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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.

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

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

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

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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 $\mathbb{N}$.

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

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