How can I show that if a set is bounded, then it's contained in a k-cell? The set is a bounded subset of R (under the Euclidean metric), and a k-cell is a set of points {x_1...x_k} such that a_j < x_j < b_j for j=1...k. 
Any ideas on how to show this?
 A: The above comments take two approaches, either use that (by definition) a bounded set is contained in a ball, or alternatively, that each coordinate is bounded. 


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*Assuming that a set $X$ is bounded (by definition) if there is a real number $M>0$ such that $||x||\le M$ for each $x\in X$ (where $x=\{x_1,...,x_k\}$ and $||x||=\sqrt{x_1^2+...+x_k^2}$ ). Then $X$ is contained in the box $a_j < x_j < b_j$ where $a_j=-M$ and $b_j=M$, for j=1...k. Indeed if some $x$ is not in this box, then there is $j$ such that either $x_j<-M$ or $x_j>M$, but then $||x||\ge \sqrt{x_j^2}>M$, so $x\not\in X$. (So, if $x\in X$ then $x$ is in the box.) 

*The above argument also shows that if $X$ is bounded, then each coordinate is bounded, that is $|x_j|\le M$ for each $j$ and for each $x=\{x_1,...,x_j,...,x_k\}\in X$. Instead of using the same bound $M$ for each coordinate, we may let $a_j=\inf\{x_j: x=\{x_1,...,x_j,...,x_k\}\in X\}$ and $b_j=\sup\{x_j: x=\{x_1,...,x_j,...,x_k\}\in X\}$. Then $-M\le a_j\le b_j\le M$ for each $j$ so the $a_j,b_j$ are well-defined finite real numbers, and clearly from their construction they determine a box (called a $k$-cell an the question) that contains $X$. 
