Intersection between the set and neighborhood of a limit point of the set is an infinite set

Question

This question is from Kasriel's Undergraduate Topology Book Exercise 7 from pg. 71

Let $S \subset R^n$ and let $z$ be a limit point of $S$. Show that for every $\epsilon > 0$, $N(z;\epsilon) \cap S$ is an infinite set.

My thought process:

Would the best way of showing this be by the proof of contradiction? If $z$ is not a limit point, then it is a isolated point, and the intersection between the two sets will be a finite set.

Will this be enough? If so, how can I show this?

Answer 1

1. It won't be enough.
2. That's not proof by contradiction; when you want to do a contradiction proof, you assume the opposite of the conclusion. What you've done is to assume the opposite of one of the hypotheses.
3. The opposite of the conclusion reads like this:

"Suppose that $z$ is a limit point of $S \subset \Bbb R^2$, and $U$ a neighborhood of $z$ such that $F = U \cap S$ is finite. Then ..."

...and then you go on to derive a contradiction.

Answer 2

It is okay to search to seek it in contradiction here, but not the way you propose.

The fact that an isolated point induces a finite set, does not prove that a limit point will induce an infinite set.

Actually you should say: assume that $N(z,\epsilon)\cap S$ is not an infinite set..., and you must proceed with deducing a contradiction.

Hint: If the set is finite then the distance of $z$ to the set $N(z,\epsilon)\cap S-\{z\}$ is positive. Now let $\epsilon'>0$ be smaller than that distance and have a look at $N(z,\epsilon')\cap S$.