I'm working through Abbott's Understanding Analysis and am stuck on Exercise 3.2.10. Here's the question:
Only one of the following three descriptions can be realized. Provide an example that illustrates the viable description, and explain why the other two cannot exist.
i) A countable set contained in $[0,1]$ with no limit points.
ii) A countable set contained in $[0,1]$ with no isolated points.
iii) A set with an uncountable number of isolated points.
I came up with a proof that (iii) cannot exist, but I don't see why both (i) and (ii) can't work. The set $\{ \frac{1}{2} \}$ would work for (i) since it's countable and doesn't have any limit points. (His definition for limit point is as follows: A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood of $x$ intersects $A$ at some point other than $x$.)
And $\mathbb{Q}\cap [0,1]$ would work for (ii). Since $\Bbb Q$ is countable, its intersection with $[0,1]$ is as well. This set has no isolated points because $\mathbb{Q}$ is dense in $\mathbb{R}$: for every $q \in \mathbb{Q} \cap [0,1]$ and for every $\epsilon > 0$ there exists some $r \in \mathbb{Q} \cap [0,1]$ such that $|r-q| < \epsilon$.
But, apparently, either (i) or (ii) cannot exist. Why is my reasoning here incorrect?
Best Answer
Got it! As pointed out by @Wojowu, the set has to be countably infinite. So (i) wouldn't work because of the Bolzano-Weierstrass Theorem. Since $A$ is countable, its elements can be listed as $a_1,...$, and in that list there exists a convergent subsequence. The limit of this sequence would be a limit point.