subspace topology

intersected topology

Let

X

be a topological space with toplogy

T

. If

Y

is a subset of

X

, then we define the the toplogy

T

intersected with

Y

as the set

T_{\cap Y} := {Y \cap U : U \in T}

intersected topology is a topology

The set

T_{\cap Y}

is a topology on

Y

We'll show that $\emptyset$ and $Y$ are elements of $T_{\cap Y}$ , this can be seen as since $\emptyset \in T$ , then $Y \cap \emptyset = \emptyset \in T_{\cap Y}$ 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 $⋃_{α \in I} (U_{α} \cap Y) = (⋃_{α \in I} U_{α}) \cap Y$ . 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

T_{\cap Y}

and call it the subspace topology and that

Y

is a subspace of

X

basis for the subspace topology

Suppose that

B

is a basis for the topology of

X

, then

M := {Y \cap B : B \in B}

is a basis and it generates

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 $T_{\cap Y}$ , 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 $B$ is a basis for $X$ then we know there is some $B \in B$ 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.