Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A family $\mathcal U$ of subsets of a space $X$ is called k-in-countable if every set $A \subset X$ with $|A|=k$ is contained in at most countably many elements of $\mathcal U$.

If $X$ is a separable space with a k-in-countable base, then is $X$ second countable?


share|cite|improve this question
Is $k$ any cardinal? – nigelvr Oct 31 '13 at 2:41
@nigelvr: Finite. If $k=1$ it’s just a point-countable base. – Brian M. Scott Oct 31 '13 at 2:43
up vote 2 down vote accepted

The answer is yes for $T_1$ spaces.

Let $\mathscr{B}$ be a $k$-in-countable base, and let $D$ be a countable dense subset of $X$. Let $D_0$ be the set of non-isolated points of $D$ and let $x\in D_0$. For each $F\in [D\setminus\{x\}]^{k-1}$ let $$\mathscr{B}(x,F)=\big\{B\in\mathscr{B}:\{x\}\cup F\subseteq B\big\}\;.$$ (For any set $S$ and cardinal $\kappa$, $[S]^\kappa$ is the set of subsets of $S$ of cardinality $\kappa$.) Let $$\mathscr{B}(x)=\bigcup\big\{\mathscr{B}(x,F):F\in[D\setminus\{x\}]^{k-1}\big\}\;.$$ Then $\mathscr{B}(x)$ is countable, and since $X$ is $T_1$, $\mathscr{B}(x)=\{B\in\mathscr{B}:x\in B\}$. Thus,

$$\big\{\{x\}:x\in D\setminus D_0\big\}\cup\bigcup_{x\in D_0}\mathscr{B}(x)$$

is a countable base for $X$.

share|cite|improve this answer
Why does $X$ being $T_1$ imply ${\scr B}(x)=\{B\in{\scr B}:x\in B\}$? Also, was something wrong with my answer? – Mario Carneiro Oct 31 '13 at 23:54
Also, your proof generalizes to $R_0$ spaces: let $I(x)=\bigcap\{B\in{\scr B}:x\in B\}$ (the set of points topologically indistinguishable from $x$), and let $D_0=\{x\in D:I(x)\text{ is not open}\}$. Then $\{I(x):x\in D\setminus D_0\}\cup\bigcup_{x\in D_0}\mathscr{B}(x)$ is a countable base. – Mario Carneiro Nov 1 '13 at 0:15
@Mario: I don’t really care about $R_0$ spaces; for me the default assumption is $T_1$, and $R_0$ is an exotic property. The fact that $X$ is $T_1$ means that every open nbhd of $x$ contains some $F\in[D\setminus\{x\}]^{k-1}$ (since in fact it contains infinitely many points of $D$). Your answer isn’t wrong: it just doesn’t address the situation in interesting spaces. – Brian M. Scott Nov 1 '13 at 7:54

Note first that if $\cal U$ is $k$-in-countable, then it is $n$-in-countable for any $n\ge k$, because for any set $|A|=n$, picking a set $B\subseteq A$ of cardinality $k$, this set is contained in countably many elements of $\cal U$, and since every set that contains $A$ contains $B$, there are countably many elements of $\cal U$ that contain $A$.

Considering first a point-countable base ($1$-in-countable), we can prove the theorem:

Let $\{x_n\}$ be a countable dense set in $X$ and define $\{U_{nm}\}_{m\in\Bbb N}$ as the set of all elements of $\cal U$ that contain $x_n$. I claim that $\{U_{nm}\}_{n,m\in\Bbb N}=\cal U$, so that the basis is in fact countable. Given $U\in\cal U$, there is some $n$ such that $x_n\in U$ because $\{x_n\}$ is dense. But then there is an $m$ such that $U=U_{nm}$, because by construction every element of $\cal U$ containing $x_n$ is in $\{U_{nm}\}_{m\in\Bbb N}$. Thus $X$ is second-countable.

For $k\ge2$, the theorem is false. Let $X=\Bbb R$, with the "almost discrete" topology: the basis is ${\cal U}=\{\{0,x\}:x\in \Bbb R\}$ (this is basically a giant star graph with $0$ at the center: every open set contains $0$). Then $\{0\}$ is a countable dense subset, so $X$ is separable, and every set of two elements either has two nonzero elements (so that no element of $\cal U$ contains it) or else is an element of $\cal U$. Thus $\cal U$ is $2$-in-countable, and by extension $k$-in-countable. But $\cal U$ is uncountable, and clearly no lesser basis will do, so $X$ is not second-countable.

share|cite|improve this answer
A point-countable base is not the same as first countable. – Paul Oct 31 '13 at 3:01
@Paul Beat me to it (was going to remark this myself). Doesn't matter for the proof though. There is implication one way; do you have a counterexample for the converse? – Mario Carneiro Oct 31 '13 at 3:02
Yes. A space with a point-countable base implies it is first countable; however the converse is not true. – Paul Oct 31 '13 at 3:12
However, I cannot find a counterexample now. Sorry. – Paul Oct 31 '13 at 3:13
You’ve completely ignored the separability condition. – Brian M. Scott Oct 31 '13 at 3:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.