Let $X\_{p} = \mathbb{Q}\_{p} / \sim $, where $\sim$ is defined by:


$x\sim 0 \Leftrightarrow x\in \mathbb{Q}$


$X\_{p}$ 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.


Is $X\_p$ contractible?



Is this quotient space of Q_p contractible?
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 $Y = \mathbb{Q}$.


