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So I am having trouble proving that if $$ K= \{0\} \cup \left\{\frac1n: n = 1,2,\ldots\right\} \cup \left\{\frac1n + \frac1m : n=m,m+1,\ldots; m=1,2,\ldots \right\} $$ then 0 and the points $\frac1m$ are the only limit points of $K$. Cooke proves this by saying that:

Since $x\geq0$ for all $x \in K$ and for any positive number $\epsilon$ there is only a finite set of number s in $K$ larger than $1+\epsilon$, it is clear that no negative number and no number larger than 1 can be a limit point of K. Hence we need only consider positive numbers $x$ satisfying $0\lt x \lt 1 $. If $x$ is such a number and $x$ is not one of the points $\frac1m$, let $p$ be such that $\frac1{1+p} \lt x \lt \frac1\epsilon$, and let $\epsilon=\frac12\min(x-\frac1{p+1},\frac1p - x)$.

The intersection of the set $K$ with the interval $(x-\epsilon,x+\epsilon)$ is contained in the set of points $$ \left\{\frac1{p+1}+\frac1k:p+1\le k \lt \frac1\epsilon\right\} \cup \left\{\frac1m+\frac1n:m\le n \lt \frac1{p+1} - \frac1{p+2}; m=p+2, \ldots,2p+2\right\}, $$ which is a finite set. Therefore $x$ cannot be a limit point of $K$.

I understand why you take $\epsilon$ to be half of $\min$, and the using that every neighborhood around a limit point contains infinitely many points, but I don't understand why the intersection of $K$ with the interval $(x-\epsilon,x+\epsilon)$ is contained in the set mentioned above, and why it is finite. Is it a typo? $$ \left\{\frac1m+\frac1n:m\le n \lt \frac1{p+1} - \frac1{p+2}; m=p+2, \ldots,2p+2\right\}, $$ here I think n is meant to be an integer in order for the finite argument to work, but how can be $$ n \lt \frac1{p+1} - \frac1{p+2}\:\:? $$ Here is how I fixed the typo to work: if $$ \left\{\frac1m+\frac1n:m\le n \lt \frac1{p+1} - \frac1{p+2}; m=p+2, \ldots,2p+2\right\} $$ is changed to $$ \left\{\frac1m+\frac1n: m\le n, \frac1n \ge \frac1{p+1} - \frac1{p+2}; m=p+2, \ldots,2p+2, n \in N\right\}, $$ then $m\le n \le (p+1)(p+2)$ which would be finite. Am I understanding this argument correctly? I think there is a typo or error in Cooke's solution, as the original version can't seem to work.

I know there are other constructions that can be understood easily, but I would like to understand Cooke's proof of $\frac1n + \frac1m$ and see if there are are any typos or errors in the original solution. Thanks a lot.

Daniele Tampieri
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freudianrudin
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1 Answers1

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As you noted, Cooke's argument is flawed.

Your proposed fix is fine.

Here's an alternate argument . . .

Using Cooke's choice of $\epsilon$ and $p$, let $A=(x-\epsilon,x+\epsilon)$.

To show $A\cap K$ is finite, we can argue as follows . . . \begin{align*} &\epsilon= \frac{1}{2} \min \!\left\{ x-\frac{1}{p+1},\frac{1}{p} - x \right\} \\[4pt] \implies\;& \begin{cases} 2\epsilon < x-{\large{\frac1{p+1}}}\\[4pt] 2\epsilon < {\large{\frac{1}{p}}} - x\\ \end{cases} \\[4pt] \implies\;&\frac{1}{p+1}+\epsilon < x-\epsilon < x+\epsilon < \frac{1}{p}-\epsilon\\[4pt] \implies\;&A\subset B,\;\text{where}\;B=\left(\frac{1}{p+1}+\epsilon,\frac{1}{p}-\epsilon\right)\\[0pt] \end{align*} If ${\large{\frac{1}{n}}}\in B$ for some positive integer $n$, then \begin{align*} &\frac{1}{n}\in B\\[4pt] \implies\;&\frac{1}{p+1}+\epsilon < \frac{1}{n} < \frac{1}{p}-\epsilon\\[4pt] \implies\;&\frac{1}{p+1} < \frac{1}{n} < \frac{1}{p}\\[4pt] \implies\;&p < n < p+1\\[4pt] \end{align*} contradiction.

Thus, $B$ contains no elements of the form ${\large{\frac{1}{n}}}$, where $n$ is a positive integer.

Next, let $X$ be the set of pairs $(m,n)$ of positive integers with $m\le n$ such that ${\large{\frac{1}{n}+\frac{1}{m}}}\in B$. \begin{align*} \text{Then}\;\;&(m,n)\in X\\[4pt] \implies\;&\frac{1}{n}+\frac{1}{m}\in B\\[4pt] \implies\;&\frac{1}{n}+\frac{1}{m} > \frac{1}{p+1}\\[4pt] \implies\;&\frac{2}{m} > \frac{1}{p+1}\\[4pt] \implies\;&m < 2(p+1)\\[4pt] \end{align*} hence there are only finitely many choices for $m$.

Suppose $X$ is infinite.

Since there are only finitely many choices for $m$, it follows that for some fixed $m$, say $m=M$, there are infinitely many positive integers $n$ such that ${\large{\frac{1}{n}+\frac{1}{M}}}\in B$.

It follows that ${\large{\frac{1}{M}}}$ is a limit point of $B$, hence \begin{align*} &\frac{1}{p+1}+\epsilon\le\frac{1}{M}\le\frac{1}{p}-\epsilon\\[4pt] \implies\;&\frac{1}{p+1} < \frac{1}{M} < \frac{1}{p}\\[4pt] \implies\;&p < M < p+1\\[4pt] \end{align*} contradiction.

Therefore $X$ is finite.

It follows that $B\cap K$ is finite.

Hence, since $A\subset B$, it follows that $A\cap K$ is finite, as was to be shown.

quasi
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