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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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    The diagonal length makes sense; see [Wikipedia](http://en.wikipedia.org/wiki/Hilbert_cube#The_Hilbert_cube_as_a_metric_space). I don't see how to make sense of the volume of the Hilbert cube.2012-11-04
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    I am now seeing in the Wikipedia article "Infinite-dimensional Lebesgue measure" that the Hilbert cube does in fact carry the product Lebesgue measure. Does this then imply what I suggested above, that $\text{Vol}(H)=1\cdot \frac{1}{2} \cdot \frac{1}{3} \cdots = 0$? This seems nicely unintuitive, since this would make H a 'shape' with a positive distance between its 'farthest away points,' yet having no volume.2012-11-08
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    You can add links in the mini-Markdown used in the comments by enclosing the linked text in square brackets, followed by the URL in parentheses; see the help text you get when you click "help" under the "Add Comment" button.2012-11-08
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    I think the problem here is that "Hilbert cube" is sometimes used to refer to $[0,1]\times[0,1/2]\times\dotso$ and sometimes $[0,1]\times[0,1]\times\dotso$. Both [the paper cited in the Wikipedia article](http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1102806462) and [the MathWorld article](http://mathworld.wolfram.com/HilbertCube.html) use the latter definition, so the product measure of the entire space in this case would be $1$. If the entire space had measure $0$, the measure would be the zero measure, which can be defined on any space, product or not.2012-11-08
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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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