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In an exercise from Munkres-Topology Article 30 the author writes that there is a very familiar space which is NOT first countable but every point is a $G_\delta $ set. What is it?

Though there are answers posted on this site to the above question, I don't find the spaces familiar to what has been taught in the book up to Article-30

I am not able to find examples either. Is any help possible on familiar examples?

Martin Sleziak
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Learnmore
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  • @BrianO; Why is a singleton set in a $T_2$ space $G-\delta$ ? It's true that a singleton is closed in $T_2$ but $G-\delta$ means it is countable intersection of open sets – Learnmore Feb 07 '16 at 05:01
  • Momentary lapse of reason, sorry. Excuses, rant: I really hate the $G_{\delta}, F_{\sigma}, F_{\sigma\delta}, ... $ notation, the terms straddle three languages. Whereas you can actually think with the more modern $\Pi_{\alpha}, \Sigma_{\alpha}$. Oh well. – BrianO Feb 07 '16 at 05:01
  • It is hard to guess what spaces are familiar. Maybe you could post a link to the answers on this site with unfamiliar examples, so we don't duplicate any of those? I'm think of examples like "any countable T$_1$ space which is not first countable" or "box product of infinitely many lines" but I don't know if any of those are familiar. – bof Feb 07 '16 at 05:21
  • @BrianO: $\sigma=$Summe, $\delta=$Durchschnitt. $F$ and $G$ are arbitrary letters. Or not? Which are the other two languages besides German? $F$ may come from French fermé I suppose, but is that a mnemonic or the actual origin of the use of the letter? – ForgotALot Feb 07 '16 at 05:27
  • @ForgotALot No they're not arbitrary, though sometimes it seems that they might as well be. $F$ is from French for fermé (closed), and $G$ is from German gebiet (area, neighborhood). And of course the subscripts are in Greek. – BrianO Feb 07 '16 at 05:56
  • See also http://math.stackexchange.com/a/780741/4280 in response to the same question (partly) – Henno Brandsma Feb 07 '16 at 11:30

3 Answers3

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Another example (perhaps a bit more familiar with readers of Munkres's text) is $\mathbb R^\omega$ with the box topology.*

  • It is not first countable because you can "diagonalise" through any countable collection of open neighbourhoods of a point.

    Given $\mathbf x = ( x_n )_n $ and a collection $\{ U_i : i \in \mathbb N \}$ is open neighborhoods of $\mathbf x$, without loss of generality we may assume that $U_i = \prod_n ( a_n^{(i)} , b_n^{(i)} )$ where $a_n^{(i)} < x_n < b_n^{(i)}$. Taking $c_n = \frac{a_n^{(n)} + x_n }{2}$ and $d_n = \frac{x_n + b_n^{(n)}}{2}$ it follows that $V = \prod_n ( c_n , d_n )$ is an open neighbourhood of $\mathbf x$, but $U_n \not\subseteq V$ for each $n$.

  • It is pretty easy to verify that points are Gδ. (Given $\mathbf x = ( x_n )_n$, set $U_i = \prod_n ( x_n - \frac{1}{i} , x_n + \frac{1}{i} )$ for each $i$, and note that $\bigcap_i U_i = \{ \mathbf x \}$.)

*This space is first explicitly mentioned on p.117 of Munkres's text, and has a separate index entry. Its non-metrizability is shown on p.132 and its disconnectedness is shown on p.151, both before the stated exercise. It is also the subject of several exercises prior to Section 30. To someone going through the text, it should be "familiar".

Taumatawhakatangihangakoauauot
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  • Yes, but how familiar is it? – bof Feb 07 '16 at 05:22
  • How familiar? I guess that depends on you.It's like saying something is obvious. That depends on who you are. BTW the value of the character of the box-product topology on $R^{\omega}$ is undecidable from ZFC. – DanielWainfleet Feb 07 '16 at 05:58
  • @bof I took being "familiar" to basically mean "is mentioned in Munkres's text" (and the exercise doesn't use "very familiar", just "familiar"). After a cursory flip through the text up to section 30 this space was the first I noticed that satisfies the requirements (admittedly, I could have missed one). It is also mentioned in several places in the text, both in worked out examples worked and in exercises. To someone going through Munkres's text, I think that it should be familiar by the time the stated exercise is encountered. – Taumatawhakatangihangakoauauot Feb 07 '16 at 06:15
  • Good. I'm not familiar with Munkres's text. Just out of curiosity, is there no example of a non-first-countable topology on $\mathbb N$ in the first 30 sections? – bof Feb 07 '16 at 06:34
  • Yeah ! I need to understand this topology quite well .Thank you very much – Learnmore Feb 07 '16 at 07:43
  • I have a question. $\mathbb R^\omega$ with product topology is first countable by Thm30.2 and any box topology is finer than a product topology so that $\mathbb R^\omega$ with box topology is first countable ; what's wrong with this argument? – Focus Aug 14 '18 at 11:04
  • @izimath Simply that "first-countability" is not inherited by finer topologies. Otherwise since the trivial topology is first-countable, and every topology is finer than the trivial topology, every topological space would be first-countable. (Neither is first-countability inherited by coarser topologies: the discrete topology is first-countable.) – Taumatawhakatangihangakoauauot Aug 14 '18 at 11:07
  • Thanks much. I thought we can use the same basis in a finer topology which is not true. – Focus Aug 14 '18 at 11:12
  • Nice example! +1. Was thinking on this counterexample for hours but was not able to come up with smth useful :) – RFZ Dec 02 '18 at 22:05
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Hint: Glue uncountably many copies of $\mathbb R$ together at the origin

William Stagner
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A very late answer: $\cal{D}(\Omega)$, the space of $C^\infty$ functions with compact support in an open $\Omega\subset \mathbb{R}^n$ with the distribution topology, seems to me (but what do I know) the most familiar space that isn't first countable. Proof that it's not first countable: Rudin Functional Analysis 6.9 tells us it's not metrizable; Rudin 1.9(b) tells us that, among topological vector spaces, a space is metrizable iff it has a countable local base.

Now, we are also told at 6.3 that sets of the type $\{\varphi\in{\cal D}(\Omega):|\varphi(x_m)|<c_m, m=1,2,3\ldots\}$ are open in $\cal{D}(\Omega)$. Enumerate a countable dense set in $\Omega$ as $\{x_m\}$, and let $$A_m = \{\varphi\in{\cal D}(\Omega):|\varphi(x_i)|<1/m{\rm\,\,for\,}i=1,2,\ldots,m\}$$ We should have $\{0\}=\bigcap A_m$ since any member of the latter intersection would be $C^\infty$ and $0$ on a dense subset of $\Omega$. Because ${\cal D}(\Omega)$ is a topological vector space, for any $\varphi\in{\cal D}(\Omega)$, we also have $\{\varphi\}=\bigcap (A_m+\varphi)$. Every point is thus a $G_\delta$.

ForgotALot
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