Topological Spaces

Topologies and Topological Spaces

Definition: Topology

A topology on a non-empty set $X$ is a collection $\tau$ of subsets of $X$ which has the following properties:

• The empty set $\varnothing$ and $X$ are in $\tau$.

• The union of any subset of $\tau$ is in $\tau$.

• The intersection of any two elements of $\tau$ is in $\tau$.

INTUITION

A topology $\tau$ on a set $X$ can be interpreted as a definition of “closeness” between elements of $X$ without using any notion of distance. Moreover, a topology provides a way for us to define what is “inside” a set, what is “outside” a set and what separates the inside of a set from its outside.

EXAMPLE

Consider the sets $X=\{1,2,3,4,5,6\}$ and $\tau =\{\varnothing ,X,\{1\},\{3,4\},\{1,3,4\},\{2,3,4,5,6\}\}$. The set $\tau$ is a topology on $X$, since it satisfies the requirements in the definition.

Definition: Topological Space

A topological space $(S,\tau )$ is a non-empty set $S$ equipped with a topology $\tau$ on it.

Note: Points

It is common to refer to the elements $s\in S$ of a topological space as points.

Open Sets

Definition: Open Subset

Let $(X,\tau )$ be a topological space.

A subset of $X$ is open iff it is an element of $\tau$.

Openness Criteria

THEOREM

Let $(X,\tau )$ be a topological space.

A subset $U\subseteq X$ is open if and only if each $x\in U$ has neighbourhoodd) $N(x)$ such that $N(x)\subseteq U$.

PROOF

If $U$ is open, then $U$ is, by definition, neighbourhoodd) of every $x\in U$.

If there exists neighbourhoodd) $N(x)$ of each $x$ such that $N(x)\subseteq U$, then inside each neighbourhood $N(x)$ there exists, by definition, an open set $O(x)\subseteq N(x)$ which contains $x$. Since $N(x)\subseteq U$, we have $O(x)\subseteq N(x)\subseteq U$. This means that we can construct $U$ as the union of these open sets $O(x)$.

$$U=\bigcup _{x\in U}O(x)$$

Since this is a union of open sets, i.e. it is a union of a subset of the topology $\tau$, we have that $U\in \tau$.

Theorem

Let $(X,\tau )$ be a topological space.

A subset $U\subseteq X$ is open if and only if for each $x\in U$ there exists an open set $V$ such that $V\subseteq U$ and $x\in V$.

PROOF

TODO

Theorem

Let $(X,\tau )$ be a topological space.

A subset $S\subseteq X$ is open if and only if it is equal to its own interior.

$$S=\mathrm{int}S$$

PROOF

We need to prove two things:

• (I) If $S$ is open, then $S=\mathrm{int}S$.
• (II) If $S=\mathrm{int}S$, then $S$ is open.

Proof of (I):

Suppose $S$ is open. Recall the definition of the interior $\mathrm{int}S$:

$$\mathrm{int}S\stackrel{\text{def}}{=}\bigcup \{O\subseteq S\mid O\text{ is open}\}$$

Since $S\subseteq S$ and $S$ is open, we know that $S\in \{O\subseteq S\mid O\text{ is open}\}$ and thus $S\subseteq \mathrm{int}S$. However, the interior is a subset of $S$. Since $S\subseteq \mathrm{int}S$ and $\mathrm{int}S\subseteq S$, we know deduce that $S=\mathrm{int}S$.

Proof of (II):

Suppose that $S=\mathrm{int}S$. Since the interior $\mathrm{int}S$ is a union of Open Sets, it is itself open. Therefore, $S$ is open.

Properties of Open Sets

Theorem: Union of Open Sets

Let $(S,\tau )$ be a topological space.

The union of any collection of open subsets is also open.

PROOF

This follows directly from the definition of a topology.

Theorem: Intersection of Open Sets

Let $(S,\tau )$ be a topological space.

The intersection of any finite collection of Open Sets is also open.

PROOF

We consider $n\ge 2$ arbitrary open subsets $U_1,\cdots ,U_n$.

For $n=2$, the definition of the topology $\tau$ tells us that $U_1\cap U_2\in \tau$.

Now suppose $U^{\prime }=U_1\cap \cdots \cap U_{n-1}\in \tau$. We have

$$\begin{array}{rl}{U}_1\cap \cdots \cap U_n& =(U_1\cap \cdots \cap U_{n-1})\cap U_n\\ U_1\cap \cdots \cap U_n& =U^{\prime }\cap U_n\end{array}$$

Since $U_1\cap \cdots \cap U_n$ is the intersection of two elements of $\tau$, it must itself be an element of $\tau$, Q.E.D.

Closed Sets

Definition: Closed Set

Let $(S,\tau )$ be a topological space.

A subset $U$ of $S$ is closed if its Complement $S\setminus U$ is an open set.

THEOREM

Let $(S,\tau )$ be a topological space.

A subset $C\subseteq S$ is closed if and only if for each $x$ in its Complement $S\setminus C$ there exists neighbourhoodd) $N$ of $x$ such that $N\subseteq S\setminus C$.

PROOF

TODO

Closedness Criteria

THEOREM

Let $(X,\tau )$ be a topological space.

A subset $S\subseteq X$ is closed if and only if it is equal to its own Closure.

PROOF

We have to prove two things:

• (I) If $S$ is closed, then $S=\overline{S}$.
• (II) If $S=\overline{S}$, then $S$ is closed.

Proof of (I):

TODO

Proof of (II):

TODO

Theorem: Limit Points of Closed Subsets

Let $(X,\tau )$ be a topological space.

A subset $S\subseteq X$ is closed if and only if it contains all of its limit points.

PROOF

We need to prove two things separately:

• (I) If $S$ is closed, then every limit point of $S$ lies in $S$.
• (II) If $S$ contains all of its limit points, then it is closed.

Proof of (I):

We prove this by contradiction. Suppose that $S$ is closed and there exists a limit point $p$ of $S$ which lies outside of $S$, i.e. $p\in X\setminus S$. We know that $X\setminus S$ is open because it is the Complement of a closed set. However, since $p$ is a limit point of $S$, every open set which contains $p$ must also contain an element of $S$. This implies that $X\setminus S$ contains an element of $S$ which is a contradiction.

Proof of (II):

Suppose that $S$ contains all of its limit points. This means that there are no points $p\in X\setminus S$ such that every open set $U$ which contains $p$ also contains another element of $S$. Alternatively, this means that each $p\in X\setminus S$ is contained in some open set $U_p$ such that $U_p\cap S=\varnothing$, i.e. $U_p\subseteq X\setminus S$. Therefore, the union ${\bigcup }_{p\in X\setminus S}{U}_p$ is a subset of $X\setminus S$ and, since it contains every $p\in X\setminus S$, it means that ${\bigcup }_{p\in X\setminus S}{U}_p=X\setminus$. Since $X\setminus S$ is a union of open sets, it is itself open and so $S$ is closed.

Properties

Theorem: Intersection of Closed Sets

Let $(S,\tau )$ be a topological space.

The intersection of any collection of closed subsets is also closed.

PROOF

TODO

Theorem: Union of Closed Sets

Let $(S,\tau )$ be a topological space.

The union of any finite collection of closed subsets is also closed.

PROOF

Let $S_1,\cdots ,S_n$ be closed sets. We need to show that $\bigcup \{S_1,\cdots ,S_n\}=S_1\cup \cdots \cup S_n$ is closed. According to the definition of a closed set, this means we must show that $S\setminus (S_1\cup \cdots \cup S_n)$ is open.

By the distributive law

$$S\setminus (S_1\cup \cdots \cup S_n)=(S\setminus S_1)\cap \cdots \cap (S\setminus S_n)$$

The sets $S_1,\cdots ,S_n$ are closed and by definition their complements $S\setminus S_1,\cdots ,S\setminus S_n$ are open. The right-hand side is thus an intersection of open sets and is, therefore, open itself. This means that the complement $S\setminus S_1\cup \cdots \cup S_n$ is open and so $S_1\cup \cdots \cup S_n$ is closed.

Clopen Sets

It is important to note that “openness” and “closedness” are neither mutually exclusive nor complete - Subsets in a Topological Spaces can be both open and closed or they can also be neither.

Definition: Clopen Set

Let $(X,\tau )$ be a topological space.

A subset of $X$ is a clopen iff it is both open and closed.

Neighborhoods

Definition: Neighborhood of a Point

Let $(X,\tau )$ be a topological space and let $x\in X$.

A subset $N\subseteq X$ is a neighbourhood of $x$ iff there exists an open set $U$ such that $U\subseteq N$ and $x\in U$.

NOTATION

$$N(x)N_x$$

Definition: Neighbourhood of a Set

Let $(X,\tau )$ be a topological space and let $S$ be a subset of $X$.

A subset $N\subseteq X$ is a neighbourhood of $S$ iff there exists an open set $U$ such that

$$S\subseteq U\subseteq N$$

NOTATION

$$N_{S}N(S)$$