Topology/Points in Sets

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Contents

1 Some Important Constructions
2 Types of Spaces
- 2.1 Perfect Spaces
3 Some Basic Results
4 Exercises

[edit] Some Important Constructions

Let $A$ be an arbitrary subset of $X$ .

[edit] Closure

A point $x$ is called a point of closure of a set $A$ if for every neighbourhood $U$ of $x$ , $U \cap A \neq \emptyset$
Define the closure of $A$ to be the intersection of all closed sets containing $A$ , denoted $Cl(A)$ (some authors use $\bar{A}$ ). The closure has the nice property of being the smallest closed set containing $A$ . Each neighborhood (nbd) of every point in the closure intersects $A$ .

[edit] Interior

We say that $x$ is an internal point of $A$ iff There is an open set $U$ , $x \in U$ and $U \subseteq A$
Define the interior of $A$ to be the union of all open sets contained inside $A$ , denoted $Int(A)$ (some authors use $A^\circ$ ). The interior has the nice property of being the largest open set contained inside $A$ . Every point in the interior has a nbd contained inside $A$ .

Note that a set $A$ is Open iff $A = I n t (A)$

[edit] Exterior

Define the exterior of $A$ to be the union of all open sets contained inside the complement of $A$ , denoted $\mathrm{Int} (X \setminus A)$ . It is the largest open set inside $X \setminus A$ . Every point in the exterior has a nbd contained inside $X \setminus A$ .

[edit] Boundary

Define the boundary of $A$ to be $\mathrm{Cl}(A)\setminus\mathrm{Int}(A)$ , denoted $Bd(A)$ (some authors prefer $\partial A$ ). The boundary is also called the frontier. It is always closed since it is the intersection of the closed set $Cl(A)$ and the closed set $X\setminus\mathrm{Int}(A)$ . It can be proved that $A$ is closed if it contains all its boundary, and is open if it contains none of its boundary. Every nbd of every point in the boundary intersects both $A$ and $X \setminus A$ . All boundary points of a set $A$ are obviously points of contact of $A$ .

[edit] Limit Points

A point $x$ is called a limit point of a set $A$ if for every neighborhood $U$ of $x$ , $(U\setminus\{x\}) \cap A \neq \emptyset$ . All limit points of a set $A$ are obviously points of closure of the set $A$ .

[edit] Isolated Points

If a neighborhood $N$ of a point $x\in S\subseteq X$ can be found such that $N\cap S = \{x\}$ , then x is called an isolated point.

[edit] Types of Spaces

We can also categorize spaces based on what kinds of points they have.

[edit] Perfect Spaces

If a space contains no isolated points, then the space is a perfect space.

[edit] Some Basic Results

For every set $A$ ; $A\subseteq \mathrm{Cl}(A)$ and $\mathrm{Int}(A)\subseteq A$
Proof:
Let $x\in A$ . If a closed set $\alpha\supseteq A$ , then $x\in\alpha$ . As $\mathrm{Cl}(A)=\displaystyle\bigcap_{\alpha\subset X} \alpha$ for closed $α$ ; we have $x\in\mathrm{Cl}(A)$ . $x\in A$ being arbitrary, $A\subseteq \mathrm{Cl}(A)$
Let $U\subseteq A$ be open. Thus, $x\in A\forall x\in U$ . As $\mathrm{Int}(A)=\displaystyle\bigcap_{U\subseteq A}U$ for open $U$ ; we have $x\in \mathrm{Int}(A)\forall x\in U$ . $U\subseteq A$ being arbitrary, we have $\mathrm{int}(A)\subseteq A$

A set $A$ is open if and only if $Int(A) = A$ .
Proof:
( $\Longrightarrow$ )
$A$ is open and $A\subseteq A$ . Hence, $A\subseteq\mathrm{Int}(A)$ . But we know that $\mathrm{Int}(A)\subseteq A$ and hence $Int(A) = A$
( $\Longleftarrow$ )
As $Int(A)$ is a union of open sets, it is open (from definition of open set). Hence $A = Int(A)$ is also open.

A set $A$ is closed if and only if $Cl(A) = A .$
Proof:
Observe that the complement of $Cl(A)$ satisfies $X\setminus \mathrm{Cl}(A)=\mathrm{Int}(X\setminus A)$ . Hence, the required result is equivalent to the statement " $X\setminus A$ is open if and only if $\mathrm{Int}(X\setminus A)=X\setminus A$ ". $A$ is closed implies that $X\setminus A$ is open, and hence we can use the previous property.

The closure $Cl(A)$ of a set $A$ is closed
Proof:
Let $α$ be a closed set such that $A\subseteq\alpha$ . Now, $\mathrm{Cl}(A)= \displaystyle\bigcup_{\alpha\subset X}\alpha$ for closed $α$ . We know that the intersection of any collection of closed sets is closed, and hence $Cl(A)$ is closed.

[edit] Exercises

Prove the following identities for subsets A,B of a topological space X:
- $\mathrm{Cl}(A\cup B)=\mathrm{Cl}(A)\cup\mathrm{Cl}(B)$
- $\mathrm{Cl}(A\cap B)\subseteq\mathrm{Cl}(A)\cap\mathrm{Cl}(B)$
- $\mathrm{Int}(A)\cup\mathrm{Int}(B)\subseteq\mathrm{Int}(A\cup B)$
- $\mathrm{Int}(A\cap B)=\mathrm{Int}(A)\cap\mathrm{Int}(B)$
Show that the following identities need not hold (i.e. give an exaple of a topological space and sets A and B for which they fail):
- $\mathrm{Cl}(A\cap B)=\mathrm{Cl}(A)\cap\mathrm{Cl}(B)$
- $\mathrm{Int}(A)\cup\mathrm{Int}(B)=\mathrm{Int}(A\cup B)$

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Category: Topology (book)

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