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Here's something sort of fun that I gave thought to a while ago, and now that I've done some maturing mathematically I'm curious to see if my musings are legitimate.

Let $H=[0,1] \times [0,\frac{1}{2}] \times [0,\frac{1}{3}] \times \cdots$ be the Hilbert cube. What are the volume and diagonal length of $H$?

If we try to calculate the volume in a fashion analogous to calculating the volume of a square (2-cube) or ordinary cube (3-cube), by multiplying the edge lengths, then $\text{Vol}(H)=1\cdot \frac{1}{2} \cdot \frac{1}{3} \cdots$ would seem to be 0 in the limit. However, the usual Lebesgue measure does not have an analogue in $\mathbb{R}^\infty$. How do we obtain an appropriate notion of volume here?

Now let's call the diagonal of $H$ the line from the point $(0,0,...)$ to the point $(1,\frac{1}{2}, \frac{1}{3}, ...)$. Then if we extend the usual Euclidean metric we obtain $\text{Length}(\text{Diag}(H))=\sqrt{\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}} = \sqrt{\dfrac{\pi^2}{6}}=\dfrac{\pi \sqrt{6}}{6}$. Is this the correct (or A correct/meaningful) way to think of this? Does this turn out to be the diameter of $H$ considered as a metric space?

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    @joriki Yes, I have seen that alternate definition. But in [Infinite-dimensional Lebesgue measure](http://en.wikipedia.org/wiki/Infinite-dimensional_Lebesgue_measure) it does link to the Wikipedia article on the Hilbert cube with the definition that I used above. This could just be an inconsistency within Wikipedia though.2012-11-08

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I did the calculation of the diameter of the hilbert cube, too. But I ended up with a different result:

$\mathrm{diam}(H) = \sup_{x \in H, y \in H} d(x,y) =\sqrt{\sum_{k=1}^\infty |x_k - y_k|^2} = \sqrt{\sum_{k=1}^{\infty} (-\frac{1}{k} - \frac{1}{k})^2} = \pi \frac{\sqrt{6}}{3}$

This should be true if my definition for the diam is right and if we define the hilbert cube as follows: $ H= \{(x_n)_{n \in \mathbb{N}} \in l_2 \mid|x_n| \leq \frac1n, n \in \mathbb{N} \}$

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    You'$v$e just taken your Hilbert cube to be twice as big as the asker's, so it's not surprising that you got twice the diameter.2013-01-27