If $\mathcal{T}$ is a topology and $B$ is a basis, does $\mathcal{T_1} \subset \mathcal{T_2} \implies \mathcal{B_1} \subset \mathcal{B_2}$?

Question

If $\mathcal{T}$ is a topology and $B$ is a basis, does $\mathcal{T_1} \subset \mathcal{T_2} \implies \mathcal{B_1} \subset \mathcal{B_2}$ ?

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Definition (Topology Generated By a Basis): If $B$ is a basis for a topology $\mathcal{T}$ con a set $X$, then $\mathcal{T} = \{ U \ | \ \forall x \in U, \exists B \in \mathcal{B} \text{ such that } x \in B \subset U$} or equivalently $\mathcal{T} = \{ U = \bigcup_{B \in \mathcal{K}} B \ | \ \mathcal{K} \subset \mathcal{B}\}$

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It is easy to prove that $\mathcal{B_1} \subset \mathcal{B_2} \implies \mathcal{T_1} \subset \mathcal{T_2}$, but proving or disproving the converse does not seem to be as easy.

This is what I did so far:

Let's assume that $\mathcal{T_1} \subset \mathcal{T_2} \not\Rightarrow \mathcal{B_1} \subset \mathcal{B_2}$. Then $\exists$ at least one $B \in \mathcal{B_1}$ such that $B \in \mathcal{B_1} \implies B \not\in \mathcal{B_2}$.

But then I got stuck, if we can show that there is a $U \in \mathcal{T_1}$ that cannot be expressed as a union of the basis elements in $\mathcal{B_2}$, then we could arrive at the needed contradiction, but the fact that $B \not \in \mathcal{B_2}$ does not (directly) prove the existence of such a $U$ since there could be other basis elements $B' \subset U$ in $\mathcal{B_2}$ whose union equals $U$.

Any hints on how to prove this?

Answer 1

$\mathcal{T}_2$ is a basis in itself so obviously this does not hold.

Explicit counter example: Let $X = \mathbb R$, $\mathcal{T}_1$ the Euclidean topology, $\mathcal{T}_2$ the discrete topology. Let $\mathcal{B}_1$ be the open intervals and $\mathcal B_2$ the singletons.

Answer 2

No. Even $\mathcal T_1=\mathcal T_2$ does not imply this.

The collection of balls of radius $1/2k$ forms a basis for the standard topology on $\mathbb R^n$, as does the collection of balls of radius $1/(2k+1)$. These are disjoint.