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I don't understand the part that I underlined with red. Thanks.

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    $\begingroup$ Uncountable subsets of $\Bbb R$ have accumulation points. The more general claim is that if $X$ is second countable and $A$ is an uncountable subset, it has uncountably many limit points in $X$. $\endgroup$ – Pedro Tamaroff Sep 15 '13 at 2:43
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    $\begingroup$ When one has infinitely many points, one has certain freedom to do things with them. In particular, we may choose three distinct points $a<b<c$, and in particular we may choose them so that $c-a<1/n$ by the definition of limit point. $\endgroup$ – Pedro Tamaroff Sep 15 '13 at 2:53
  • $\begingroup$ De nada. ${}{}{}$ $\endgroup$ – Pedro Tamaroff Sep 15 '13 at 2:56
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The author’s argument is very badly stated. In fact he means that if $J_{rn}$ is infinite, it must have an accumulation point (since it’s bounded). Take an open interval of length $\frac1n$ about that accumulation point: it contains infinitely many points of $J_{rn}$ and therefore certainly contains three points $a,b$, and $c$ such that $a<b<c$ and $c-a<\frac1n$. The author should not have used $a$ and $b$, here, however, since those already denote the endpoints of the interval on which $f$ is defined.

He could in fact have avoided accumulation points altogether and shown directly that $J_{rn}$ is actually finite.

Suppose that $u,v,w\in J_{rn}$ with $u<v<w$ and $v-u,w-v<\frac1n$. Then $r<f(v)$, since $u\in J_{rn}$ and $u<v<u+\frac1n$, and $f(v)<r$, since $w\in J_{rn}$ and $w-\frac1n<v<w$. This is impossible, so no such points $u,v,w\in J_{rn}$ can exist. In particular, if $w-u\le\frac1n$, then automatically $v-u,w-v<\frac1n$, so $J_{rn}$ cannot contain three points $u,v$, and $w$ such that $u<v<w\le u+\frac1n$. In other words, if $u,w\in J_{rn}$ and there is a point $v$ of $J_{rn}$ between $u$ and $w$, then $|u-w|>\frac1n$.

For some positive integer $m$ there are points $x_k$ for $k=0,\ldots,m$ such that

$$a=x_0<x_1<x_2<\ldots<x_m=b$$

and $x_{k+1}-x_k\le\frac1n$ for $k=0,\ldots,m-1$. Each subinterval $[x_k,x_{k+1}]$ contains at most two points of $J_{nr}$, so $J_{nr}$ is finite.

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  • $\begingroup$ @SuperTroll: Your attempted edit was in fact correct, and I’ve made the change. $\endgroup$ – Brian M. Scott Sep 15 '13 at 4:10
  • $\begingroup$ @SuperTroll You want \mathbb. And yes, that argument is quite nice. $\endgroup$ – Pedro Tamaroff Sep 15 '13 at 4:18
  • $\begingroup$ @SuperTroll: You’re welcome. $\endgroup$ – Brian M. Scott Sep 15 '13 at 4:20
  • $\begingroup$ @SuperTroll: Yes, that works fine. Or, since you’ve already handled the bounded case, you could simply use it to say that $J_{rn}\cap[m,m+1]$ is finite for each $m$ and therefore $J_{rn}$ is countable. $\endgroup$ – Brian M. Scott Sep 15 '13 at 19:05

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