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Can someone give me an example of a compact topological space that is not first-countable such that there is a sequence $(x_n)_{n \in \mathbb{N}}$, with the property that for every subsequence $(x_{n_p})_{p \in \mathbb{N}}$ of $(x_n)_{n \in \mathbb{N}}$, the subsequence wouldn't converge?

In our topology course we learned that in a compact space, every net has a convergent subnet. Now a sequence is of course also a net, so it has to have a convergent subnet. But generally a subnet of a sequence need not be a sequence; but could it still be possible in compact space to choose as subnet of a sequence another sequence? It is of course important that this space should not be first countable, because if it were we wouldn't have to deal with the cumbersome nets. Compactness and first-countable would imply that every sequence contains a convergent subsequence, so my question would be meaningless.

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    $\begingroup$ According to Stephen Willard (problem 17G.1 is his $General\ Topology$, an uncountable product of copies of $[0,1]$ is compact but not sequentially compact (it's not first countable either). So, there is an an example in this space (I don't see how to get a particular sequence verifying that it is not sequentially compact, though). $\endgroup$ – David Mitra Feb 17 '12 at 18:01
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    $\begingroup$ @DavidMitra: You don't have to use $...$ to get italic, and if you do, it has improper spacing anyway. Note how the letter spacing in "$some\ text$" isn't normal. This website uses MarkDown which lets you use *...* for italics: "some text". $\endgroup$ – kahen Feb 18 '12 at 4:55
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Take $X = \{0,1\}^{[0,1]}$, thought of as the set of all functions from the unit interval $[0,1]$ to the two-point set $\{0,1\}$. Equip $X$ with the product topology; $X$ is compact by Tychonoff's theorem. In the product topology, a sequence $f_n$ converges iff it converges pointwise, i.e. if the $\{0,1\}$-valued sequence $f_n(x)$ converges (i.e. is eventually constant) for every $x \in [0,1]$.

Define $f_n : [0,1] \to \{0,1\}$ to be the function such that $f_n(x)$ is the $n$th bit in the binary expansion of $x$. (If $x$ has more than one binary expansion, take $f_n(x) = 0$ for all $n$; it doesn't really matter here.) Then given any subsequence $f_{n_m}$, you should be able to produce an $x \in [0,1]$ such that $f_{n_m}(x)$ does not converge. Hence $\{f_n\}$ has no convergent subsequence.

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  • $\begingroup$ For instance, take $x\in[0,1]$ as the number whose $n_k$'th binary digit is 0 for $n_k$ odd and 1 for $n_k$ even. $\endgroup$ – David Mitra Feb 17 '12 at 18:35
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    $\begingroup$ A slightly neater presentation (from my point of view) is to let $X=\{0,1\}^{\wp(\mathbb{N})}$ and $$f_n:\wp(\mathbb{N})\to\{0,1\}:A\mapsto\chi_A(n)\;.$$ If $A\subseteq\mathbb{N}$ is infinite, let $B\subseteq A$ be such that $|B|=|A\setminus B|=\omega$; then $\langle f_n(B):n\in A\rangle$ does not converge. $\endgroup$ – Brian M. Scott Feb 18 '12 at 3:28
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My favorite example is $\beta\omega$, in which the sequence $\langle n:n\in\omega\rangle$ has no convergent subsequence. To see this, let $A$ be any infinite subset of $\omega$, and suppose that $\langle n:n\in A\rangle\to p$ for some $p\in\beta\omega$. Clearly $p\in\beta\omega\setminus\omega$, so $p$ is a free ultrafilter on $\omega$. Partition $A$ into two infinite subsets $A_0$ and $A_1$. Since $p$ is an ultrafilter, exactly one of $A_0$ and $A_1$ belongs to $p$, say $A_0$. But then $A_0\cup\{q\in\beta\omega\setminus\omega:A_0\in q\}$ is an open nbhd of $p$ that misses infinitely terms of $\langle n:n\in A\rangle$, so $\langle n:n\in A\rangle$ that does not converge to $p$ after all.

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  • $\begingroup$ And just to add a comment: Using an extension of the idea presented by Brian, it can be shown that no one-to-one sequence in $\beta \omega$ converges. $\endgroup$ – user642796 Feb 18 '12 at 4:54
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    $\begingroup$ This is really clever,but I prefer Nate's example since it uses spaces that are bit more familiar to most mathematics students.These set theoretic constructs are a bit more alien to most of us and it takes a few minutes to decode it.But it's a matter of preference,of course-it's a perfectly valid counterexample. $\endgroup$ – Mathemagician1234 Feb 18 '12 at 5:34

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