Re: Accumulation point of a sequence that is not a subsequential limit

On Wed, 16 Jul 2008 fjblurt@xxxxxxxxx wrote:

I'm trying to develop some intuition about infinite dimensional spaces
and I ran into the issue that in non-first-countable spaces, a
sequence can have an accumulation point to which no subsequence
converges.

A much easier example is S = I^I with product topology.
This is given by Steen's "Counterexamples in Topology", p125.

Since I = [0,1] is compact, so is I^I.
For all n in N, define f_n:I -> I, f_n(x) = nth diget of
the binary expansion of x. For all n, f_n in S.

Since S is compact, { f_n | n in N } has an accumulation point.
Assume (f_nj)_j is subsequence that converges to f.

Then for all x in R, (f_nj(x))_j -> f(x), since the product
topology is the pointwise convergent topology.

Pick some p with f_nj(p) = 0 if j even, = 1 if j odd.
Thus (f_nj(p))_j doesn't converge, which is contradiction.

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