Topological spaces that resemble the space of irrationals - MathOverflow

Topological spaces that resemble the space of irrationals

2010-07-30T22:11:20Z

(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.)

Answer by Stefan Geschke for Topological spaces that resemble the space of irrationals

2010-07-30T22:57:03Z

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.

Answer by Richard Borcherds for Topological spaces that resemble the space of irrationals

2010-07-30T22:57:08Z

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 $J\times J$.

Answer by Tony Huynh for Topological spaces that resemble the space of irrationals

2010-07-30T23:09:03Z

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

Any zero-dimensional, separable, nowhere compact, and complete metric space is homeomorphic to J.

Answer by Stefan Geschke for Topological spaces that resemble the space of irrationals

2010-07-30T23:40:07Z

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.

Answer by Henno Brandsma for Topological spaces that resemble the space of irrationals

2010-07-31T06:17:02Z

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.

Answer by ethan akin for Topological spaces that resemble the space of irrationals

2010-08-09T17:53:29Z

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.