What is an example of a non-regular, totally path-disconnected Hausdorff space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:48:14Z http://mathoverflow.net/feeds/question/23430 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23430/what-is-an-example-of-a-non-regular-totally-path-disconnected-hausdorff-space What is an example of a non-regular, totally path-disconnected Hausdorff space? David Roberts 2010-05-04T14:14:52Z 2010-05-04T17:38:14Z <p>I need this for a counterexample: the multiplication in the fundamental group $\pi_1(\Sigma X_+)$, when it is equipped with the topology inherited from $\Omega \Sigma X_+$, fails to be continuous for the sort of space in the question, by a result from</p> <blockquote> <p>J. Brazas, The topological fundamental group and hoop earring spaces, 2009, arXiv:0910.3685</p> </blockquote> <p>but the author doesn't supply an explicit example of such a space.</p> http://mathoverflow.net/questions/23430/what-is-an-example-of-a-non-regular-totally-path-disconnected-hausdorff-space/23435#23435 Answer by Nate Eldredge for What is an example of a non-regular, totally path-disconnected Hausdorff space? Nate Eldredge 2010-05-04T14:42:09Z 2010-05-04T14:42:09Z <p>There are several such examples in Steen and Seebach, <em>Counterexamples in Topology</em>. The first one I saw is number 60, the "relatively prime integer topology", consisting of the set $\mathbb{Z}^+$ of positive integers, with a basis of sets of the form {b + na} where (a,b)=1 and n is an integer.</p> http://mathoverflow.net/questions/23430/what-is-an-example-of-a-non-regular-totally-path-disconnected-hausdorff-space/23461#23461 Answer by Jeremy Brazas for What is an example of a non-regular, totally path-disconnected Hausdorff space? Jeremy Brazas 2010-05-04T17:38:14Z 2010-05-04T17:38:14Z <p>One of the easiest examples is the rational numbers with the subspace topology of the real line with the K-topology. Total path disconnectedness is not entirely necessary for multiplication of $\pi_{1}(\Sigma X_{+})$ to fail to be continuous. It just makes the path component space of $X$ equal to $X$, greatly simplifying complications.</p>