Topology

Topology on a set

X

A topology on $X$ is a collection $𝒯$ of subsets of $X$ , that is: it is a subset of the $P (X)$ , with the following properties:

$\emptyset, X \in 𝒯$
For any $𝒜 \subseteq 𝒯$

⋃ 𝒜 \in 𝒯

For any $𝒞 \subseteq 𝒯$ such that $𝒞$ is finite

⋂ 𝒞 \in 𝒯

The set

X

along with

𝒯

satsifying the above conditions is called a topological space and is denoted by

(X, 𝒯)

Open set

Suppose

(X, 𝒯)

is a topological space, if

U \in 𝒯

then we say that

U

is open with respect to

X

a set filled with open sets is open

Let

X

be a topological space, and

A \subseteq X

. Suppose that for each

x \in A

there is an open set

U

containing

x

such that

U \subseteq A

. Show that

A

is open in

X

Since for each $x \in A$ , there is an open set $U_{x}$ such that $x \in U_{x} \subseteq A$ , then $A$ is covered by subsets, therefore it is a union, and we can write $A = ⋃_{x \in A} U_{x}$

Since each $U_{x}$ was assumed to be open with respect to $X$ , then an arbitrary union of them is also open with respect to $X$ , in other words $A$ must be open.

The Finite Complement Topology

Let

X

be a set and define the set

F C = {U \subseteq X : | X ∖ U | < \infty} \cup {\emptyset}

, then

F C

is a topology and we denote it by

𝒯_{F C}

TODO: Add the proof here.

The Countable Complement Topology

Let

X

be a set and define the set

C C = {U \subseteq X : | X ∖ U | \leq ℵ_{0}} \cup {\emptyset}

where we've used

ℵ_{0}

then

C C

is a topology and we denote it by

𝒯_{C}

We'll note that $X \in C C$ as $X ∖ X = \emptyset$ is certainly finite, on the other hand $\emptyset \in C C$ by definition

Now let $𝒜 \subseteq C C$ , then $X ∖ ⋃ 𝒜 = X ∖ ⋃_{A \in 𝒜} A = ⋂_{A \in 𝒜} (X ∖ A)$ therefore we have an intersection of countable sets and therefore is countable

Now let $𝒞 \subseteq C C$ , where $𝒞$ is finite, then $X ∖ ⋂ 𝒞 = X ∖ ⋃_{C \in 𝒞} C = ⋃_{C \in 𝒞} (X ∖ A)$ so we have a finite union of countable sets, and therefore it is countable. Therefore the set is closed under arbitrary unions and finite intersections and thus is a topology.

Note that $𝒯 = {X} \cup {U \subset X : X - U is infinite} .$ is not a topology so long as $X$ is infinite. For example pick some $p \in X$ then every singleton ${q}$ where $q \neq p$ is open in $X$ because $X ∖ {q}$ is infinite, if it were to be a topology then we would know that $⋃_{q \neq p} {q} = X ∖ {q}$ is open, but $X ∖ (X ∖ {q}) = {q}$ which is finite, thus a contradiction, so it cannot be a topology.

The Intersection of Topologies Is a Topology

𝔗

is a collection of topologies then

⋂ 𝔗

is a topology.

Since for any

𝒯 \in 𝔗

we know that

\emptyset, X \in 𝒯

then

\emptyset, X \in ⋂ 𝔗

, then suppose that

𝒜 \subseteq ⋂ 𝔗

then for any

𝒯 \in 𝔗

we know

⋃ 𝒜

is a union of elements in

𝒯

so that

⋃ 𝒜 \in 𝒯

, therefore

⋃ 𝒜 \in ⋃ 𝔗

, similarly if

𝒞 \subseteq ⋂ 𝔗

is finite then for any

𝒯 \in 𝔗

we know that

⋂ 𝒞 \in 𝒯

therefore

⋂ 𝒞 \in 𝔗

, therefore

⋂ 𝔗

is a topology.

The Intersection of a Collection of Sets That Are Supersets of a Given Set and Satisfy a Property Is the Smallest Set Which Satisfies the Property and Is Still a Superset of the Given Set

Let

X

be a set and

Q

a predicate then suppose that

𝒞

is a collection of sets such that for any

C \in 𝒞

we have

X \subseteq C

and

Q (C)

then if

Q (⋂ 𝒞)

holds true then it is the smallest such set where it holds true

Suppose that there were a smaller set, which is to say we had some set $S$ where $X \subseteq S \subset ⋂ 𝒞$ and $Q (S)$ , since $X \subseteq S$ and $Q (S)$ then $X \in 𝒞$ therefore $⋂ 𝒞 \subseteq S$ , a contradiction, and thus there is no strictly smaller set.

Now we show that this set is unique, as if $B$ is another set such that $Q (B)$ , $X \subseteq B$ and $B \subseteq ⋂ 𝒞$ then the first two imply that $B \in 𝒞$ and therefore $⋂ 𝒞 \subseteq B$ , so that $𝒞 = B$

Given a Family of Topologies There Is a Unique Smallest Topology Containing All of Them

Suppose that

𝔗

is a collection of topologies on

X

, and let

𝔉

be the collection of topologies such that for any

ℱ \in 𝔉

, we know that for every

𝒯 \in 𝔗

we have

𝒯 \subseteq ℱ

, then

⋂ 𝔉

is the unique smallest topology on

X

that contains all the topologies in

𝔗

We know

⋂ 𝔉

is a topology. Additionally it must be the smallest such topology that contains all topologies in

𝔗

for if there was another such one

𝒴

that did, then we would know that

𝒴 \in 𝔉

and therefore

⋂ 𝔉 \subseteq 𝒴

so that

⋂ 𝔉 = 𝒴

, note that the final equality shows that it is uniquely the smallest.

Given a Family of Topologies There Is a Unique Largest Topology Contained in Every Topology in the Family

Suppose that

𝔗

is a collection of topologies on, then

⋂ 𝔗

is the unique largest topology that is contained in

𝒯

for each

𝒯 \in 𝔗

We know that it is a topology, to see that it is the largest, suppose that there was a larger topology $𝒴$ that is contained in $𝒯$ for each $𝒯 \in 𝔗$ but this implies that $𝒴 \subseteq ⋂ 𝔗$ which is a contradiction, so $⋂ 𝔗$ is the largest.

It is unique as given another topology $𝒵$ that is contained in each $𝒯$ then as above we know that $𝒵 \subseteq ⋂ 𝔗$ but sinze $𝒵$ is the largest topology with this property then we know that $⋂ 𝔗 \subseteq 𝒵$ so that $⋂ 𝔗 = 𝒵$ as needed.

Example of the Largest and Smallest Topology

With

X = {a, b, c}

𝒯_{1} = {\emptyset, X, {a}, {a, b}}

and

𝒯_{2} = {\emptyset, X, {a}, {a, b}}

, find the smallest topology containing both, and the largest one contained in both.

By the above proposition we know the smallest topology is the intersection of all topologies that contain all open sets from $𝒯_{1}$ and $𝒯_{2}$ so any such topology contains at least $\emptyset, X, {a}, {a, b}, {b, c}$ due to the closure under unions and intersection, then each such topology contains at least ${\emptyset, X, {a}, {b}, {a, b}, {b, c}}$ since the above is a topology, then the intersection of all such topologies must be the above set (the smallest possible set, is included in the intersection, thus it equals it).

The largest topology contained in both $𝒯_{1}, 𝒯_{2}$ is $𝒯_{1} \cap 𝒯_{2} = {\emptyset, X, {a}}$

The Power Set is a Topology

Given a set

X

, then

P (X)

is a topology on it.

We note that $\emptyset \subseteq X$ and $X \subseteq X$ therefore we know that $\emptyset, X \in P (X)$ .

Now suppose that $𝒜$ is a collection of sets from $P (X)$ . Thus if we look at the union it is of the form $⋃_{A \in 𝒜} A$ and since we know that each $A \in P (X)$ then $A \subseteq X$ and therefore it is a union of elements from $X$ and so $⋃_{A \in 𝒜} A \subseteq X$ therefore $⋃_{A \in 𝒜} \in P (X)$

Now suppose that $𝒞$ is a finite collection of elements from $P (X)$ , then similarly we find that $⋂ 𝒞 \subseteq X$ .

finer and coarser topologies

suppose that $𝒯$ and $𝒯^{'}$ are two topologies on a given set $X$ . If $𝒯 \subseteq 𝒯^{'}$ , then $𝒯^{'}$ is finer than $𝒯$ . If the reverse inclusion is true, then we say that $𝒯^{'}$ is coarser than $𝒯$ , there are also strict variations of these definitions for the strict inclusions.

comparable topologies

given two topologies, they are comparable if at least one is finer than the other