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I have touched the Examples-section at sequentially compact topological space:
moved the detailed discussion of the compact space $\{0,1\}^{[0,1]}$ which is not sequ compact to the examples-section at compact topological space, and left a pointer to it,
added pointers (just pointers for the moment) to two detailed discussions of examples of sequ compact spaces that are not compact.
I have spelled out the standard counter-example of a compact space that is not sequentially compact, here
There’s something I find slightly odd about the description of the example of #2. It seems to me to be more sensible and less confusing to use $2^\mathbb{N}$ (the set of functions $f: \mathbb{N} \to 2$) as the exponent, than the half-open interval $[0, 1)$ that’s in the article. The only use made of $[0, 1)$ is that its elements have binary representations, but this creates a distraction for the reader: do we have to worry about the ambiguity at dyadic rationals? And why a half-open interval; is there some subtle reason why $[0, 1]$ isn’t used instead? Also, the spatial picture we have of $[0, 1)$ as a half-open interval is irrelevant to the construction.
If we use $Disc(\{0, 1\})^{2^\mathbb{N}}$ with underlying set $2^{2^\mathbb{N}}$ instead, then the desired sequence $(x_n)_{n: \mathbb{N}}$ is simply the double-dual embedding $\mathbb{N} \to 2^{2^\mathbb{N}}$, defined by $(x_n)_f = f(n)$.
The rest of the proof is more or less followable, but personally I find it helpful to be very definite about the choice of open set (rather than switch from $r$ to $r'$ midstream, as it were). So I might write it like this. Suppose some subsequence $x_{n_k}$ converged to some $x$. Now choose any $f: \mathbb{N} \to 2$ that is not eventually constant on the subsequence $(n_k)_{k: \mathbb{N}}$; for example, define $f: \mathbb{N} \to 2$ by $f(n_k) = k\; mod\; 2$, else $f(n) = 0$ if $n$ does not appear in the subsequence. Consider the open set $U_f = \{x_f\} \times \prod_{g: g \neq f} \{0, 1\}$. In order to have $x_{n_k} \in U_f$ for all $k \geq k_0$, we’d have to have $f(n_k) = x_f$ for all $k \geq k_0$, in other words $f$ would be eventually constant on the subsequence $n_k$. Contradiction.
Hi Todd,
if you have the energy, please feel invited to improve/rewrite the example.
Thanks, Urs. Done.
Thanks for the alert.
The passage in question was added in revision 11 by Andrew Stacey (who is no longer active here), over 10 years back.
Since you just looked into it, could you just fix/rewrite the proof?
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