Bases of topologies of finite weight

Question

The weight of a topology $τ$ is defined as $\min\{\text{card}(B) : B\text{ is a base of }τ\}$.

Can someone give me an hint to proof of this:

Given a topology $τ$ of finite weight $n$ and a base $B$ of $τ$ such that $\text{card}(B) = n$, prove that any base of $τ$ contains $B$.

Answer 1

If $B'$ is another base of $\tau$, then every element $b\in B$ is a union of elements of $B'$, which are in turn unions of elements of $B$.

If $B'$ does not contain $B$, this means that $b$ is a union of elements of $B$ other that $b$ itself. Therefore...

Answer 2

If there is a finite base for $(X,\mathcal{T})$, the topology $\mathcal{T}$ is finite too (maximum of $2^k$ unions from $k$ subsets etc.).

Then define $\mathcal{B}_m = \{B_x: x \in X\}$ where $B_x = \bigcap \{O \in \mathcal{T}: x \in O\}$ is open as a finite intersection of open sets for each $x$. (There will be cases where $B_x = B_y$ for distinct points, if $|X|$ is infinite, e.g. or larger than $w(X)$, but this is immaterial in sets, we just mention the same set more than once. $\{\emptyset, \emptyset, \emptyset\} = \{\emptyset\}$ etc.)

If then $\mathcal{B}$ is any open base, and then consider any $B_x$. This must be a union of base elements so for some $B \in \mathcal{B}$ we have $x \in B \subseteq B_x$. But then $B$ is an open set containing $x$, so by definition of $B_x$: $B_x \subseteq B$ and so we have $B = B_x$. So all $B_x$ are in $\mathcal{B}$ and hence $\mathcal{B}_m \subseteq \mathcal{B}$ for any base.

It follows that the unique minimal base equals $\mathcal{B}_m$ and its the only base of size $n$ if $w(X) = n$.