Is this quotient space of Q_p contractible? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:18:36Z http://mathoverflow.net/feeds/question/42580 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42580/is-this-quotient-space-of-q-p-contractible Is this quotient space of Q_p contractible? David Cohen 2010-10-18T04:36:03Z 2010-10-18T09:27:03Z <p>Let $X_{p} = \mathbb{Q}_{p} / \sim$, where $\sim$ is defined by:</p> <p>$x\sim 0 \Leftrightarrow x\in \mathbb{Q}$</p> <p>$X_{p}$ is path-connected, because (unless I'm making some horrible mistake,) for any $x\in X_{p} \backslash \mathbb{Q}$, we have that $\lbrace x,0\rbrace$ under the subspace topology is path-connected.</p> <p>Is $X_p$ contractible?</p> http://mathoverflow.net/questions/42580/is-this-quotient-space-of-q-p-contractible/42597#42597 Answer by S. Carnahan for Is this quotient space of Q_p contractible? S. Carnahan 2010-10-18T08:34:24Z 2010-10-18T08:34:24Z <p>The answer is yes. Whenever you crunch a dense subspace $Y$ of a topological space $Z$ to a point $q$ in the quotient $Z/Y$, you have the following contracting homotopy: For any $x \in Z/Y$ and any $t \in [0,1]$, set $f(x,0) = x$ and $f(x,t) = q$ for $t > 0$. Here, $q$ is the equivalence class of zero, $Z = \mathbb{Q}_p$, and $Y = \mathbb{Q}$.</p> http://mathoverflow.net/questions/42580/is-this-quotient-space-of-q-p-contractible/42605#42605 Answer by Ryan Reich for Is this quotient space of Q_p contractible? Ryan Reich 2010-10-18T09:27:03Z 2010-10-18T09:27:03Z <p>For any n, we have $\mathbb{Q} + p^n \mathbb{Z}_p = \mathbb{Q}_p$. Thus, every coset of $p^n \mathbb{Z}_p$ has a rational representative. Thus, the only nonempty open set in $\mathbb{Q}_p$ which is closed under $\mathbb{Q}$-translates is the whole space. Thus, $X_p$ has the indiscrete topology. Thus, as Scott says in his answer, any "homotopy" of the identity map with a constant map is continuous (thus a <em>homotopy</em>).</p>