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Let $(M,d)$ be a metric space, where $A\subset M$. If $x$ is an accumulation point of $A$, by definition $\forall r > 0,\; B(x,r)\setminus \{x\}\cap A \neq \emptyset$, so $\forall r > 0,\; B(x,r)\cap A \neq \emptyset$. Any hint?

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What is the connection between $\mathbb{Q}$ and $M$? – copper.hat Dec 4 '12 at 19:36
Corrected, I thought I were working on $\mathbb{R}$ because of another proof. – user50554 Dec 4 '12 at 19:38
up vote 2 down vote accepted

Suppose $r>0$ and suppose $B(x,r)\setminus\{x\}\cap A = \{x_1,\dots,x_n\}$ (It is nonempty, because $x$ is an accumulation point). Choose $r'=\min\left\{\frac{d(x,x_i)}{2}\mid i=1\dots n\right\}$. Then there exists $x' \in B(x,r')\setminus\{x\}\cap A$. But $B(x,r')\subset B(x,r)$, so $x'=x_i$ for one $i$. Thats a contradiction.

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Here is a hint:

Choose $r_1=\frac{1}{2^1}$, and $x_1 \in B(x,r_1)$. Now let $r_2 = \min(\frac{1}{2}d(x,x_1), \frac{1}{2^2})$, and pick $x_2 \in B(x,r_2)$. Continue the process. What can you say about $r_k$ and hence about $x_k$?

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As long as you keep cutting the minimum distance in half, you don’t actually need to minimize against $2^{-k}$ as well. – Brian M. Scott Dec 4 '12 at 21:52
Thanks Brian; I wanted the hint to force $r_k \downarrow 0$ without any additional reasoning. – copper.hat Dec 4 '12 at 22:02
I thought that that was probably your intent, but it might have been a genuine oversight. – Brian M. Scott Dec 4 '12 at 22:06

Let $A \subseteq M$, and let $x \in A$ be a point of accumulation of $A$, i.e. for all $r>0$ $B_r(x) \cap A \neq \emptyset$. Let $r_1 = 1$. Then chose $x_1 \in B_r(x) \cap A$. We construct an infinite sequence of elements of $A$ inductively. Suppose $x_1, x_2, \ldots, x_k$ have all been chosen in $A$ where each $x_i$ for $1 \leq i \leq k$ are distinct. Now let $r = \min\{d(x_1,x),d(x_2,x),\ldots,d(x_k,x)\}/2$. Now let $x_{k+1} \in B_r(x)$. Now $x_{k+1}$ is distinct from $x_1,x_2, \ldots,x_k$. Hence, we get a mapping $\mathbb{N} \to A$, $k \mapsto x_k$. This is injective, and we're done.

Also note that if we take $r=\min\{\frac{1}{k+1}, d(x_1,x), \ldots,d(x_k,x)\}/2$ then we get that $x_k \to x$ for free.

Hope this helps.

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You get convergence even without throwing in $\frac1{k+1}$, though it’s easier to see if you index your $r$’s: if $r_k\le 2^{-k}$ for $k\le n$, then $d(x_k,d)<2^{-n}$ for $k\le n$, and $r_{n+1}\le2^{-(n+1)}$. – Brian M. Scott Dec 4 '12 at 21:48

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