11
$\begingroup$

(This question actually arose in some research on number theory.)

I once learned that any countable dense subspace of any Euclidean space ℝn is homeomorphic to the rationals ℚ.

Now I wonder if something similar is true for the irrationals J := ℝ - ℚ (with the subspace topology from ℝ).

Let c denote the cardinality of the continuum.

I. Is each cartesian power Jn homeomorphic to J ?

Also, how far can this be pushed?

II. Let X be a dense totally disconnected subspace of ℝn such that every neighborhood of each point of X contains c points. Is X homeomorphic to J ?

What about for such subspaces of fairly nice subspaces of ℝn ?

IIa. Let X be any subspace of ℝn as described in II., and let B denote any subspace of ℝn homeomorphic to [the open unit ball in ℝn union any subset of its boundary]. Then is X ∩ B homeomorphic to J ?

And what about greater generality ?

III. Is there a simple set of conditions that describe exactly all spaces (or subspaces of ℝn) that are homeomorphic to J ? What about Jn ? (Perhaps the word homogeneous or metric needs to be included.)

(I found nothing relevant via Google, in MathSciNet, or here on MathOverflow.)

$\endgroup$
0
20
$\begingroup$

The space of irrationals is homeomorphic to the Baire space $\mathbb N^{\mathbb N}$ of functions from $\mathbb N$ to $\mathbb N$. Here $\mathbb N$ gets the discrete topology and the power gets the product topology. In particular, every finite or countably infinite power of the space of irrationals is homeomorphic to the irrationals.

The Baire space is very well studied in descriptive set theory. See the book by Kechris, Classical Descriptive Set Theory.

$\endgroup$
3
  • $\begingroup$ Hi Stefan, do you have any idea to prove that $\omega^{\omega}$ is a homogeneous space? $\endgroup$ – Gabriel Medina Feb 27 at 1:18
  • $\begingroup$ @GabrielMedina Let $A=(\alpha_i)_{i\in\mathbb{N}}$ be any sequence of permutations of $\omega$. Then consider the map $$f_A:\omega^\omega\rightarrow\omega^\omega: (c_i)_{i\in\mathbb{N}}\mapsto(\alpha_i(c_i))_{i\in\mathbb{N}}.$$ It's easy to check that this is an autohomeomorphism of $\omega^\omega$. Now given any two elements of Baire space $v=(v_i)_{i\in\mathbb{N}},w=(w_i)_{i\in\mathbb{N}}$, let $\alpha_i$ be the permutation of the naturals which switches $v_i$ and $w_i$ and fixes every other element of $\mathbb{N}$. Then the associated map $F_{(\alpha_i)_{i\in\mathbb{N}}}$ swaps $v$ and $w$. $\endgroup$ – Noah Schweber Mar 28 at 19:58
  • $\begingroup$ @Noah Schweber Thanks a lot. $\endgroup$ – Gabriel Medina Mar 29 at 16:19
20
$\begingroup$

The space $J$ of irrationals is homeomorphic to the Baire space $N^N$ of sequences of natural numbers (this follows easily from the continued fraction expansion). In particular it is homeomorphic to $J\times J$.

$\endgroup$
9
$\begingroup$

Concerning II and IIa, every subspace of $\mathbb R^n$ that is completely metrizable is in fact a $G_\delta$ set, i.e., a countable intersection of open sets.
If you are not $G_\delta$, you are not homeomorphic to the irrationals.

That completely metrizable subspaces of $\mathbb R^n$ are $G_\delta$ was shown by E. Čech in: On bicompact spaces. Annals of Math. 38 (1937), 823–844.

$\endgroup$
8
$\begingroup$

Regarding III, the Alexandrov-Urysohn Theorem gives sufficient conditions.

Any zero-dimensional, separable, nowhere compact, and completely metrizable space is homeomorphic to J.

$\endgroup$
4
  • 1
    $\begingroup$ Since $J$ is not complete in its given metric, maybe a better way to phrase the result is "completely metrizable". Also, does "nowhere compact" mean that no nonempty open set has compact closure? (If not, what?) $\endgroup$ – Pete L. Clark Jul 31 '10 at 1:09
  • $\begingroup$ @Pete: Yes, it would be nice to rephrase to get necessary and sufficient conditions. Indeed, nowhere compact means what you think it means. $\endgroup$ – Tony Huynh Jul 31 '10 at 1:46
  • $\begingroup$ @TonyHuynh : When you say "necessary and sufficient" are you looking for a theorem that the above list of conditions cannot be weakened? I recently learned of a nice result by Mel Currie ("A Metric Characterization of the Irrationals Using a Group Operation", Topology and Its Applications 21 (1985), 223-236) that if the word "completely" is dropped, then there are uncountably many non-homeomorphic examples. In particular we can take any metric space $(S,d)$ satisfying $\forall x\in S \forall r\in\mathbb{R}^+ \exists ! y\in S : d(x,y) = r$. Currie calls these "spyc spaces". $\endgroup$ – Timothy Chow Mar 21 '19 at 22:58
  • $\begingroup$ @TimothyChow Yes, that's what I meant when I said ''necessary and sufficient''. Thanks for the reference! $\endgroup$ – Tony Huynh Apr 5 '19 at 9:07
2
$\begingroup$

As regards Q (your first remark), it is true that all countable metrisable spaces without isolated points are homeomorphic to Q. If you want to omit metrisable, replace it by T_3 and second countable. One then notes that a dense subset of R^n doesn't have isolated points, and is metrisable.

$\endgroup$
4
  • $\begingroup$ Neat. In particular, for any $p$, the rationals endowed with the $p$-adic metric are homeomorphic to the rationals endowed with the Euclidean metric. (This came up in an offhand way in an answer I gave here some months ago. In a comment, I sketched an argument that is certainly more complicated than this.) $\endgroup$ – Pete L. Clark Jul 31 '10 at 8:44
  • $\begingroup$ I guess your comment about replacing metrizable by regular (Hausdorff) second countable is just for those who don't know Urysohn's Metrization Theorem? $\endgroup$ – Pete L. Clark Jul 31 '10 at 8:45
  • $\begingroup$ It's more for those who want pure topological properties. Metrisable refers to an external object R, and some like this less as a property. I don't really care either way, now that we have a complete characterisation of metrisability since the fifties. $\endgroup$ – Henno Brandsma Aug 1 '10 at 9:09
  • $\begingroup$ It's also amusing that given n, the complement of any countable dense subset of R^n is homeomorphic to the complement of any other such. $\endgroup$ – Daniel Asimov Aug 7 '10 at 18:56
2
$\begingroup$

Several answers point out the following:

The space of irrationals is homeomorphic to the Baire space $\mathbb N^{\mathbb N}$ of functions from $\mathbb N$ to $\mathbb N$.

No one gave an explicit homeomorphism. [PS: It's now pointed out that the answer by Richard Borcherds did so, if very tersely. I skimmed too fast.]

Let $a_1,a_2,a_3,\ldots$ be a sequence of positive integers. Then $$ a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {a_4 +\ddots}}} \in \mathbb R \smallsetminus\mathbb Q. $$ This number cannot be rational since an expansion of a rational number in this way must terminate because of well-ordering of $\mathbb N.$

That gives you the positive irrationals. To see that that is homeomorphic to the space of all irrationals, recall a fact proved by Georg Cantor: any two countable densely linearly ordered sets without endpoints are order-isomorphic to each other. ("Densely" means only that between any two elements there is another (so no knowledge of topology is needed to understand that word).) And an order isomorphism of those two sets of rationals will give you an order isomorphism of those two sets of irrationals.

$\endgroup$
3
  • $\begingroup$ To make everything explicit, rather use an explicit (obvious) way of splitting $\mathbf{N}^\mathbf{N}$ into two open parts homeomorphic to itself? $\endgroup$ – YCor Mar 28 at 19:50
  • $\begingroup$ Note that the continued fraction expansion was evoked in Richard Borcherds' answer, although not written down explicitly. $\endgroup$ – YCor Mar 28 at 19:53
  • $\begingroup$ @YCor : I see. I've now up-voted that one. $\endgroup$ – Michael Hardy Mar 28 at 20:03
1
$\begingroup$

Hello, Dan: Two countable dense subsets of the reals are order isomorphic and this extends to a homeomorphism of the reals. In particular, two countable dense subsets are homeomorphic via the restriction of a homeomorphism of the reals and this yields a homeomorphism of the complements.

$\endgroup$
1
  • $\begingroup$ Hello, Ethan. Yes, indeed -- as you may recall, I proved that (as well as an n-dimensional version) in a class of yours on PL topology around 1970. $\endgroup$ – Daniel Asimov Aug 10 '10 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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