Smallest/Minimal bases of a topological space

Question

1. The smallest possible cardinality of a base is called the weight of the topological space. I was wondering if all minimal bases have the same cardinality, and if every base contains a subset whose cardinality is the weight of the topological space?

2. What aspects are common between a (smallest) base of a topology and a base of a vector space, besides the following similarity (open subset <-> vector, union <-> linear combination):

   • every open subset is the union of some members in the base;

   • every vector is the linear combination of some members in the base.

   Note that a base in a vector space is also a base in the linear matroid. Not sure if we can have some nice structure like matroid for a topological space to understand its (smallest) bases.

Thanks and regards!

Answer 1

Let $\langle X,\tau\rangle$ be a topological space. While there are exceptions, in general there is no such thing as a minimal base for $\tau$: if $\mathscr{B}$ is a base for $\tau$, in general some proper subset of $\mathscr{B}$ is also a base for $\tau$.¹ However, among bases for $\tau$ there are bases of minimal cardinality, and that minimal cardinality of a base for $\tau$ is $w(X)$, the weight of $X$.²

Yes, it is true that every base for $\tau$ has a subset of cardinality $w(X)$ that is also a base for $X$. Here’s a proof.

Let $\mathscr{B}$ be a base for $\tau$, and let $\mathscr{W}$ be a base for $\tau$ such that $|\mathscr{W}|=w(X)$. For each $W\in\mathscr{W}$ let $\mathscr{B}(W)=\{B\in\mathscr{B}:B\subseteq W\}$; clearly $\bigcup\mathscr{B}(W)=W$. Let $$\mathscr{W}_W=\{V\in\mathscr{W}:V\subseteq B\text{ for some }B\in\mathscr{B}(W)\}\;;$$ clearly $\bigcup\mathscr{W}_W=\bigcup\mathscr{B}(W)=W$, and $|\mathscr{W}_W|\le|\mathscr{W}|=w(X)$. For each $V\in\mathscr{W}_W$ let $B(V)\in\mathscr{B}(W)$ be such that $V\subseteq B(V)$, and let $$\mathscr{B}_0(W)=\{B(V):V\in\mathscr{W}_W\}\;;$$ $|\mathscr{B}_0(W)|\le|\mathscr{W}_W|\le w(X)$, and $$\bigcup\mathscr{W}_W\subseteq\bigcup\mathscr{B}_0(W)\subseteq\bigcup\mathscr{B}(W)=\bigcup\mathscr{W}_W\;,$$ so $\bigcup\mathscr{B}_0(W)=W$.

Now let $$\mathscr{B}_0=\bigcup_{W\in\mathscr{W}}\mathscr{B}_0(W)\subseteq\mathscr{B}\;.$$

$\mathscr{B}_0$ is the union of $w(X)$ subsets of $\mathscr{B}$, each of which has cardinality at most $w(X)$, so $|\mathscr{B}_0|\le w(X)$. Moreover, each $W\in\mathscr{W}$ is the union of members of $\mathscr{B}_0$, and $\mathscr{W}$ is a base for $\tau$, so $\mathscr{B}_0$ is also a base for $\tau$. Since $w(X)$ is the minimum cardinality of a base for $\tau$, it follows that $|\mathscr{B}_0|=w(X)$.

As Asaf already pointed out in the comments, there is little connection between this notion of base and the notion of basis in vector spaces. What connection there is does not go beyond the fact that both are in some sense small families (of open sets and vectors, respectively) from which the entire topology or vector space can be generated in some natural way.

¹ Specifically, if $\mathscr{B}$ is a partition of $X$, then $\mathscr{B}$ is the unique minimal base for the topology $\tau$ on $X$ generated by $\mathscr{B}$. If $\langle X,\tau\rangle$ is $T_1$, this means that $\tau$ is the discrete topology, and $\mathscr{B}=\big\{\{x\}:x\in X\big\}$. Added 25 July 2023: Here is another exceptional case. Suppose that $\tau$ is a finite topology on $X$. For each $x\in X$ let $B(x)=\bigcap\{U\in\tau:x\in U\}$; clearly the sets $B(x)$ are open, and $B(x)$ is the smallest open set containing $x$; it’s not hard to check that $\mathscr{B}=\{B(x):x\in X\}$ is a minimal base for $\tau$.

² The weight of $X$ is actually defined to be $\omega+\min\{|\mathscr{B}|:\mathscr{B}\text{ is a base for }X\}$, so the weight of any finite topology is $\omega$.