Real Analysis – Rudin Exercise 2.13: Last Part Explained

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.