# A question on second countable space

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?

Thanks.

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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

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 $X$.

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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.

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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