intersected topology

Let $$ X$$ be a topological space with toplogy $$ \mathcal{T}$$. If $$ Y$$ is a subset of $$ X$$, then we define the the toplogy $$ \mathcal{T}$$ intersected with $$ Y$$ as the set

$$ {T}_{\cap Y}:=\{Y\cap U:U\in \mathcal{T}\}$$

intersected topology is a topology

The set $$ {T}_{\cap Y}$$ is a topology on $$ Y$$

We'll show that $$ \mathrm{\varnothing}$$ and $$ Y$$ are elements of $$ {T}_{\cap Y}$$, this can be seen as since $$ \mathrm{\varnothing}\in T$$, then $$ Y\cap \mathrm{\varnothing}=\mathrm{\varnothing}\in {T}_{\cap Y}$$ and also since $$ Y\subseteq X$$, then $$ Y\cap X=Y$$ so $$ Y\in {T}_{\cap Y}$$

Now we want to prove that $$ {T}_{\cap Y}$$ is closed under arbitrary unions, which we know because $$ \bigcup _{\alpha \in I}({U}_{\alpha}\cap Y)=\left(\bigcup _{\alpha \in I}{U}_{\alpha}\right)\cap Y$$. Similarly we can see that it's closed under finite intersections as $$ ({U}_{1}\cap Y)\cap \dots \cap ({U}_{n}\cap Y)=({U}_{1}\cap \dots \cap {U}_{n})\cap Y$$.

Therefore we know that $$ {T}_{\cap Y}$$ is a topology.

subspace topology

Since $$ {T}_{\cap Y}$$ forms a topology, we will denote it by $$ {\mathcal{T}}_{\cap Y}$$ and call it the subspace topology and that $$ Y$$ is a subspace of $$ X$$

basis for the subspace topology

Suppose that $$ \mathcal{B}$$ is a basis for the topology of $$ X$$, then $$ M:=\{Y\cap B:B\in \mathcal{B}\}$$ is a basis and it generates $$ {\mathcal{T}}_{\cap Y}$$

To prove this statement true, we'll use the handy basis criterion

First note that given any element $$ Y\cap B\in M$$, we know that it is open with respect to $$ {\mathcal{T}}_{\cap Y}$$, because $$ B$$ is open with respect to $$ X$$

Now let $$ Y\cap U$$ be an open set of $$ Y$$ which means that $$ U$$ is open in $$ X$$. Let $$ p\in Y\cap U$$ , since $$ p\in U$$ , $$ U$$ is open in $$ X$$ and $$ \mathcal{B}$$ is a basis for $$ X$$ then we know there is some $$ B\in \mathcal{B}$$ such that $$ p\in B\subseteq U$$ if that's the case then we know that $$ p\in Y\cap B\subseteq Y\cap U$$, noting that $$ Y\cap B\in M$$, therefore we know that $$ M$$ is a basis and it is for $$ Y$$

open in a subspace implies open

Suppose that $$ Y$$ is a subspace of $$ X$$ and also that $$ Y$$ is open in $$ X$$, then any set that's open in $$ Y$$ is also open in $$ X$$

Suppose that $$ U$$ is open in $$ Y$$, thus there is some $$ V$$ open in $$ X$$ such that $$ U=Y\cap V$$, but since $$ Y$$ is open in $$ X$$, then as a finite intersection of open elements, then $$ U$$ is also open in $$ X$$, as needed.

TODO

Suppose that $$ A$$ is a subspace of $$ X$$ and $$ B$$ is a subspace of $$ Y$$, then the product topology on $$ A\times B$$ is the same as the topology $$ A\times B$$ has as a subspace of $$ X\times Y$$

We will show that the two topologies are equal by showing that the basis that generate them are equal.

Given the subspace $$ A$$ of $$ X$$ and $$ B$$ of $$ Y$$, then the product topology of $$ A\times B$$ has a basis of elements of the form $$ (A\cap U)\times (B\cap V)$$ where $$ U,V$$ are open in $$ X,Y$$ respectively.

On the other hand ,an open set in $$ X\times Y$$ is of the form $$ U\times V$$ where $$ U,V$$ are open in $$ X,Y$$ respectively, thus a basis element for $$ A\times B$$ (as a subspace of $$ X\times Y$$)is a set of the form $$ (A\times B)\cap (U\times V)$$.

Now to connect the two recall thqt $$ (A\cap U)\times (B\cap V)=(A\times B)\cap (U\times V)$$. So given a set, it is in the first basis if and only if it is in the second basis, meaning that the two bases are equal, and thus they generate the same topology.