21
$\begingroup$

Let me start by reminding two constructions of topological spaces with such exotic combination of properties:

1) The elements are non-zero integers; base of topology are (infinite) arithmetic progressions with coprime first term and difference.

2) Take $\mathbb{R}^{\infty}\setminus \{0\}$ with product-topology and factorize by the relation $x\sim y \Leftrightarrow x=ty$ for some $t>0$ (infinite-dimensional sphere). Then consider only points with rational coordinates, all but finitely of them vanishing.

The first question is whether are these two examples homeomorphic or somehow related.

The second is an historical one. I've heard that the first example of such space belongs to P. S. Urysohn. What was his example?

$\endgroup$
  • $\begingroup$ Just to be sure I get the example (2): would it be the same starting from $\mathbb{Q}^\infty$ with the product topology; then taking the subspace of sequences with all but finitely many vanishing coord.; then the set of all rays in it with the quotient topology? $\endgroup$ – Pietro Majer Nov 23 '10 at 12:17
  • $\begingroup$ @Pietro: yes, it is the same (except isolated point 0 maybe) $\endgroup$ – Fedor Petrov Nov 23 '10 at 12:56
  • $\begingroup$ Sorry for the late comment. Connected metrizable spaces cannot be countable. This means that infinite dimensional sphere with rational coordinates is not metrizable. How is that possible? Any two points lie on some $S^n$ and we can define the distance between them. Why isn't it a distance function for this topology? $\endgroup$ – Mihail Oct 22 '17 at 13:00
  • 1
    $\begingroup$ @Mihail this defines another topology! $\endgroup$ – Fedor Petrov Oct 22 '17 at 13:42
16
$\begingroup$

First let us fix the terminology.

The space (1) is known in General Topology as the Golomb space. More precisely, the Golomb space $\mathbb G$ is the set $\mathbb N$ of positive integers, endowed with the topology generated by the base consisting of arithmetic progressions $a+b\mathbb N_0$ where $a,b$ are relatively prime natural numbers and $\mathbb N_0=\{0\}\cup\mathbb N$.

Let us call the space (2) the rational projective space and denote it by $\mathbb QP^\infty$.

Both spaces $\mathbb G$ and $\mathbb QP^\infty$ are countable, connected and Hausdorff but they are not homeomorphic. A topological property distinguishing these spaces will be called the oo-regularity.

Definition. A topological space $X$ is called oo-regular if for any non-empty disjoint open sets $U,V\subset X$ the subspace $X\setminus(\bar U\cap\bar V)$ of $X$ is regular.

Theorem.

  1. The rational projective space $\mathbb QP^\infty$ is oo-regular.

  2. The Golomb space $\mathbb G$ is not oo-regular.

Proof. The statement 1 is relatively easy, so is left to the interested reader.

The proof of 2. In the Golomb space $\mathbb G$ consider two basic open sets $U=1+5\mathbb N_0$ and $V=2+5\mathbb N_0$. It can be shown that $\bar U=U\cup 5\mathbb N$ and $\bar V=V\cup 5\mathbb N$, so $\bar U\cap\bar V=5\mathbb N$.

We claim that the subspace $X=\mathbb N\setminus (\bar U\cap\bar V)=\mathbb N\setminus 5\mathbb N$ of the Golomb space is not regular.

Consider the point $x=1$ and its neighborhood $O_x=(1+4\mathbb N)\cap X$ in $X$. Assuming that $X$ is regular, we can find a neighborhood $U_x$ of $x$ in $X$ such that $\bar U_x\cap X\subset O_x$.

We can assume that $U_x$ is of basic form $U_x=1+2^i5^jb\mathbb N_0$ for some $i\ge 2$, $j\ge 1$ and $b\in\mathbb N\setminus(2\mathbb N_0\cup 5\mathbb N_0)$.

Since the numbers $4$, $5^j$, and $b$ are relatively prime, by the Chinese remainder Theorem, the intersection $(1+5^j\mathbb N_0)\cap (2+4\mathbb N_0)\cap b\mathbb N_0$ contains some point $y$. It is clear that $y\in X\setminus O_x$.

We claim that $y$ belongs to the closure of $U_x$ in $X$. We need to check that each basic neighborhood $O_y:=y+c\mathbb N_0$ of $y$ intersects the set $U_x$. Replacing $c$ by $5^jc$, we can assume that $c$ is divisible by $5^j$ and hence $c=5^jc'$ for some $c'\in\mathbb N_0$.

Observe that $O_y\cap U_x=(y+c\mathbb N_0)\cap(1+4^i5^jb\mathbb N_0)\ne\emptyset$ if and only if $y-1\in 4^i5^jb\mathbb N_0-5^jc'\mathbb N_0=5^j(4^ib\mathbb N_0-c'\mathbb N_0)$. The choice of $y\in 1+5^j\mathbb N_0$ guarantees that $y-1=5^jy'$. Since $y\in 2\mathbb N_0\cap b\mathbb N_0$ and $c$ is relatively prime with $y$, the number $c'=c/5^j$ is relatively prime with $4^ib$. So, by the Euclidean Algorithm, there are numbers $u,v\in\mathbb N_0$ such that $y'=4^ibu-c'v$. Then $y-1=5^jy'=5^j(4^ibu-c'v)$ and hence $1+4^i5^ju=y+5^jc'v\in (1+4^i5^jb\mathbb N_0)\cap(y+c\mathbb N_0)=U_x\cap U_y\ne\emptyset$. So, $y\in\bar U_x\setminus O_x$, which contradicts the choice of $U_x$.


Remark. Another well-known example of a countable connected space is the Bing space $\mathbb B$. This is the rational half-plane $\mathbb B=\{(x,y)\in\mathbb Q\times \mathbb Q:y\ge 0\}$ endowed with the topology generated by the base consisting of sets $$U_{\varepsilon}(a,b)= \{(a,b)\}\cup\{(x,0)\in\mathbb B:|x-(a-\sqrt{2}b)|<\varepsilon\}\cup \{(x,0)\in\mathbb B:|x-(a+\sqrt{2}b)|<\varepsilon\}$$ where $(a,b)\in\mathbb B$ and $\varepsilon>0$.

It is easy to see that the Bing space $\mathbb B$ is not oo-regular, so it is not homeomorphic to the rational projective space $\mathbb QP^\infty$.

Problem 1. Is the Bing space homeomorphic to the Golomb space?

Remark. It is clear that the Bing space has many homeomorphisms, distinct from the identity.

So, the answer to Problem 1 would be negative if the answer to the following problem is affirmative.

Problem 2. Is the Golomb space $\mathbb G$ topologically rigid?

Problem 3. Is the Bing space topologically homogeneous?

Since the last two problems are quite interesting I will ask them as separate questions on MathOverFlow.

Added in an edit. Problem 1 has negative solution. The Golomb space and the Bing space are not homeomorphic since

1) For any non-empty open sets $U_1,\dots,U_n$ in the Golomb space (or in the rational projective space) the intersection $\bigcap_{i=1}^n\bar U_i$ is not empty.

2) The Bing space contain three non-empty open sets $U_1,U_2,U_3$ such that $\bigcap_{i=1}^3\bar U_i$ is empty.

Added in a next edit. Problem 2 has a partial affirmative solution: 1 is a fixed point of any homeomorphism of $\mathbb G$. This implies that $\mathbb G$ is not homeomorphic to the Bing space or the rational projective space (which do not have such a fixed point).

$\endgroup$
  • $\begingroup$ I was about to reflexively fix your non-TeX (oo → $\infty$) before I realised that I was missing the point. :-) $\endgroup$ – LSpice Nov 8 '17 at 20:53
  • $\begingroup$ @LSpice I am also hesitating about this new term oo-regularity; oo is chosen to symbolize two disjoint open sets. I also considered something like "rim-regular", "corim-regular", but "rim-regular" should mean having a base of the topology consisting of sets with topologically regular boundaries; corim-regular is too long. So, no better idea than oo-regularity :( By the way, the Bing space is an example of a non-regular rim-regular space. $\endgroup$ – Taras Banakh Nov 8 '17 at 20:58
  • $\begingroup$ For the record, I wasn't criticising it; I think it's on the right side of cuteness. (The example of "fascist functor" in place of "cofree functor" probably shows that there is also a wrong side of cuteness.) I just misread it at first, and fortunately caught myself before 'correcting' it. $\endgroup$ – LSpice Nov 8 '17 at 21:26
11
$\begingroup$

Urysohn's example of a countable connected Hausdorff space with a countable base was published in his last paper «Über die Mächtigkeit der zusammenhängenden Mengen», Math Annalen 94 (1925), 262—295.

Urysohn's original description of the space occupies about 4 pages so I would rather refrain from reproducing it here. This is probably the most complicated construction of its kind. Simpler examples were later obtained by Bing, Hewitt, Stone and others («Countable connected spaces» by Miller contains many relevant references).

There is also a Russian translation of Urysohn's paper in his collected works

  • Урысон П.С. Труды по топологии и другим областям математики [Том 1], ГИТТЛ, 1951, стр. 177-214.
$\endgroup$
  • 1
    $\begingroup$ In example (2) the space is a quotient of a products of real lines, then the phrase "only points with rational coordinates" isnt well definited. May be you considering the point of the product with all but a finite vanish, and have distance $1$ from ${0}$? (i.e . the point on the 1-sphere by rational coordinates) Anyway here there is a (very nice) example of Bing: ams.org/journals/proc/1953-004-03/S0002-9939-1953-0060806-9/… $\endgroup$ – Buschi Sergio Nov 22 '10 at 20:37
  • $\begingroup$ @Sergio: well, we may consider say "points with rational ratios of coordinates", it is maybe more rigorous, but actually does not matter. Thanks for the link! $\endgroup$ – Fedor Petrov Nov 22 '10 at 20:58
  • 2
    $\begingroup$ by the way, how may translation of 34-pages paper consist of 5 pages? $\endgroup$ – Fedor Petrov Nov 22 '10 at 21:04
  • $\begingroup$ @Fedor: I have edited the answer. The example itself is described in Section 3 of the paper, pp. 189-193 in the Russian translation. $\endgroup$ – Andrey Rekalo Nov 22 '10 at 21:24

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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