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Here's the proof I was given:

Proposition.- If $A$ is countable then $\mathbb{R} \setminus A $ is dense.

Proof: Suppose otherwise, then there exists real numbers $a$ and $b$, with $a < b$, such that there is no $a < x < b$ with $x \in \mathbb{R} \setminus A $. To put it in another way, if $a < x < b$ then $x \in \mathbb{R} \setminus A$, as $ \mathbb{R} \setminus A$ is the complement of $A$, we see that if $a < x < b$ then $x \in A$. Thus,$(x,y) \subseteq A$. But $A$ is countable and $(x,y)$ is not, this leads to a contradiction. QED.

Is it correct to assume that $(x,y)$ is uncountable because it is not dense? That all countable sets are dense? (That non-dense sets are uncountable?)

What would be the rule that governs why $(x,y)$ is not countable? Is there anything wrong with this proof?

Thank you

Asaf Karagila
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Forever_A_Student
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  • What do you mean by $(x,y)\subset A$ ? What's $y$ ? And no, countable sets are not necessarily dense, I'm sure you know several counter-examples. – Augustin May 18 '15 at 20:07
  • Did you mean $(a,b)\subseteq A$ instead $(x,y)\subseteq A$? – Ángel Mario Gallegos May 18 '15 at 20:22
  • The "rule" that $(x,y)$ is not countable is that every nonempty interval has cardinality $\left|\mathbb{R}\right|$. – graydad May 18 '15 at 20:34
  • As the proof was given to me it said (x,y). In either case, assuming y $\in \mathbb{R}$ OR, if it should have been (a,b) what are the implications? – Forever_A_Student May 18 '15 at 20:51
  • @Augustin This is the only proof I have (out of textbooks and class notes) proving this specific proposition. Rereading it I can assume y $\in$ A and share the same properties as x. I understand through the point in the proof "x $\in $ A" and I understand afterwards that any uncountable set cannot be a subset of a countable set. I don't know why an open interval was brought into the proof? Perhaps because A $\subseteq \mathbb{R}$, so since intervals are a characteristic of $ \mathbb{R} $ they can be applied to A as well? – Forever_A_Student May 18 '15 at 21:03

2 Answers2

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No open interval is countable. A well-known proof of that is the so called Cantor's diagonal argument. See this.

Not every countable set is dense. Take $\Bbb N$, for example. But others, are: $\Bbb Q$, for example.

ajotatxe
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The set $\{\frac1n|n\in\mathbb N\}$ is a countable set which is not dense.
The Cantor set is an uncountable set which is not dense.
Hence denseness is not related to countability, but we are sure that every dense set on the real line must be infinite.
There is a general result in mathematics known as Baire category theorem which asserts that

If $A$ is a set of first category in the real line, then $\mathbb R/A$ is dense.

And your statement is an special case of the above theorem, because every countable set is of first category.

Keivan
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  • I have not learned about Baire categories yet, and I need to stick to what I've been told, but I look forward to learning about it I the future. – Forever_A_Student May 18 '15 at 21:06