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As the title suggests, I want a hyper-cube in half of one dimension. If you could provide an image, that would be superb. I have no idea how I would do this; I can't connect hyper-cubes of a lower level to make this. Help much appreciated.

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    Is there a reason you think there should be such a thing? – pjs36 Nov 21 '15 at 00:07
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    Well, you can have fractional dimensions called fractal dimensions, so, as a matter of fact, yes. – Simply Beautiful Art Nov 21 '15 at 00:11
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    There are various notions of fractional dimensions, or more precisely, notions of "dimension" that can take on fractional values. It does not follow that such a thing as "hyper-cube in half of one dimension" exists. You should define this for us. – hardmath Nov 21 '15 at 00:13
  • See... some of us can't define this for you because we are not good enough. You could edit my question to make it more precise if you wish. – Simply Beautiful Art Nov 21 '15 at 00:14
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    I think "Does the notion of a fractional-dimensional cube (or any polytope) make sense?" would be a much better question. Fractional dimensions are certainly something I know $\epsilon$ about, but I think the comments are merely pointing out that assuming such a thing exists is a pretty big assumption - perhaps lowering your expectations would make for a better question with a more interesting answer. But as you have an upvoted answer, you'd have to ask another question. – pjs36 Nov 21 '15 at 00:32

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One way to make this precise would be to define the $n$-cube as $I^n$, where $I=[0,1]$ as a topological space or metric space. The $1/2$-cube would then be a topological space $X$ such that $X^2$ is homeomorphic to $I$. This is impossible, as shown in https://mathoverflow.net/questions/60375/is-mathbb-r3-the-square-of-some-topological-space. The idea is that $X\times X=I$ means that $X$ is path connected, from which it follows that $(X\times X)\setminus\{p\}$ is path connected for all points $p$. This is not true for $I$, so we have a contradiction.