An example of a bounded countably infinite subset of the real numbers.

Question

I've been trying to think of an example of bounded, countably infinite subset of the real numbers. However, knowing that countably infinite means can be put into 1-1 correspondence with the naturals, this doesn't seem intuitively obvious.

Thanks in advance.

score 27 · Accepted Answer · answered Jan 07 '13 at 22:12

27

$$\left\{\frac1n:n\in\Bbb Z^+\right\}$$

answered Jan 07 '13 at 22:12

Brian M. Scott

616,228

3

Why am I so dumb? – user55511 Jan 07 '13 at 22:14
12

@user55511: It’s fine. Don’t be so hard on yourself. Everyone has his or her own moments. :) – Haskell Curry Jan 07 '13 at 22:16
5

@user55511: We’ve all known that feeling a time or two; don’t worry about it too much. – Brian M. Scott Jan 07 '13 at 22:21
Any particular reason why choose positive integers rather than natural numbers? – Dahn Jan 08 '13 at 01:19
@DahnJahn well $1/0$ would be a problem, wouldn't it? – Philip JF Jan 08 '13 at 01:41
@PhilipJF Are you implying that $0 \in \mathbb{N}$? – Christian Mann Jan 08 '13 at 03:27
1

@ChristianMann yes. Otherwise $\mathbb{N}$ is not a monoid under addition. For me, $\mathbb{N}$ is the smallest set such that $0 \in \mathbb{N}$ and $\forall n \in \mathbb{N}, n + 1 \in \mathbb{N}$. It is only natural for the "point" here to be $0$ and not $1$, since addition is primitively recursive. – Philip JF Jan 08 '13 at 04:27
Alright, if that is the definition of natural numbers you use, then it makes absolute sense, thanks. – Dahn Jan 08 '13 at 09:20
@user55511: Maybe lack of sleep? =P It certainly affects me! – user541686 Jan 08 '13 at 10:27
@DahnJahn: $0\in\Bbb N$, and I didn’t want to clutter up the example by writing $\left{\frac1{n+1}:n\in\Bbb N\right}$. – Brian M. Scott Jan 08 '13 at 13:36

Haskell Curry · Answer 2 · 2013-01-08T08:39:24.573

The set of rationals contained in $ [0,1] $ is another example.

Addendum

If you start off with any countably infinite set $ S \subseteq (- \infty,\infty) $ that is unbounded, then there is a quick way out. The inverse tangent function $ \tan^{-1} $ maps $ (- \infty,\infty) $ bijectively to the bounded interval $ \left( - \dfrac{\pi}{2},\dfrac{\pi}{2} \right) $, so the image $ {\tan^{-1}}[S] $ is a bounded and countably infinite set. :)

score 10 · Answer 3 · edited Jan 07 '13 at 22:22

10

$$\mathbb{Q} \cap [a,b]$$ where $a,b \in \mathbb{R}$, $a<b$.

edited Jan 07 '13 at 22:22

Cameron Buie

102,994

answered Jan 07 '13 at 22:13

score 7 · Answer 4 · answered Jan 07 '13 at 22:12

7

Let $A = \{ 1/n : n \in \mathbb{N} \}$ where $\mathbb{N} = \{ 1, 2, 3, ... \}$.

$A$ is bounded between 0 and 1, and has an obvious bijection with $\mathbb{N}$.

answered Jan 07 '13 at 22:12

zrbecker

4,048

score 5 · Answer 5 · answered Jan 07 '13 at 22:13

5

$\left\{\frac{1}{n} | n \in \mathbb{N}\right\}$

It's bounded in $(0,1]$
It corresponds to $\mathbb{N}$ by $\varphi(n)=\frac{1}{n}$

answered Jan 07 '13 at 22:13

Alfonso Fernandez

4,047

An example of a bounded countably infinite subset of the real numbers.

5 Answers5