# Topology/Points in Sets

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## Some Important Constructions

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

### 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 $\mathrm{Cl}(A)$ (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 $A$.

### 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 $\mathrm{Int}(A)$ (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 $A = Int(A)$

### 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 $X \setminus A$.

### Boundary

• Define the boundary of $A$ to be $\mathrm{Cl}(A)\setminus\mathrm{Int}(A)$, denoted $\mathrm{Bd}(A)$ (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 $\mathrm{Cl}(A)$ 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 $A$.

### 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 $A$.

### 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.

### Density

Definition: A subset $A$ of a topological space $X$ is called dense if any point $x\in X$ is in $A$, or if the point $x$ is a limit point of $A$.

Definition: In a topological space $X$, $A\subseteq X$ is dense if $Cl(A)=X$.

Example: The set of rational numbers is dense in the set of real numbers.

Definition: In a topological space $X$, a set $A\subseteq X$ is nowhere dense if $Cl(A)$ has no nonempty open sets.

Example: The set of natural numbers is nowhere dense in the set of real numbers.

Definition: Suppose X is a topological space. Then for $A\subseteq X$, A is dense in X if $\bar{A} = X$.

Definition: Suppose X is a topological space. Then for $A\subseteq X$, A is nowhere dense in X if and only if $int (\bar{A}) = \emptyset$.

Definition: A Gσ set is a subset of a topological space that is a countable intersection of open sets.

Definition: An Fσ set is a countable union of closed sets.

Theorem

(Hausdorff Criterion) Suppose X has 2 topologies, r1 and r2. For each $x \in X$, let B1x be a neighbourhood base for x in topology r1 and B2x be a neighbourhood base for x in topology r2. Then, $r_1\subseteq r_2$ if and only if at each $x \in X$, if $(B^1 \in B^1_x)(\exists (B^2 \in B^2_x)( B^2 \subseteq B^1).$

Theorem

In any topological space, the boundary of an open set is closed and nowhere dense.

Proof:
Let A be an open set in a topological space X. Since A is open, int(A) = A. Thus, σA ( or the boundary of A) = $\bar{A}/ int (A)$. Note that $\bar A/A = \bar A \cap A^c$. The complement of an open set is closed, and the closure of any set is closed. Thus, $\bar A \cap A^c$ is an intersection of closed sets and is itself closed. A subset of a topological space is nowhere dense if and only if the interior of its closure is empty. So, proceeding in consideration of the boundary of A.

The interior of the closure of the boundary of A is equal to the interior of the boundary of A.
Thus, it is equal to $int (\bar{A} \cap A^c)$.
Which is also equal to $int (\bar{A}) \cap int (A^c)$.

And, $int (A^c) = \bar{[(A^c)^c]}^c$. So, the interior of the closure of the boundary of A = $int (\bar{A}) \cap int \bar{(A^c)}$., and as such, the boundary of A is nowhere dense.

## Types of Spaces

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

### Perfect Spaces

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

## 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 $\alpha$; 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 $\mathrm{Int}(A)=\displaystyle\bigcup_{U\subseteq A}U$ 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 $\mathrm{Int}(A)=A$.
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 $\mathrm{Int}(A)=A$
($\Longleftarrow$)
As $\mathrm{Int}(A)$ is a union of open sets, it is open (from definition of open set). Hence $A=\mathrm{Int}(A)$ is also open.

• A set $A$ is closed if and only if $\mathrm{Cl}(A)=A.$
Proof:
Observe that the complement of $\mathrm{Cl}(A)$ 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 $\mathrm{Cl}(A)$ of a set $A$ is closed
Proof:
Let $\alpha$ be a closed set such that $A\subseteq\alpha$. Now, $\mathrm{Cl}(A)= \displaystyle\bigcap_{\alpha\subset X}\alpha$ for closed $\alpha$. We know that the intersection of any collection of closed sets is closed, and hence $\mathrm{Cl}(A)$ is closed.

## Exercises

1. Prove the following identities for subsets $A,B$ of a topological space $X$:
• $\mathrm{Cl}(A\cup B)=\mathrm{Cl}(A)\cup\mathrm{Cl}(B)$
• $\mathrm{Cl}(A\cap B)\subseteq\mathrm{Cl}(A)\cap\mathrm{Cl}(B)$
• $\mathrm{Int}(A)\cup\mathrm{Int}(B)\subseteq\mathrm{Int}(A\cup B)$
• $\mathrm{Int}(A\cap B)=\mathrm{Int}(A)\cap\mathrm{Int}(B)$
2. 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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