# If a basis of a topology has order no greater than the weight, is it contained in all other bases?

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?

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What is $\text{w}$ in your question? –  Stijn Mar 8 '11 at 21:12
What is $w(X)$? –  Rasmus Mar 8 '11 at 21:18
I would assume it is the weight of the space, the minimum cardinality of a basis. –  Apollo Mar 8 '11 at 21:37
I think that, as stated, the question is incorrect; you could have the true weight be much smaller than $n$ and the cardinality of $B_1$, and so have a base that is smaller than $B$. For example, if we took $X$ to be a set with $k$ elements endowed with the discrete topology, $B_1=\mathcal{P}(X)$, and $B$ the set of singletons, then set $n=2^k$; we have $w(X)\leq n$, $|B_1|\leq n$, both $B_1$ and $B$ are bases for $X$, but $B_1$ is not contained in $B$. –  Arturo Magidin Mar 8 '11 at 23:48

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.

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So, you are interpreting the inequality signs as equality signs, right? –  Arturo Magidin Mar 8 '11 at 23:56
@Arturo: indeed. The $\leq$s do not make any sense... –  Mariano Suárez-Alvarez Mar 8 '11 at 23:57

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:

• The base elements cover $X$.
• Let $B$ be a base. Suppose $x_1, x_2 \in B$. Let $l = x_1 \cap x_2$. Then for each $x \in l$, there is a base element $x_3$ containing $x$ and contained in $l$.
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Two two points are, essentially, the definition of what being a basis means. It is not surprising that they should come up in proving something about bases! :) –  Mariano Suárez-Alvarez Mar 8 '11 at 23:51