My book states that

if $X$ is compact, then $w(X) \leq |X|$.

This leads me to wonder, is there a nice example of a (non-compact) topological space where $w(x) > |X|$ holds?

The weight of a topological space is the smallest cardinality of a basis of the topology.

  • 2
    $\begingroup$ You want to remind us the definition of weight ? $\endgroup$ Nov 13, 2017 at 18:55
  • $\begingroup$ @ReneSchipperus : The weight of a topological space is the smallest cardinality of a basis of the topology. $\endgroup$ Nov 13, 2017 at 18:59
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    $\begingroup$ But is it polite to question a topology about its weight $\endgroup$
    – user541686
    Nov 13, 2017 at 22:09
  • $\begingroup$ here I give a proof the first fact, if you're interested. $\endgroup$ Nov 13, 2017 at 22:31

2 Answers 2


The maximal weight of a space $X$ is $2^{|X|}$ and in this answer I give some examples (with proofs) of spaces that realise this maximum. E.g. a countable dense subset of $\{0,1\}^{\mathbb{R}}$ in the product topology has weight $|\mathbb{R}|$.

The bound $w(X) \le |X|$ for compact spaces follows from the equality $nw(X) =w(X)$ for (locally) compact spaces (and it also holds for metrisable spaces and orderable spaces), so many standard examples don’t work.

  • $\begingroup$ +1 - this is a better answer than mine, since it provides examples of maximal weight. (I'll leave mine up anyways though.) $\endgroup$ Nov 13, 2017 at 19:32
  • $\begingroup$ Interesting. So even though the weight of the discrete topology on $X$ is $|X|$, and the weight of the anti-discrete topology on $|X|$ is 1 (I think), there are topologies that are "in between" these, yet have a larger weight than both? $\endgroup$
    – theQman
    Nov 13, 2017 at 20:11
  • $\begingroup$ @theQman by convention the weight of the anti-discrete topology is $\aleph_0$ (weight is infinite by convention). But yes, the max lies in between. $\endgroup$ Nov 13, 2017 at 20:20
  • $\begingroup$ 1) Why is weight infinite by convention? What if the space is finite? And how is this convention implied by the definition " minimum cardinality of a basis"? 2) What do you mean by "the max lies in between"? $\endgroup$
    – theQman
    Nov 13, 2017 at 20:28
  • $\begingroup$ @theQman That’s the way it’s defined (so some counting arguments go through) in “cardinal functions in topology” and its sequel, which form a basis for the theory. Finite spaces have weight $\aleph_0$ as well, but in papers that only deal with finite spaces sometimes finite cardinals are used. In that case the maximal weight is also at most $|X|$ as well, of course. Max in between refers to that the maximal weight lies between indiscrete and discrete. $\endgroup$ Nov 13, 2017 at 20:37

I believe the club-generated topology on $\omega_1$ is an example (a set is open in this topology iff it is a union of sets which are unbounded and closed in the order topology - this does indeed form a topology, since the intersection of two clubs is a club). Given any $\omega_1$-sequence of open sets $C_\eta$, I can build a club (hence open) set which doesn't contain any of them, as follows:

  • We begin at stage $0$ with $D_0=\emptyset$.

  • At stage $\alpha+1$, pick some $\beta\in C_\alpha$ greater than $\sup(D_\alpha)$, and let $D_{\alpha+1}=D_\alpha\cup\{\beta\}$ (so since we keep adding bigger ordinals, we'll keep $\beta$ out of the club we're building). This is possible since $C_\alpha$, being a union of clubs, is unbounded.

  • At stage $\lambda$ limit, let $D_\lambda=(\bigcup_{\alpha<\lambda} D_\alpha)\cup\{\sup(\bigcup_{\alpha<\lambda} D_\alpha)\}$.

Let $D=\bigcup_{\eta<\omega_1} D_\eta$. $D$ is club, hence open, and doesn't contain any of the $C_\eta$s. So $\{C_\eta:\eta<\omega_1\}$ was not a base for the club-generated topology on $\omega_1$.

It's worth noting that this argument does use a bit of choice: in the successor steps of our construction, we're assuming that there are in fact ordinals bigger than $\sup(D_\alpha)$. This uses the fact that $\omega_1$ is not a countable union of countable sets, which surprisingly is not provable in ZF alone. Interestingly, ZF does prove that $\omega_2$ is not a countable union of countable sets. So that's nice.

EDIT: As Asaf points out, if $\omega_1$ has countable cofinality then "club" doesn't really make sense (every singleton is now the intersection of two clubs, so the topology is discrete).

  • $\begingroup$ What would be the club topology on $\omega_2$ if it had cofinality $\omega$? You need some choice, I guess, to even make sense of the idea of a club topology. $\endgroup$
    – Asaf Karagila
    Nov 13, 2017 at 19:38
  • $\begingroup$ @AsafKaragila Oh yeah, good point. (I think you mean $\omega_1$ ...) $\endgroup$ Nov 13, 2017 at 19:57
  • $\begingroup$ (No, I meant what I meant, but also $\omega_1$, yes.) $\endgroup$
    – Asaf Karagila
    Nov 13, 2017 at 19:58
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    $\begingroup$ Is it known what the weight of this topology on $\omega_1$ is? $\aleph_2$ provable in ZFC? $\endgroup$ Nov 13, 2017 at 20:24
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    $\begingroup$ @Henno: Without being too fussy about it, I'd guess that (or its negation) would be equivalent to some saturation or other combinatorial property of the non-stationary ideal, and would therefore be equiconsistent with some large cardinal properties. But that's a very nice question! $\endgroup$
    – Asaf Karagila
    Nov 13, 2017 at 20:51

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