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  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!

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Don't let the word "basis" confuse you. Much like "normal distribution" and "normal function" (in set theory) have nothing to do with one another, the word "basis" is just a common word, like "normal" or "regular". –  Asaf Karagila Dec 31 '12 at 23:52
    
Thanks @AsafKaragila! Happy new year! –  Tim Dec 31 '12 at 23:57

1 Answer 1

up vote 3 down vote accepted

Let $\langle X,\tau\rangle$ be a topological space. 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 $\tau$. 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 $X$.

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.

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+1 Thanks, @Brain! First let me wish you a happy new year! You have shown that every base admits a subset which is a smallest base. A question would be: for every two bases with different cardinalities, will it be possible to move an open subset away from the bigger base (with the aid of the smaller base), so that it remains a base but only smaller? The motivation of the question is that I suspect the collection of all bases of a topology has some nice structure, which is just opposite to a matroid. –  Tim Jan 1 '13 at 0:36
    
@Tim: Let $\mathscr{B}$ be any base for $X$; I showed that there is a base $\mathscr{B}_0\subseteq\mathscr{B}$ of cardinality $w(X)$, so if $|mathscr{B}|>w(X)$, you can remove all of $\mathscr{B}\setminus\mathscr{B}_0$ and still have a base. (Happy New Year!) –  Brian M. Scott Jan 1 '13 at 0:47

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