Rudin Real Analysis exercise 2.13, question about last part.

Question

Self studying real analysis from baby rudin and stuck at this proof:

Trying to understand that proof for hours and stuck at the last part where the points of
$$ K \ \cap \ (x - \epsilon, x + \epsilon) $$ are contained in $$ \left\{\frac{1}{p+1} + \frac{1}{k} : p+1 \leq k < \frac{1}{\epsilon }\right\} \cup \left\{ \frac{1}{m} + \frac{1}{n} : m \leqslant n < \frac{1}{p+1} - \frac{1}{p+2} ;m = p+2, ... ,2p+2 \right\} $$

Is there a way to make this part intuitive or it's just algebra? I saw similar examples like this and got confused every time so general help about this kind of proofs (where we choose an interval and prove with that) would be helpful.

Answer 1

In my opinion there is no really intuitive argument, you have to do some calculations. But very simple steps are sufficient to show that $J = (x-\epsilon, x+\epsilon)$ contains only finitely many elements of $K$.

Certainly no number $\frac 1 n$ is contained in $J$. So let us consider the numbers $a_{m,n} = \frac 1 m + \frac 1 n$ with $n \ge m$. By definition $\frac 1 m < a_{m,n} \le a_{m,m} = \frac 2 m$.

1. For $m \le p$ we have $a_{m,n} > \frac 1 m \ge \frac 1 p > x+\epsilon$. Therefore only for $m > p$ it is possible that some $a_{m,n} \in J$.

2. For $m \ge 2(p+1)$ we have $a_{m,n} \le \frac 2 m \le \frac{1}{ p+1}< x-\epsilon$. Therefore only for $m < 2(p+1)$ it is possible that some $a_{m,n} \in J$.

3. Let us consider the finitely many $m$ with $p < m < 2(p+1)$. For each such $m$ we have $\frac 1 m \le \frac{1}{p+1} < x-\epsilon$, hence all but finitely many $n$ satisfy $a_{m,n} = \frac 1 m + \frac 1 n < x -\epsilon$. In other words, at most finitely many $a_{m,n}$ can be contained in $J$.

Clearly 1. - 3. imply that $K \cap J$ is finite.

Answer 2

The set $K$ is the union of $\{0\}$ and all sets of the form $$ K_n=\Big\{\frac{1}{n}+\frac{1}{n}, \frac{1}{n}+\frac{1}{n+1}, \frac{1}{n}+\frac{1}{n+2}, \dots,\frac{1}{n}\Big\} $$ for $n=1,2,3,...$. Every $1/n$ is a limit point of $K_n$, hence of $K$. Also $0$ is a limit point of $K$. We need to show that $K$ has no other limit points.

It suffices to consider $x>0$. If $x\neq 1,1/2, 1/3,\cdots$, then we can pick $\epsilon>0$ such that $x-\epsilon>2/N$ for some large enough natural number $N$. If $n\geq N$ and $y\in K_n$, then $$ y\leq 2/n\leq 2/N<x-\epsilon, $$ so $B_\epsilon=(x-\epsilon,x+\epsilon)$ is disjoint from $K_N, K_{N+1},\dots$ (and from $\{0\}$ as well). Roughly speaking, the sets $K_N,K_{N+1},\dots$ are to the left of $B_\epsilon$.

We can improve $\epsilon$ so that $$ \epsilon<\min\{|x-1|,|x-1/2|,\dots,|x-1/N|\}. $$ Given $n=1,\dots,N-1$, there are two possibilities:

1. If $x<1/n$, then $x+\epsilon<1/n$, so $B_\epsilon$ is disjoint from $K_n$. Roughly speaking, such $K_n$ are to the right of $B_\epsilon$.
2. If $x>1/n$, then $x-\epsilon>1/n$. Pick a natural number $M\geq n$ large enough, such that $$ \frac{1}{n}+\frac{1}{m}<x-\epsilon $$ for all $m\geq M$. Then $B_\epsilon$ is disjoint from the subset $$ \Big\{\frac{1}{n}+\frac{1}{M}, \frac{1}{n}+\frac{1}{M+1}, \frac{1}{n}+\frac{1}{M+2}, \dots,\frac{1}{n}\Big\} $$ of $K_n$, so $$ K_n\cap B_\epsilon\subseteq\Big\{\frac{1}{n}+\frac{1}{n}, \dots,\frac{1}{n}+\frac{1}{M-1}\Big\} $$ is finite. Roughly speaking, almost all members of such $K_n$ are to the left of $B_\epsilon$.

We conclude that $$ B_\epsilon\cap K=B_\epsilon\cap(K_1\cup\cdots\cup K_{N-1})=(B_\epsilon\cap K_1)\cup\cdots\cup(B_\epsilon\cap K_{N-1}) $$ is a finite union of finite sets, hence finite. This shows that $x$ is not a limit point of $K$.

I suggest a modification to make the matter much clearer. Consider thet set $$ K'=\Big\{1-\frac{1}{2},1-\frac{1}{4},\dots,1\Big\}\cup\big\{\frac{1}{2}-\frac{1}{4},\frac{1}{2}-\frac{1}{8},\dots,\frac{1}{2}\big\}\cup\cdots\cup\{0\}, $$ or in binary representations, \begin{align} K'=\{0.1,0.11,\dots,1\}\cup\{0.01,0.011,\dots,0.1\}\cup\cdots\cup\{0\}. \end{align} The limit points are $1,0.1,0.01,\dots,0$.