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

\( T_{\cap Y} := \left \lbrace Y \cap U : U \in \mathcal{T} \right \rbrace \)

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 \( \bigcup_{\alpha \in I} \left ( U_{\alpha} \cap Y \right ) = \left ( \bigcup_{\alpha \in I} U_{\alpha} \right ) \cap Y \). Similarly we can see that it's closed under finite intersections as \( \left ( U_{1} \cap Y \right ) \cap \ldots \cap \left ( U_{n} \cap Y \right ) = \left ( U_{1} \cap \ldots \cap U_{n} \right ) \cap Y \).

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 \( M := \left \lbrace Y \cap B : B \in \mathcal{B} \right \rbrace \) 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 \( \left ( A \cap U \right ) \times \left ( B \cap V \right ) \) 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 \( \left ( A \times B \right ) \cap \left ( U \times V \right ) \).

Now to connect the two recall thqt \( \left ( A \cap U \right ) \times \left ( B \cap V \right ) = \left ( A \times B \right ) \cap \left ( U \times V \right ) \). 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.