Source of most content: Gunhee Cho and Lecture note by Matsumura
Closed set
Def. Given T.S. $(X, T_X)$, $A \subseteq X$ is closed if $A^c$ is open.
NB1; Let $X = [ 0,1 ] \cup (2,3) \subseteq (\mathbb{R}, T_{std})$. Consider $(X, T_{subsp})$. Then,
- $(2,3) = X \setminus [0 ,1]: (2, 3)$ is complement of open set, i.e., $(2,3)$ is closed set.
- $(2,3) = Y \cap (1.5, 3.5)$, where $(1.5, 3.5)$ is open set on $(\mathbb{R}, T_{std})$, i.e., $(2,3)$ is open set.
This implies that open and closed are not mutually exclusive!
Propositions
\[\text{Given T.S. } (X, T_X). \text{ Then, following holds: } \nonumber\]
- $\emptyset$ and $X$ is closed.
- Arbitrary intersection of closed sets is closed.
- Finite unions of closed sets is closed.
Proof
- $\emptyset^c = X$ and $X^c = \emptyset$ are open.
- if $F_i$ is closed $\forall i \in J$, then $(\cap_{i \in J} F_i)^c = \cup_{i\in J} F_i^c$ is open.
- if $F_1, \cdots, F_n$ is closed for $n < \infty$, then, $(\cup_{i=1}^n F_i)^c = \cap_{i=1}^n F_i^c$ is open.
\[\text{Given T.S. } (X, T_X) \text{ and its subspace T.S. } (Y\subseteq X, T_{subsp}). \nonumber\] \[\text{ Let } A \subseteq Y. \text{ Then, } A \text{ is closed in } Y \text{ iff } A = F \cap Y \text{ for some closed } F \in T_X. \nonumber\]
Proof
- $(\Rightarrow)$ $A^c$ is open in $Y \Leftrightarrow A^c = U \cap Y$, where $U \in T_X \Leftrightarrow A = Y - A^c = Y - [Y \cap U] = Y \cap U^c$, i.e., if $A$ is closed in $Y$, then $\exists$ closed $F$ in $X$ s.t. $A= F \cap Y$.
- $(\Leftarrow)$ Suppose $A=F\cap Y$ for closed $F$ in $X$. Then, $A^c = Y - A = Y - (F\cap Y) = Y \cap F^c$. Since $F^c$ is closed in $X$, $A^c$ is open in $Y$.
NB2; From $A = F \cap Y$, if $Y$ is closed in $X$, then $A$ is closed in $X$.
Interior and Closure
Def. Given T.S. $(X, T_X)$, let $A \subseteq X$. Define the interior of $A$, $\mathrm{Int} A := \cup_{\text{open } U \subseteq A} U$, i.e., union of all open sets in $A$
Def. Given T.S. $(X, T_X)$, let $A \subseteq X$. Define the closure of $A$, $\bar{A} := \cap_{A \subseteq \text{closed }F} F$, i.e., intersection of all closed sets containing $A$.
NB3; $\forall A \subseteq X, \mathrm{Int} A \subseteq A \subseteq \bar{A}$.
NB4; $A$ is open in $X \Leftrightarrow A= \mathrm{Int} A$.
NB5; $A$ is closed in $X \Leftrightarrow A= \bar{A}$.
NB6; $\mathrm{Int} A$ is the biggest open set in $A$.
NB7; $\bar{A}$ is the smallest closed set contatining $A$. \
Def. Given T.S. $(X, T_X)$, $U$ is called a neighborhood (nbhd) of $x \in X$ if $U \in T_X$ and $x \in U$.
Def. Given $A \subseteq (X, T_X)$, $x\in X$ is called a limit point of $A$ if $\forall$ nbhd $U$ of $x$, $U \cap (A - \{ x\}) \ne \emptyset$ and denote $A’ = \{ x\in X : x \in \overline{A - \{ x \}} \}$.
Propositions
\[\text{ Let } (Y, T_{subsp}) \text{ be a subspace of } (X,T_X) \text{ and } A \subseteq Y. \text{ Then, } \bar{A}^Y =\bar{A}^X \cap Y. \nonumber\]Here, $\bar{A}^X, \bar{A}^Y$ implies closure of $A$ w.r.t. $T_X$ and $T_Y$, respectively.
Proof
Let $F_X$ and $F_Y$ denote the closed set in $X$ and $Y$ respectively. Then, $\bar{A}^Y = \cap_{A \subseteq F_Y} F_Y \stackrel{( * )}{=} \cap_{A \subseteq Y \cap F_X} \stackrel{A \subseteq Y}{=} \cap_{A \subseteq F_X} Y \cap F_X$ $= Y \cap \bigcap_{A \subseteq F_X} F_X = Y \cap \bar{A}^X$. $(*)$ follows from above Prop.
\[\text{ Let } A \subseteq (X, T_X) \text{ and } \mathcal{B} \text{ be a basis of } T_X. \text{ Then, following holds:} \nonumber\]
- $x \in \bar{A} \Leftrightarrow \forall $ nbhd $U$ of $x$, $U \cap A \ne \emptyset$.
- $x \in \bar{A} \Leftrightarrow \forall $ $B \in \mathcal{B}$ s.t. $x \in B$, $B\cap A \ne \emptyset$.
Proof
It suffices to show 1. since open set can be express in terms of bases. And prove the contrapositive statement.
- $(\Leftarrow)$ Let $x \not\in \bar{A}$, i.e., $x \in (\bar{X})^c = ( \cap_{A \subseteq F} F)^c = \cup_{A \subseteq F} F^c =: U_0$ is open in $X$. (union of open sets) Here, $A \subseteq F$, i.e., $F^c \cap A \ne \emptyset$. Thus, $\cup_{A \subseteq F} (F^c \cap A) = U_0 \cap A = \emptyset$. Hence, $\exists$ nbhd $U$ of $x$ s.t. $U \cap A = \emptyset$.
- $(\Rightarrow)$ Suppose $\exists$ nbhd $U$ of $x$ s.t. $U \cap A = \emptyset \Leftrightarrow A \subseteq U^c$. Then, $x \not\in U^c =: F_0$, i.e., $ x \not\in \bar{A} = \cap_{A \subseteq F} F$.
\[\text{ Let } A \text{ be a subset of T.S. } (X, T_X). \text{ Then, } \mathrm{Int} A = X - (\overline{X-A})\text{ and }\bar{A} = A \cup A' . \nonumber\]
Proof
- Let $U$ be open sets and $F$ be closed sets. Then, $A= \cup_{U \subseteq A} U$. Then, $X- \mathrm{Int} A = X - \cup_{U \subseteq A} U = \cap_{U \subseteq A} (X \cap U^c) = \cap_{X\cap A^c \subseteq F} F = \overline{X-A}$.\
- $\forall$ nbhd $U$ of $x$, $U \cap A \ne \emptyset \Leftrightarrow x \in \bar{A}$. Since $U \cap [ (A \cap \{x \}) \cup ( A - \{ x \}) ] \ne \emptyset$, this implies $U \cap (A \cap \{x \}) \ne \emptyset \ (x \in A )$ or $U \cap (A - \{x \}) \ne \emptyset \ (x \in A’ )$.