I have to prove or disprove the following:
Theorem 1? Every bounded set of reals consisting of isolated points is finite.
My intuition is that it is false, since the problem set continues with a stronger statement:
Theorem 2? Every compact set of reals consisting of isolated points is finite.
A possible counterexample to Theorem 1 would be the bounded set $S=\left\{ \frac{1}{n} | n \in \mathbb N \right\}$. Since every point in $S$ can be contained within an open set (under the standard topology of $\mathbb R$) which intersects no other points of $S$.
However, the contrapositive of the $\mathbb R$-version of Bolzano-Weierstrass is:
$$
\text{Let $S\subseteq \mathbb R$. If $S$ has no accumulation points, then either $S$ is unbounded or finite.}
$$
Thus $-$ in line with Theorem 1's assumption $-$ let $S \subseteq \mathbb R$, such that $S$ is bounded and consists exclusively of isolated points. Since a point is an isolated point iff it is not an accumulation point, we use Bolzano-Weierstrass to deduce that $S$ is either unbounded or finite. But, by assumption, $S$ is bounded, so by a simple disjunctive syllogism, it must be that $S$ is finite, thus "proving???" Theorem 1.
Where is the mistake? Is my statement of Bolzano-Weierstrass wrong? Is the counterexample to Theorem 1 incorrect?
Best Answer
I've already said the gist of it in the comments, but just so there is an actual answer:
The problem here was that you were interpreting the accumulation points in your contrapositive of Bolzano-Weierstrass as accumulation points in $S$ as opposed to accumulation points in $\mathbb{R}$. The difference can feel subtle but it exists nonetheless, and the notion of accumulation points in the subset itself is used in, for example, the identity theorem for holomorphic functions (which may or may not be something you've seen or are going to see for now, but that's the first example that comes to mind), thus the two concepts do coexist.
In the context of Bolzano-Weierstrass, the theorem says (once rephrased with sets like you've done) that for a countable bounded set $S$, $S$ has at least one accumulation point in $\mathbb{R}$: if it were "in $S$", then your example $S := \left\{\frac1n \mid n \in \mathbb{N}^*\right\}$ ($\mathbb{N}^*$ means $\mathbb{N} \setminus \{0\}$ for us French) would disprove Bolzano-Weierstrass since its only overall accumulation point is $0 \notin S$, and generally it's never a good sign when you go against a well-established theorem... :^)
Note that technically the exact contrapositive of the usual statement of Bolzano-Weierstrass would have the conclusion "uncountable" alongside the options "unbounded" and "finite" but you've probably noticed yourself - or whatever material you used might have made you noticed with some remark or corollary - that if $S$ is bounded uncountable then taking a countable sequence does the trick to find an accumulation point, of course. Anyway, I'm just rambling at this point.