# If a topological space can be partitioned into finitely many second countable subspaces, is it second countable? [Collecting examples]

A topological space is second countable if its topology has a countable basis.

Let $(X,\tau)$ be a topological space.

Suppose there exists $S_1,\ldots, S_n$ is a finite partition of $X$ such that, for each $1\leq i \leq n$, the subspace $(S_i,\tau|_{S_i})$ is second countable. Does it follow that $(X,\tau)$ is second countable?

A negative answer to this question has already been given, but I'd be interested knowing more counterexamples.

David Hartley has suggested this follow-up question:

If $(X,\tau)$ is first countable and it can be partitioned into finitely many second countable subspaces, is $(X,\tau)$ second countable?

Let X be the result of glueing countably infinitely many copies of [0,1] together at 0. It is not second countable, it even fails to be first countable at 0. But {0} and X - {0} are each second countable.

(ETA) Suppose Y is an uncountable, second countable, T1 space and Z a countable, discrete space, with Y and Z disjoint. Let X be their union with the topology with a basis comprising the open sets of Y and sets of the form C∪{z} with z∈Z and C a cofinite subset of Y. X is not first countable as no point of Z has a countable neighbourhood basis in X, but it splits into Y and Z which are both second countable.

Both examples depend on X being not first countable, which suggests the follow-up question: If a first countable space can be partitioned into two second countable subspaces, is it second countable?

• @CarlosJiménez The partition suggested has $2$ elements. – drhab Jul 12 '17 at 9:16
• To the O.P.: It's a good A but it's not a Hedgehog space. Hedgehog spaces are metrizable. In the A, take a countable infinity of copies of [$0,1]$ glued together at $0$ with a topology such that the intersection of a nbhd of $0$ with each copy of $[0,1]$ is co-finite in that copy. – DanielWainfleet Jul 13 '17 at 1:07
• i.e. consider $[0,1] \cup [2,3]$, let $\tau_e$ be the euclidean topology restricted to $[2,3]$ and consider the basis for a topology $\tau_e \cup \{ ( (x-\varepsilon, x+\varepsilon) \cap [0,1] ) \cup ( (x+2-\varepsilon, x+2+\varepsilon)\cap [2,3] \setminus \{x+2\} ) : \varepsilon>0, x\in [0,1] \}$. Let $\tau$ be the topology generated by this basis, then $\tau$ is first countable and not second countable but $\tau|_{[0,1]}$ and $\tau|_{[2,3]}$ are both second countable. – Anguepa Jul 15 '17 at 21:28
• Yes, that's a good improvement. Another variation would be to use pairs of open discs, radius $\epsilon$, centres $\{(x \pm \epsilon, 0) \}$ plus $(x,0)$. Still regular but slightly coarser than your topology. (I like it because of the similarity to the Moore plane (whole plane variety); just twist the nbhds through $90°$.) Are either of these topologys normal? – David Hartley Jul 16 '17 at 16:25
• Answering the question at the end of my previous comment, yes, they are normal. Indeed, any space which is the union of finitely many - or even countably many - second countable subspaces is Lindeloef, and so if regular Hausdorff it will be paracompact and normal. – David Hartley Jul 21 '17 at 11:18

In General Topology by R.Engelking this example is given to show that the image $X_{/E}$ of a quotient mapping $f:X\to X_{/E}$ with closed equivalence classes can fail to be $1$st-countable even when $X$ is $2$nd-countable:

Let $X=\mathbb R$ have the usual topology. Take $p\not \in X.$ For $x,y \in \mathbb R$ let $xEy$ iff $(x=y\lor \{x,y\}\subset Z).$ Let $Y=(X$ \ $\mathbb Z) \cup \{p\}$. The quotient map $f:X\to Y$, where $f(x)=x$ if $x\not \in Z$ and $f(x)=p$ if $x\in \mathbb Z,$ is called the identification of $\mathbb Z$ to a point.

The sub-space topology on $S_1= Y$ \ $\{p\} =\mathbb R$ \ $\mathbb Z$ is just the usual topology on $\mathbb R$ \ $\mathbb Z,$ which is $2$nd-countable. And the sub-space $S_2=\{p\}$ is (trivially) $2$nd-countable .

To show that the character of $p$ in $Y$ is uncountable, let $\{U_m:m\in \mathbb Z\}$ be a family of nbhds of $p.$ For each $m$ take $f_m:\mathbb Z\to (0,1/2]$ such that $$U_m\supset \{p\}\cup \{(-f_m(n)+n,f_m(n)+n)\;):n\in \mathbb Z\} \;\backslash \;\mathbb Z.$$ Let $g(n)= f_n(n)/3$ for each $n\in \mathbb Z.$ Then $$V=\{p\}\cup \{(-g(n)+n,g(n)+n)\;):n\in \mathbb Z\}\; \backslash \; \mathbb Z$$ is a nbhd of $p.$ And $U_m\not \subset V$ for any $m\in \mathbb Z$ because $m+ 2f_m(m)/3 \in U_m$ \ $V$.

• This is the same as glueing countably infinitely many copies of [0,1] together at 0 and 1 (every 0 to every 1); a wedge sum of circles. The first answer is a wedge sum of closed intervals. Any wedge sum of countably many copies of a second countable space will work, as long as the base point does not have a minimum neighbourhood. – David Hartley Jul 13 '17 at 8:54