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

Question

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 exitence 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...

Answer 1

$(1,0,0,\ldots), (0,1,0,\ldots)$ 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.

Answer 2

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.)