I need some help continuing the proof for the following proposition.

A set $A \subset \mathbb{R}^d$ is bounded $\Leftrightarrow$ there exists a cube $Q_s=[-s,s]^d=\left\{ (x_1,\dots,x_d) \: : \: -s \leq x_j \leq s, j=1,\dots,d \right\}$ such that $A \subset Q_s$.

For $\Rightarrow$, I have started by stating that if $A \subset \mathbb{R}^d$ is bounded $\Rightarrow$ there exists a closed ball $\overline{B}$ such that $\vec x \in \overline{B} \:\: \forall \vec x \in A$.

I'm not really sure where to go from here, some hints would be great!


The ball has a center $m=(m_1,\dotsc,m_d)$ and a radius $r$, choose $$s:= r+\max_{1\leq i\leq d} |m_i|$$ Then we have $A\subseteq \bar{B}\subseteq Q_s$.

For the other implication choose a Ball with radius $r=\sqrt{ds^2}$ around 0.

  • $\begingroup$ I understand the first implication, but I'm struggling to see how setting the radius to be $r=\sqrt{ds^2}$ helps. $\endgroup$ – Will Feb 15 '16 at 19:18
  • $\begingroup$ It's the pythagorean theorem. You have to get the diagonal of the cube in your ball. The choosen $r$ is just the length of the diagonal. $\endgroup$ – user302982 Feb 15 '16 at 19:28
  • $\begingroup$ Could you explain a bit further? $\endgroup$ – Will Feb 16 '16 at 10:19

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