Education: Sets in topological space.

04/11/2014

Sets in topological space.

Sets in topological space

Let

X

be a topological space. There are many types of sets we can define on

X.

The complement of a set A in X, denoted by $A^C$ , is $A^C = X \setminus A$ (that is, the entire space except for A).
A subset $C$ is called closed if the set $C^C$ is open. Notice that the intersection of any non-zero number of closed sets is closed and the union of finitely many closed sets is closed.
Note also that a set can be both closed and open. The trivial examples are the empty set $\emptyset$ and the entire set $X$ , each of which is both closed and open. By definition, $\emptyset$ is open, so its complement, $X$ , is closed. But $X$ , by definition, is an open set, so $X$ is both open and closed.
A set $N$ is called a neighborhood of a point $x \in X$ , if there is an open set $U$ such that $x\in U\subseteq N.$

We now investigate some commonly occurring sets in the study of Topology.

Definition

In a topological space, a $G_{\delta }$ set is a countable intersection of open sets. A $F_{\sigma }$ set is a countable union of closed sets.

Theorem

The complement of a

F_{\sigma }

set is

G_{\delta }

, and vice versa.

Proof:
Let A be a

F_{\sigma }

set and let

n\in \mathbb{N}

. Then A is a countable union of closed sets,

\bigcup\limits_{n}^{{}}{A_{n}}

such that

A_{n}

is closed for all n. Then

A^{c}=\bigcap\limits_{n}^{{}}{\left( A_{n} \right)^{c}}

. Since

A_{n}

is closed,

\left( A_{n} \right)^{c}

is open, so we have a countable intersection of open sets. Hence

A^{c}

is

G_{\delta }

.

The entirely similar proof of the other implication is left to the reader.

Theorem

In any metric space, a closed set is a

G_{\delta }

set.
Proof:
Let X be a metric space and let

A\subseteq (X,d)

.
Define

O_{n}=\bigcup\limits_{n}{\left\{ \beta_{1/n}(x)~\left| ~x\in A \right. \right\}}

. Observe that

O_{n}

is open for any n, and hence the union is open. Now our goal is to show that

\bar{A}=\bigcap\limits_{n}{O_{n}}

to show that a closed set is the intersection of countably many open sets.

\subseteq

:
Let

x\in \bar{A}

. Then

\beta_{1/n}(x)

intersects A at some

x_{0}

which implies

x\in \beta _{1/n}(x_{0})\subseteq O_{n}.

. This is true for any n so

x\in \bigcap\limits_{n}{O_{n}}

.

\supseteq

:
Let

x\in \bigcap\limits_{n}{O_{n}}

and

\varepsilon >0

. Then

\exists ~n\in \mathbb{N}

such that

1/n<\varepsilon

. So

x\in O_{n}\Rightarrow \exists ~x_{0}

in A such that

x\in \beta _{1/n}(x_{0})

, which implies

x\in \beta _{\varepsilon }(x_{0})

. Thus

x\in \bar{A}

.

Therefore

\bar{A}=\bigcap\limits_{n}{O_{n}}

and is a

G_{\delta }

set.

Theorem

In usual

\mathbb{R}

,

\mathbb{Q}

is a

F_{\sigma }

set.

Proof:
Since

\mathbb{Q}

with the usual topology is a metric space, every singleton such that

x\in \mathbb{Q}

is closed. Thus, we have a countable union of closed sets, and hence

\mathbb{Q}

is a

F_{\sigma }

set.