Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
Could you make your definition more precise? Is $\sim$ the equivalence relation generated by $x \sim 0$ for $x \in \mathbb{Q}$? Thus you consider the quotient group? –  Martin Brandenburg Oct 18 '10 at 8:18

2 Answers 2

up vote 7 down vote accepted

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 $Y = \mathbb{Q}$.

share|improve this answer

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 homotopy).

share|improve this answer
This not exactly the OP quotient, where only one (dense) $\mathbb{Q}$-coset is crushed to a point $q$. But as any nonempty open set of the quotient contains $q$, Scott's homotopy is indeed continuous. –  BS. Oct 18 '10 at 10:11
I see, I misunderstood the question. –  Ryan Reich Oct 18 '10 at 10:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.