p \in \bar{A}, U\cap A \ne \emptyset \leadsto \exists x \in U\cap A$. $\leadsto f(x) \in f(U) \subseteq V$ and $f(x) \in f(A)$. Thus, $f(x) \in V \cap f(A) \ne \emptyset$, i.e., $f(p) \in \overline{f(A)}$.
- ($3 \Rightarrow 4$) Take any closed $F$ in $Y$. Thus, we need to show $\overline{f^{-1}(F)} = f^{-1}(F)$, where $\supseteq$ is clear.
Let $p \in \overline{f^{-1}(F)}$. By Asm (3), let $A = f^{-1}(F) \subseteq X$. Then, $f(\bar{A}) \subseteq \overline{f(A)}$, i.e., $f(\overline{f^{-1}(F)}) \subseteq \overline{f(f^{-1}(F))}$. $\leadsto f(p) \in f(\overline{f^{-1}(F)}) \subseteq \overline{f(f^{-1}(F))} \subseteq \bar{F} =F$. Thus, $f(p) \in F$, i.e., $p \in f^{-1}(F)$.