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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?

Anguepa
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2 Answers2

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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?

David Hartley
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  • What is X? The big unknown? – William Elliot Jul 12 '17 at 09:14
  • @WilliamElliot the result of the suggested gluing, I guess. – drhab Jul 12 '17 at 09:15
  • But, the question is about a finite many copies of second countable spaces. Your example is infinity. – Carlos Jiménez Jul 12 '17 at 09:15
  • @drhab the result of the gluing is stated to be not second countable, while $X$ (whatever it is) is stated to be second countable – Alessandro Codenotti Jul 12 '17 at 09:16
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    @CarlosJiménez The partition suggested has $2$ elements. – drhab Jul 12 '17 at 09:16
  • @AlessandroCodenotti I think it must be ${0}$ and $X-{0}$. – drhab Jul 12 '17 at 09:17
  • @drhab You have all the reason. Sorry for my misunderstanding. – Carlos Jiménez Jul 12 '17 at 09:17
  • @drhab ah, I see, that'd make sense, thanks – Alessandro Codenotti Jul 12 '17 at 09:17
  • Sorry. Yes, that should be {0} not X. Corrected now. – David Hartley Jul 12 '17 at 10:26
  • You've described the $\aleph_0$-Hedgehog space (https://en.wikipedia.org/wiki/Hedgehog_space). Thank you. I won't accept your answer right away to see if more examples come up. If you can think of another one feel free to post it. – Anguepa Jul 12 '17 at 11:47
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    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 01:07
  • Really good follow-up Q. – DanielWainfleet Jul 13 '17 at 01:08
  • @DanielWainfleet thank you for pointing that out, although I'm not sure that's what the first example in the A was either. Rather the quotient map of countably many infinite copies of $[0,1]$ glued at $0$. So a neighborhood of $0$ must satisfy that its intersection with each interval $(0,1]$ is open in the euclidean topology. – Anguepa Jul 14 '17 at 09:51
  • Your second example (or scheme of examples) is never Hausdorff. Just wanted to note that if you don't care about Hausdorffness your follow-up question can be answered negatively. – Anguepa Jul 15 '17 at 21:22
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    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
  • @Anguepa Very nice! I think we can get a Hausdorff example by combining your idea with that of the Moore plane. Let $X$ be the Euclidean plane and $A$ the x-axis. $X-A$ has the usual topology. Each $(x,0)$ has a nbhd basis of sets of the form $B - L$ where $B$ is an open box $(x- \epsilon, x + \epsilon) \times (-\epsilon,\epsilon)$ and $L = {(x,y)|y \in \mathbb{R, y \neq 0} }$. $X$ is first countable and Hausdorff but not second countable. $A$ and $X - A$ are second countable. X is not regular, each L is a closed set which cannot be separated from the corresponding $(x,0)$ by neighbourhoods. – David Hartley Jul 16 '17 at 00:35
  • This is perfect. It actually answers negatively a conjecture I was currently working on. We might even be able to make the example regular by letting $L$ be, instead of a line, any triangle on the upper half of the plane, with a vertex at $(x,0)$ and intersecting ${(x,y) : y\neq 0}$, union its reflection on the $x$-axis, minus the point $(x,0)$ . I would suggest editing this into your answer. – Anguepa Jul 16 '17 at 12:04
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    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
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    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
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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 $m+ 2f_m(m)/3 \in U_m$ \ $V$.

DanielWainfleet
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    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 08:54