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Could someone please show that an open ball is open where the definition of "open" is: A set is open if for each $x$ in $U$ there is an open rectangle $A$ such that $x$ in $A$ is contained in $U$. Where an open rectangle is $(a_1,b_1)\times\ldots\times(a_n,b_n)$. I also realize that one can use rectangles or balls, but I would like to see the proof using rectangles, as this is the definition used in Spivak's calculus on manifolds. So for example, the solution located at An open ball is an open set is unacceptable.

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you do exactly that proof, and then, put a cube inside the ball.... –  N. S. Apr 20 '12 at 15:26
    
that would only be a(n) (infinite) regression. If all I have to do is put a cube (rectangle) inside that ball, then I would have the same problem: How to put the cube inside the ball that's inside the large ball. I want a rigorous proof that it can be done. –  Squirtle Apr 20 '12 at 16:40
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@user29507: the farthest point of a cube of side $\delta$ from the center is $\sqrt n \frac \delta{2^n}$ So if you can put an $\epsilon$ ball in, you can put a $\delta$ cube in for small enough $\delta$. –  Ross Millikan Apr 20 '12 at 16:48
    
.... right, that seems very clever, but unfortunately I don't "see" \sqrt n \frac \delta{2^n} very well. Its probably just some basic n-dim geometry, but as I'm just beginning at this, I don't understand. –  Squirtle Apr 20 '12 at 22:09
    
actually the value that I compute in R^2 and R^3 is sqrt(n)*delta/2. Not divided by 2^n. I don't know how to do the computation in R^4 because I can't visualize that space. Any other ideas would be much appreciated. –  Squirtle Apr 21 '12 at 0:42

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