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When to find the answer Constructing a set with exactly three limit points , Find a solution http://minds.wisconsin.edu/handle/1793/67009 to constructing a set with 3 limit points

set $A = \{\frac 1 n : n \in \Bbb{Z}^+\} = \{1, 1/2, 1/3, 1/4, ...\}$. This has exactly one limit point, namely $0$. I try to prove more formal. we take a x not $0$ .

  1. let $d = |x-0|$

  2. construct a set $W = \{ y : |y - x| < \frac d 2\}$ Here we say W has at most finite number of points A

  3. let set $V = (1, 1 + \frac d 2)$ then V has all the points of A except possibly the finite set of $\frac 1 n$ for which $n \leq \frac 2 d$

  4. So there is at most a finite set $T$ of point p which p in $A$ and p in $W$, let $g =min(\frac d 2, |x-p|)$
  5. Then we can find $U = \{ y : |y - x| < \frac g 2\}$, Set $U$ has no point in A ,

so x is not limit point.

My questions is in $2$ step, Why $W$ has at most finite points of A . Is this equal to if a subset of Countable set has Upper bound and Lower bound, then it is finite . In 2 step it mean the number of points of A in interval $(x - \frac 2 d， x + \frac 2 d)$ is infinte.

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jiamo
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    In general the statement that a subset of a countable linearly ordered set that is bounded below and above is finite is false. Take the subset $[0,1]\cap \mathbb{Q}$ which clearly is bounded below by $0$ and above by $1$ but is nevertheless infinite. Working with $\mathbb{N}$ or $\mathbb{Z}$ does work. – Hayden Sep 07 '14 at 15:33
  • so , I am wrong . Why W has at most finite points of A is not the same as I mentioned in the title? But here (1/2, 1/3, ..., 1/n) when (1 /x , 1/y ) when x , y is Integer ， the point in (1/x, 1/y) is infinte . – jiamo Sep 07 '14 at 15:44

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I can show you that it is impossible for A to have a non-zero limit point. Suppose for the sake of contradiction that $A$ has a non-zero limit point, we'll call it $p$. Then every open neighborhood $N_{\epsilon}(p)=(p-\epsilon,p+\epsilon)$ of radius $\epsilon >0$ about the point $p$ must contain ALL but finitely many points of A. Clearly $p$ can't be negative, because the open neighborhood $N_{\epsilon}(-p)$ of radius $\epsilon=\frac{|p|}{2}$ would contain no points of A. Similarly, $p$ can't be greater than $1$ because the neighborhood $N_{(p-1)/2}(p)$ would contain $p$ but not contain any elements of $A$. We now consider only $p \in (0,1]$. One nice property of the natural numbers is that given $p \in (0,1]$ we can find some $m \in \mathbb{N}$ such that $\frac{1}{m}<p$. Now let $\epsilon' = \frac{1}{2}(\frac{1}{m}+\frac{1}{m+1})$. This choice of $\epsilon'$ may seem weird, but convince yourself that what I have done is choose a point directly between $\frac{1}{m}$ and $\frac{1}{m+1}.$ By choice of $\epsilon$' we now have an open neighborhood about $p$ that contains $\frac{1}{m}$, but for all $n>m$ we know $\frac{1}{n} \notin N_{\epsilon'}(p)$. This is tantamount to saying that the infinite subset $A' \subset A$ where $A'=\{\frac{1}{m+1},\frac{1}{m+2},\frac{1}{m+3},... \}$ is a list of points of A not contained in our neighborhood of specially chosen $\epsilon'$. At best, our neighborhood contains ONLY some or all of the points $\{1, \frac{1}{2},...\frac{1}{m} \}$. Since we have found a neighborhood about $p$ that does NOT contain all but finitely many points of $A$, we deduce that $p \neq 0$ is not a limit point of $A$.

graydad
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