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 subcollection of $\tau$ is in $\tau$.

• The intersection of any finite subcollection of $\tau$ is in $\tau$.

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.

Definition: 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 $\tau$.

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: Union of Open Sets

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

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

PROOF

This follows directly from the definition of a topological space.

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

This follows directly from the definition of a topological space.

Closed Sets

Definition: Closed Set

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

A subset $C\subseteq S$ is closed if its complement $S\setminus C$ is an open set.

Theorem: Intersection of Closed Sets

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

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

PROOF

TODO

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 $S\subseteq X$ is a clopen if 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 $x$ if 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 $X$.

We say that a subset $N\subseteq X$ is a neighbourhood of $S$ if there exists an open set $U$ such that $S\subseteq U\subseteq N$.

NOTATION

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

Theorem: Openness from Neighborhoods

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

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

PROOF

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

If there exists neighborhood $N(x)$ of each $x$ such that $N(x)\subseteq U$, then inside each neighborhood $N(x)$ 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: Closedness and Neighborhoods

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

A subset $C\subseteq S$ is closed if and only if each $x$ in its complement $S\setminus C$ has a neighborhood $N(x)$ such that $N\subseteq S\setminus C$.

PROOF

TODO