# A curious compactness confusion: space filling curves in the hilbert cube that contradict a bona fide theorem?

Now, I may have only slept two hours last night and would currently struggle to discern a 'proof' by induction of FLT from a piece of genuine mathematics, but that doesn't stop mathematics from bugging me. At present I am puzzled by something I saw on MO this morning...

The linked question concerns the Hilbert cube $[0,1]^\mathbb{N}$ (an infinite product of intervals) and the existence of space filling curves thereof- that is: continuous images of the unit circle that are surjections on the Hilbert cube. The accepted answer, together with another answer (which actually constructs such a map) and various comments, seems to allude toward an answer in the affirmative. However, the linked theorem (the 'Hahn–Mazurkiewicz theorem') which states:

A nonempty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.

seems in direct contradiction to this since (and I may be mistaken for reasons explained above):

• The Hilbert cube is a subset of a normed space and hence a metric space
• The sequence $(1,0,0...), (0,1,0...), (0,0,1,...)$ has no convergent subsequence
• So the Hilbert cube is not sequentially compact, therefore non-compact (the two are equivalent in metric spaces).

Which seems at odds with the only if portion of the theorem's statement. Maybe this is wikipedia taking me for a ride. Maybe I am just hallucinating a portion of this argument. Either way, this is annoying me. Thanks in advance for clearing this up...

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[Slaps palm to forehead] aaah... thanks @Samuel, @Pete: I knew I had dropped a clanger somewhere! I think, in fairness, this has as much to do with me skipping Hilbert space lectures in second year (once I got the general idea [stuck?- see \ell^2], it seemed pointless attending...) as my tiredness, and I believe you have cleared up a genuine misconception of mine so +lots to you both! :) -ps Accepting @Samuel's answer for being fastest off the mark. – Tom Boardman Aug 11 '10 at 15:53
You also may want to recall Tychonoff theorem, stating that the product of compact spaces is compact; this is easy to remember and surely clears that what fails is your argument is that the Hilbert cube is compact. By the way, if you know that the Hilbert cube is the image of an interval, there is no need appeal to Hahn-Mazurkiewicz theorem, since trivially the image of a compact space has to be compact. – Andrea Ferretti Aug 11 '10 at 16:00
+1 For Tychonoff, now not thinking of that is down to being tired! – Tom Boardman Aug 11 '10 at 16:03
Why has this question been voted down?- I know it's not exactly fascinating and I knew that it was going to be a dumb question when I asked it. But I don't think it's useless or unclear... is it? – Tom Boardman Aug 11 '10 at 17:45
no, I don't see anything wrong with it. If it makes you feel any better -- and it should; on SE sites, (+1) + (-1) = +8 -- I have just upvoted the question. – Pete L. Clark Aug 12 '10 at 3:57

$(1,0,0,\ldots), (0,1,0,\ldots)$ converges to $(0,0,0,\ldots)$, so your example doesn't contradict the compactness of the Hilbert cube.

$[0,1]^{\mathbb N}$ is homeomorphic to $[0,1]\times[0,1/2]\times[0,1/3]\times\cdots$ with the $\ell_2$-norm. So the norm of $(a_1,a_2,a_3,\ldots)$ is $\sum\left(\dfrac{a_n}{n}\right)^2$, and therefore your sequence converges to $(0,0,0,\ldots)$ since the norm of $(0,\ldots,0,1,0,\ldots)$ (with the 1 in the $n$th position) is $1/n^2$, which tends to 0.

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The sequence you give converges to $0$ in the product topology. For instance, convergence in the product topology is equivalent to pointwise (or coordinatewise) convergence, and your sequence converges coordinatewise to $0$. See e.g.

http://en.wikipedia.org/wiki/Pointwise_convergence

(Added: Samuel's answer proceeds differently from mine -- via an explicit metric on the "Hilbert cube" -- and is also correct.)

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