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

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.

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

• Why am I so dumb? – user55511 Jan 7 '13 at 22:14
• @user55511: It’s fine. Don’t be so hard on yourself. Everyone has his or her own moments. :) – Haskell Curry Jan 7 '13 at 22:16
• @user55511: We’ve all known that feeling a time or two; don’t worry about it too much. – Brian M. Scott Jan 7 '13 at 22:21
• Any particular reason why choose positive integers rather than natural numbers? – Dahn Jan 8 '13 at 1:19
• @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 8 '13 at 4:27

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

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. :)

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

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 $\mathbb{N}$.

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

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