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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 19th 2017

I have touched the Examples-section at sequentially compact topological space:

1. 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,

2. added pointers (just pointers for the moment) to two detailed discussions of examples of sequ compact spaces that are not compact.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeMay 10th 2017

I have spelled out the standard counter-example of a compact space that is not sequentially compact, here

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeJul 26th 2018

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.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJul 26th 2018

Hi Todd,

if you have the energy, please feel invited to improve/rewrite the example.

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeJul 26th 2018

Tried to clarify example 4.1.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeJul 26th 2018

Thanks, Urs. Done.

• CommentRowNumber7.
• CommentAuthorGuest
• CommentTimeOct 16th 2020
The proof that a countable product of sequentially compact spaces is again sequentially compact contains an error. There is no proof that (\alph_{l_m})_{m\in \N} is a subsequence. In fact, the construction in the proof does not ensure that it is a subsequence. For example, if the original sequence actually converges then in construction given in the proof one can take A_m=\N for each m and then min\{ A_l \}= 1 for each l. In this case, one does not get a subsequence (rather one gets the constant sequence alpha_1, alpha_1, alpha_1, etc).
• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeOct 17th 2020

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?

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeDec 21st 2020

I have rewritten the proof that a countable product of sequentially compact spaces is again sequentially compact.

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeDec 21st 2020

I have rewritten the proof that a countable product of sequentially compact spaces is again sequentially compact.