Understanding sets with an uncountable number of isolated points in $\mathbb{R}$.

Question

Can someone please verify my proof? Thanks!

Prove that a set with an uncountable number of isolated points does not exist in $\mathbb{R}$.

(Added: June 2, 2020)

Proof: Let $B = \{x_{1}, x_2, \dots\}$ be the set of isolated points of some set with an uncountable number of isolated points. Since each $x_i$ is an isolated point, $\exists \epsilon_i > 0 \textrm{ s.t. } N(x_i , \epsilon_i) \cap B = \{x_i\}$ which implies $N^*(x_i, \epsilon_i) \cap B = \emptyset$

We claim that for any two distinct $x_a, x_b \in A$, $N^{*}(x_a , \frac{\epsilon_a}{2}) \cap N^{*}(x_b , \frac{\epsilon_b}{2}) = \emptyset$. Assume, to obtain a contradiction, that $\exists z$ satisfying $z \in N^{*}(x_a ; \frac{\epsilon_a}{2}) \cap N^{*}(x_b ; \frac{\epsilon_b}{2})$. Without loss of generality, suppose $\epsilon_a \geq \epsilon_b$. Then, \begin{equation*} \left|x_a - x_b\right| = \left|(x_a -z) + (z - x_b)\right| \leq \left|x_a -z \right| + \left|z - x_b\right| < \epsilon_{a/2} + \epsilon_{a/2} = \epsilon_{a} \end{equation*} Thus, $\left|x_a - x_b\right| < \epsilon_{a} \implies x_b \in N^{*}(x_a ; \epsilon_a)$ which contradicts $N^{*}(x_a ; \epsilon_a) \cap B = \emptyset$.

Then, by the density of rationals in reals, we can find a $q_i \in \mathbb{Q}$ s.t. $q_i \in N^*(x_i ; \frac{\epsilon_i}{2})$. This means that we can draw a $1-1$ correspondence between each $q_{i}$ and $x_{i}$ which is a contradiction since there are only a countable number of $q_{i}$ and an uncountable number of $x_{i}$.

Answer 1

$A$ (the isolated points of $U$) is a subspace of $\Bbb R$ which is second countable, so $A$ too is second countable. But $A$ is discrete in the subspace topology and a discrete space is second countable iff it is countable. QED.

Answer 2

As D. Brogan has pointed out in his answer, there is no need to argue by contradiction.

As Henno Brandsma's answer shows, there is no need to introduce the indexing set $U.$

So your argument can be simplified.

Simplifications aside, it is valid up to a point, but then, as you suspected, it goes wrong.

There are two problems that I can see:

The first problem is a matter of expression. You write, "This means that we can draw a 1-1 correspondence between each $q_u$ and $x_u$", but that doesn't really mean anything.

It is clear enough what you are getting at, viz. that there is a 1-1 correspondence between $U$ and a set of rationals. On that understanding, I thought at first that your argument was valid; but in fact, just as you suspected, it is "not entirely accurate"; and Aryaman Maithani's comment had already identified the gap.

If $A$ is the set of isolated points of a set $B \subseteq \mathbb{R},$ then every point $x \in A$ is contained in an open interval $I_x$ that is disjoint from $B \setminus \{x\},$ therefore disjoint from $A \setminus \{x\}.$

For your argument to be valid (or indeed for my own argument to be valid, when I tried to post a simplified version of your argument in a comment!), you need the intervals $\{I_x \colon x \in A\}$ to be pairwise disjoint: then you can choose a distinct rational in each interval, and deduce that $A$ is countable. But as Aryaman Maithani pointed out, they might not be pairwise disjoint.

Do not despair! Your proof can be patched up. Also, major surgery is not needed, just one more idea.

Answer 3

Looks good to me. A personal thing I have here is just that proof by contradiction is not necessary here. What you've shown is that the indexing set $U$ injects into the rationals, hence is at most countably infinite. No need to assume that $U$ is uncountable!