# Closed Sets and Limit Points

## Definition (Closed)

A subset $A$ of a topological space $X$ is said to be closed if the set $X-A$ is open.

## Theorem 17.1

Let $(X,\mathfrak{T})$ be a topological space. Then the following conditions hold

(1) $\emptyset$ and $X$ are closed.

(2) Arbitrary intersections of closed set is closed.

(3) Finite union of closed sets are closed. i.e.,

<proof>

(1) $\emptyset$ and $X$ are closed since their complements are $X$ and $\emptyset$, which are open.

(2) Let $\{F_\alpha\}_{\alpha\in\Lambda}$ be a collection of closed set. Then

\begin{align*} X-\bigcap_{\alpha\in\Lambda}F_\alpha &= X\cap\left(\bigcap_{\alpha\in\Lambda}F_\alpha\right)^c \\ &=X \cap \left(\bigcup_{\alpha\in\Lambda}F^c_\alpha \right) \\ &=\bigcup_{\alpha\in\Lambda}\left(X\cap F^c_\alpha \right) \\ &=\bigcup_{\alpha\in\Lambda}\left(X - F_\alpha \right) \end{align*}

Since $X-F_\alpha$ is open and union of them are also open. Thus the arbitrary intersections of closed sets are closed.

(3) Let $F_1,\ldots, F_n$ be closed sets.

\begin{align*} X-\bigcup_{i=1}^nF_i &= X\cap\left(\bigcup_{i=1}^nF_i \right)^c \\ &=X\cap \left(\bigcap_{i=1}^nF^c_i \right) \\ &=\bigcap_{i=1}^n(X\cap F^c_i) \\ &=\bigcap_{i=1}^n(X - F_i) \end{align*}

Since $X-F_i$ is open and finite intersections of open sets are open. Thus, the finite union of closed sets are closed.

$\tag*{\square}$

## Theorem 17.2

Let $(Y,\mathfrak{T}_Y)$ be a subspace of $(X,\mathfrak{T}_X)$. Then a set $A$ is closed in $Y$ if and only if $A=F\cap Y$ for some closed $F$ in $X$.

<proof>

Suppose that $A$ is closed in $Y$. Then $Y-A$ is open in $Y$, i.e., $Y-A=U\cap Y$ for some open $U$ in $X$. We want to show that $A= (X-U)\cap Y$. We know that

\begin{align*} Y - (Y-A) &= Y\cap(Y\cap A^c)^c \\ &=Y\cap (Y^c \cup A) \\ &= (Y\cap Y^c)\cup (Y\cap A) \\ &=Y\cap A \\ &=A \quad (\because A\subset Y) \end{align*} \begin{align*} A &= Y-(Y-A) \\ &=Y- (U\cap Y) \\ &=Y\cap (U\cap Y)^c \\ &=Y\cap(U^c\cup Y^c) \\ &=(Y\cap U^c)\cup (Y\cap Y^c) \\ &=Y\cap U^c \\ &=Y\cap(X-U) \end{align*}

Since $U$ is open in $X$, $X-U$ is closed in $X$. Thus, $A$ is equals to the intersection of a closed set of $X$ with $Y$.

Conversely, suppose that $A=F\cap Y$ for some closed $F$ in $X$. We want to show that $Y-A$ is open in $Y$.

\begin{align*} Y-A &= Y-(F\cap Y) \\ &=Y\cap(F^c\cup Y^c) \\ &=(Y\cap F^c)\cup (Y\cap Y^c) \\ &=Y\cap F^c \\ &=Y \cap (X-F) \end{align*}

Since $F$ is closed in $X$, $X-F$ is open in $X$. Thus, $Y-A$ is open in $Y$, i.e., $A$ is closed in $Y$.

$\tag*{\square}$

## Reference

• James Munkres, Topology, Pearson

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