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 $x_{i}$.

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    $\begingroup$ Why is it not possible that a $q_u$ may belong to $V_{\epsilon_{u'}}(x_{u'})$ for some $u' \neq u$? $\endgroup$ May 15, 2020 at 15:38
  • $\begingroup$ @AryamanMaithani Because I am trying to argue that there must be a rational in $V_{\epsilon_u}$ $\endgroup$ May 15, 2020 at 15:41
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    $\begingroup$ (After the edit which substantially changed your attempt..) How does that "without loss of generality" work? Using the same logic, I could equally well argue that because $\mathbb R$ has no isolated point, then "without loss of generality" the same is true of any subset of $\mathbb R$. $\endgroup$
    – user169852
    May 16, 2020 at 5:05
  • $\begingroup$ @CalumGilhooley I do not have enough reputations to perform a "rollback" unfortunately. The issue is that I don't have a copy of my original attempt. Neither do I remember that attempt very clearly. I'm sorry about that. $\endgroup$ Jun 2, 2020 at 18:24
  • $\begingroup$ That's OK, it's just a technical problem. If neither of us can sort it out, I'm sure someone else can. First, have you clicked on the link with a label that says something like "edited 1 hour ago"? It may require some more clicking, but you should be able to see all previous versions of the question. Some will be practically unreadable, because the system attempts to display differences between successive versions, and the default inline display of differences works best with small changes. I think it is possible to make things more readable by clicking on "side by side display", or something. $\endgroup$ Jun 2, 2020 at 19:08

3 Answers 3


$A$ (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.

  • $\begingroup$ This answer is irrelevant to the question asked. $\endgroup$
    – Paul
    May 15, 2020 at 22:23
  • $\begingroup$ (+1), but by $U$ you mean $B.$ $\endgroup$ May 15, 2020 at 22:26
  • $\begingroup$ @CalumGilhooley I was referring to the question before the OP edited it, there were sets $U$ and $A$ there. $\endgroup$ May 16, 2020 at 8:00
  • $\begingroup$ @HennoBrandsma Yes, I remember it well! If you refer to the edit history, you will see that the subset of $\mathbb{R}$ is named $B,$ while $U$ is the name of an index set. $\endgroup$ May 16, 2020 at 11:17

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.

  • $\begingroup$ Thanks for the answer! I am not familiar with topology except the very basic concepts, so I, unfortunately, don't fully follow your answer. Can you please reveal what "one more idea" I need, preferably using elementary concepts? I am guessing it might be Axiom of Completeness? $\endgroup$ May 16, 2020 at 0:17
  • $\begingroup$ Don't throw away your set of neighbourhoods $V_{\epsilon_u}(x_u).$ As has been pointed out, they may overlap. Can you think of a simple fix that gives you a set of neighbourhoods of the $x_u$ that definitely don't overlap? It might help to draw a picture (I did). No topological ideas are needed beyond what you've already used. $\endgroup$ May 16, 2020 at 1:01
  • $\begingroup$ I added a better intuitive proof. Is it correct? $\endgroup$ May 16, 2020 at 4:53
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    $\begingroup$ I'm afraid not. More importantly, though, you should not have deleted the original question, or, after it had been restored, deleted its entire text and replaced it with an entirely new argument, because this makes nonsense of the comments that have been made and the answers that have been given. Also, it disrespects your own work on the problem! Your original question was perfectly fine. There is nothing wrong, of course, in coming up with an entirely new argument; but I strongly suggest that you undo the changes, and post your new argument as an entirely new question with a link to this one. $\endgroup$ May 16, 2020 at 6:23
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    $\begingroup$ As you're so nearly there now, and as we're running out of space for comments: the only part of that idea that you need is the first part, i.e. divide all the $\epsilon_u$'s by $2.$ I don't think it's trivial or obvious that this idea works, so you should write out a proof carefully, using the Triangle Inequality to show that the half-size intervals are disjoint. $\endgroup$ May 16, 2020 at 14:39

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!

  • $\begingroup$ (+1) What's more, as Henno Brandsma's answer shows, there's no need for the indexing set $U,$ either! $\endgroup$ May 15, 2020 at 22:28

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