# How can a set be bounded and countably infinite at the same time?

There is a theorem in my textbook that states,

Let $E$ be a bounded measurable set of real numbers. Suppose there is a bounded countably infinite set of real numbers $\Lambda$ for which the collection of translates of $E$, $\{\lambda + E\}_{\lambda \in \Lambda}$, is disjoint. Then $m(E) = 0$.

I'm a little confused about this theorem, because it's saying that a set is bounded and countably infinite at the same time. But if a set is bounded, isn't it supposed to be finite?

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No: $[0,1]$ is a bounded set that isn’t even countable, let alone finite. An unbounded set of real numbers is necessarily infinite, but a bounded set can be of any size up to and including the cardinality of the entire set of real numbers.

To prove this, use the well-known fact, proved in several places on this site, that there is a bijection $\varphi:\Bbb R\to(0,1)$. Let $A$ be any subset of $\Bbb R$; then $\varphi[A]$ is a subset of $(0,1)$ of exactly the same cardinality, and every subset of $(0,1)$ is bounded.

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Hint: Consider $\Bbb Q$ intersected with any bounded set, finite or infinite. Since $\Bbb Q$ is countable, the new set is at most countable, and clearly can be made infinite; for example, $[0,1]\cap\Bbb Q$ is bounded and countable.

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The set $\{2^{-k}\ |\ k \in \mathbb{Z}^+\}$ is bounded and countably infinite.

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