Let $X$ be a space of weight $w(X)=\kappa$. Suppose $q:X\rightarrow Y$ is a quotient map. If $q$ is open, or if $Y$ is compact, then $w(Y)\leq w(X)$. In general it is possible for $w(Y)$ to be larger than $w(X)$ (consider $\mathbb{R}/\mathbb{Z}$).

How much larger than $w(X)$ can $w(Y)$ be?

Really what I'm looking for is a machine to produce countexamples for any fixed infinite cardinal $\kappa$. I'm particularly interested in the case that $X$ is second-countable.

  • $\begingroup$ Take $Y$ to be a single point. I'm not sure I follow your example for where the inequality is reversed: both are second countable. Maybe I'm misunderstanding the definition of weight. $\endgroup$ Jan 7, 2021 at 19:31
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    $\begingroup$ @RobertBell the weight of a space $X$ is the minimal (infinite) cardinality of a base (of open sets) of $X$. I am looking for a quotient map $q:X\rightarrow Y$ such that $w(Y)>w(X)$ by arbitrarily large amounts. $\endgroup$
    – Tyrone
    Jan 7, 2021 at 19:36
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    $\begingroup$ @RobertBell: I originally had your same objection regarding second countability, because I was interpreting $\mathbb{R}/\mathbb{Z}=S^1$. That is, I was thinking of the Lie theoretic quotient. I belive Tyrone means the topological quotient, so $\mathbb{R}/\mathbb{Z}$ is the countable wedge of circles. $\endgroup$ Jan 7, 2021 at 19:42
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    $\begingroup$ Yes! $\mathbb{R}/\mathbb{Z}$ is a countable wedge of circles and is not first-countable at the wedge point, so cannot be second-countable. Writing $\mathbb{R}/\mathbb{N}$ gives more or less the same example but may be less/more notationally confusing. $\endgroup$
    – Tyrone
    Jan 7, 2021 at 19:46

1 Answer 1


Let $T_X$ be the topology on $X.$ Then $|T_X|\le 2^{w(X)}.$

Now $q:X\to Y$ is a continuous surjection. So if $B$ is a base for $Y$ then $A=^{def}\{q^{-1}b: b\in B\}\subset T_X. $

So $|A|\le 2^{w(X)}.$

So $|B|=|\{q[a]:a\in A\}\le |A|\le 2^{w(X)}.$

  • $\begingroup$ The weight of $\Bbb R /\Bbb N$ can easily be shown to be equal to a set-theoretic value $b$ called the Bounding Number. If ZFC is consistent then so are (ZFC $\land b<2^{\aleph_0})$ and (ZFC $\land b=2^{\aleph_0}$) $\endgroup$ Jan 8, 2021 at 1:43

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