Topological spaces that resemble the space of irrationals - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T10:53:42Z http://mathoverflow.net/feeds/question/33947 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals Topological spaces that resemble the space of irrationals Daniel Asimov 2010-07-30T22:11:20Z 2010-08-09T17:53:29Z <p>(This question actually arose in some research on number theory.)</p> <p>I once learned that any countable dense subspace of any Euclidean space ℝ<sup>n</sup> is homeomorphic to the rationals ℚ.</p> <p>Now I wonder if something similar is true for the irrationals <strong>J</strong> := ℝ - ℚ (with the subspace topology from ℝ).</p> <p>Let <strong>c</strong> denote the cardinality of the continuum.</p> <blockquote> <p><strong>I</strong>. Is each cartesian power <strong>J</strong><sup>n</sup> homeomorphic to <strong>J</strong> ?</p> </blockquote> <p>Also, how far can this be pushed?</p> <blockquote> <p><strong>II</strong>. Let X be a dense totally disconnected subspace of ℝ<sup>n</sup> such that every neighborhood of each point of X contains <strong>c</strong> points. Is X homeomorphic to <strong>J</strong> ?</p> </blockquote> <p>What about for such subspaces of fairly nice subspaces of ℝ<sup>n</sup> ?</p> <blockquote> <p><strong>IIa</strong>. Let X be any subspace of ℝ<sup>n</sup> as described in <strong>II</strong>., and let B denote any subspace of ℝ<sup>n</sup> homeomorphic to [the open unit ball in ℝ<sup>n</sup> union any subset of its boundary]. Then is X ∩ B homeomorphic to <strong>J</strong> ?</p> </blockquote> <p>And what about greater generality ?</p> <blockquote> <p><strong>III</strong>. Is there a simple set of conditions that describe exactly all spaces (or subspaces of ℝ<sup>n</sup>) that are homeomorphic to <strong>J</strong> ? What about <strong>J</strong><sup>n</sup> ? (Perhaps the word <em>homogeneous</em> or <em>metric</em> needs to be included.)</p> </blockquote> <p>(I found nothing relevant via Google, in MathSciNet, or here on MathOverflow.)</p> http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/33954#33954 Answer by Stefan Geschke for Topological spaces that resemble the space of irrationals Stefan Geschke 2010-07-30T22:57:03Z 2010-07-30T22:57:03Z <p>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.</p> <p>The Baire space is very well studied in descriptive set theory. See the book by Kechris, Classical Descriptive Set Theory.</p> http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/33955#33955 Answer by Richard Borcherds for Topological spaces that resemble the space of irrationals Richard Borcherds 2010-07-30T22:57:08Z 2010-07-30T22:57:08Z <p>The space $J$ of irrationals is homeomorphic to the <a href="http://en.wikipedia.org/wiki/Baire_space_(set_theory)" rel="nofollow">Baire space</a> $N^N$ of sequences of natural numbers (this follows easily from the continued fraction expansion). In particular it is homeomorphic to $J\times J$.</p> http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/33956#33956 Answer by Tony Huynh for Topological spaces that resemble the space of irrationals Tony Huynh 2010-07-30T23:09:03Z 2010-07-30T23:09:03Z <p>Regarding III, the Alexandrov-Urysohn Theorem gives sufficient conditions. </p> <p>Any zero-dimensional, separable, nowhere compact, and complete metric space is homeomorphic to J. </p> http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/33959#33959 Answer by Stefan Geschke for Topological spaces that resemble the space of irrationals Stefan Geschke 2010-07-30T23:40:07Z 2010-07-31T05:37:54Z <p>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.<br> If you are not $G_\delta$, you are not homeomorphic to the irrationals.</p> <p>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.</p> http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/33982#33982 Answer by Henno Brandsma for Topological spaces that resemble the space of irrationals Henno Brandsma 2010-07-31T06:17:02Z 2010-07-31T06:17:02Z <p>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.</p> http://mathoverflow.net/questions/33947/topological-spaces-that-resemble-the-space-of-irrationals/35027#35027 Answer by ethan akin for Topological spaces that resemble the space of irrationals ethan akin 2010-08-09T17:53:29Z 2010-08-09T17:53:29Z <p>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. </p>