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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?

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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
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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 $Y = \mathbb{Q}$.

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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).

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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
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