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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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    $\begingroup$ The diagonal length makes sense; see Wikipedia. I don't see how to make sense of the volume of the Hilbert cube. $\endgroup$
    – joriki
    Commented Nov 4, 2012 at 22:29
  • $\begingroup$ 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. $\endgroup$
    – Alex Petzke
    Commented Nov 8, 2012 at 4:26
  • $\begingroup$ 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. $\endgroup$
    – joriki
    Commented Nov 8, 2012 at 5:48
  • $\begingroup$ 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 and the MathWorld article 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. $\endgroup$
    – joriki
    Commented Nov 8, 2012 at 5:53
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    $\begingroup$ @joriki Yes, I have seen that alternate definition. But in 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. $\endgroup$
    – Alex Petzke
    Commented Nov 8, 2012 at 19:21

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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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    $\begingroup$ You've 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. $\endgroup$
    – user856
    Commented Jan 27, 2013 at 16:00

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