If $w(X)\leq n$ ($n$ is finite), and if $B_1$ is a base of $X$ such that $|B_1|\leq n$, then for any base $B$ of $X$ we have $B_1$ is contained in $B$.
Can you help me in proving this fact?
If $w(X)\leq n$ ($n$ is finite), and if $B_1$ is a base of $X$ such that $|B_1|\leq n$, then for any base $B$ of $X$ we have $B_1$ is contained in $B$.
Can you help me in proving this fact?
Assuming $w(X)$ is the minimun cardinality of a basis...
If a space $X$ has a finite basis, then it has finitely many open sets, and then every intersection of open sets is an open set. It follows that from this that for each $x\in X$ there is a minimal open set $U_x\subseteq X$ such that $x\in U$, and it is easy to see that every basis must contain the set $\mathcal B=\{U_x:x\in X\}$. As one can check that $\mathcal B$ is actually a basis, it is then the unique basis of cardinal $w(B)$ and it is contained in every other basis.
Let $B_1$ be a base for $X$. Consider another base $B$ of $X$. Suppose $w(X) \leq n$ and $|B_1| \leq n$. Suppose for contradiction that $B \subset B_1$. You can probably use the following facts to come up with a contradiction: