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I am trying to show the following: every countable subset of $\Bbb R$ with at least two points is disconnected.

My attempt: let $D$ be such subset. Then take $a \in D$ and define $A=\{ a\}$ and $B = D - \{ a\}$. I want to use the result that a set is disconnected if and only if there exists mutually separated sets whose union is the set. I tried to show that $A$ and $B$ are mutually separated sets using $\overline{A} = \{ a \}$ and $\overline{B}$ but $\overline{B} = \overline{D - \{a \}} = D - \{a\}^{\circ} = D$. We have $\overline{A} \cap B = \emptyset$ but $A \cap \overline{B} = \{a\} \neq \emptyset$.

I'm not seeing another line of attack.

Mark Fantini
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2 Answers2

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Your approach seems to imply that every point in a countable set is isolated. Namely that a countable is synonymous with "discrete" in the case of the real numbers.

This is not true, of course. The rational numbers are countable but not discrete.

Instead, let me give you the following hint:

If $D$ is a countable subset of $\Bbb R$, show that there is some $x\in\Bbb R$ such that $x\notin D$ and neither $D\cap(-\infty,x)$ nor $D\cap(x,\infty)$ are empty.

Asaf Karagila
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  • I've always been confused by this: what divides countable from discrete? – Mark Fantini Jun 25 '14 at 11:19
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    Countable is a condition on the cardinality of the set; discrete is a topological condition on the set (every point is isolated). In the real numbers discrete implies countable, but not vice versa. – Asaf Karagila Jun 25 '14 at 11:21
  • Using your hint as guidance I've tried this: since $D$ has at least two points, namely $a,b$, we can assume $a<b$. By the real numbers there exists $a<x<b$ such that $x \notin D$. Therefore $a \in D \cap (- \infty,x)$ and $b \in D \cap (x, +\infty)$. However, I'm having trouble showing that they are mutually separated. – Mark Fantini Jun 25 '14 at 11:31
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    What do you know about this partition of $D$? What topological property does the parts have? – Asaf Karagila Jun 25 '14 at 11:35
  • It is countable and therefore has empty interior? – Mark Fantini Jun 25 '14 at 11:41
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    No, you are just guessing now. Look at the sets, what sort of sets did you intersect with $D$? – Asaf Karagila Jun 25 '14 at 14:46
  • I intersected $D$ with open sets, therefore they are disjoint nonempty open sets whose union is $D$ and $D$ is disconnected, correct? – Mark Fantini Jun 25 '14 at 19:56
  • Yes, that is the idea. – Asaf Karagila Jun 25 '14 at 20:00
  • There's a nontrivial flaw in Asaf's reasoning, and that is when we decide to pick randomly elements of $\mathbb{R}$ for which it's not an element of $D$, there's a possibility that this goes on and on infinitely, especially if you give this duty to a robot to do the job even with FTL speed in counting, note that this reasoning is inductive, and whenever we decide to stop the process of "picking", then the number of steps done is a "natural" number, and unfortunately $D$ is countable. (can have countably infinite elements in it) – Ashkan Ranjbar Jan 04 '17 at 21:30
  • @AshkanRanjbar. All Asaf says is $\exists a,b \in D ;(a<b \land \exists x\in (a,b)$ \ $D)).$ We're not picking anything. We're just observing that there must exist $a,b,x$ with some stated property – DanielWainfleet Jan 05 '17 at 01:06
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Note that a subset of $\mathbb{R}$ is connected iff it's an interval.

Now let $D\subset\mathbb{R}$ be atmost countable with at least two elements, supposing it to be connected will exhibit a contradiction, since an interval has uncountably infinite elements (except $\space \{a\}\space \forall a\in \mathbb{R}$)

Warning:$(a,b)_Q$ for $a,b\in\mathbb{Q}$ denoting the induced topology on $\mathbb{Q}\cap(a,b)$ by $\mathbb{R}$ is not an interval of $\mathbb{R}$, in most books, it's not even considered to be an interval at all!

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Ashkan Ranjbar
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