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Let $I = [0,1]$ be the compact unit interval and $T = I^I$ the Tychonoff cube.

It is pretty standard to exhibit a sequence in $T$ with no convergent subsequence. It is also fairly standard to show that $T$ is separable (e.g. the polynomials with rational coefficients are dense). A simple cardinality argument can be used to show that this sequence of polynomials has cluster points which are not the limit of any of its subsequences.

Now, I believe, but cannot prove, that we may be able to exhibit a sequence in $T$ that has no convergent subsequence, but that its closure is all of $T$. In particular, I am interested constructing such a sequence $(f_n(t))_{n=1}^\infty$ and for each point $f \in T$ exhibiting a subnet of $(f_n(t))_{n=1}^\infty$ that converges to $f$.

Any ideas?

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    $\begingroup$ What leads you to believe such a sequence exists? $\endgroup$
    – Math1000
    Dec 24, 2019 at 2:16
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    $\begingroup$ @Math1000 A combination of Tychonoff's theorem and the empirical evidence that a high cardinality space can have a (relatively) low density character. We can exhibit sequences with no convergent subsequences which, by compactness, cluster somewhere, and we can exhibit sequences that cluster everywhere. I'm simply curious to see whether both can be achieved simultaneously. $\endgroup$ Dec 24, 2019 at 17:07
  • $\begingroup$ @JoseAvilez I can't read "A simple cardinality argument can be used to show that this sequence of polynomials admits cluster points for which it has no convergent subsequence." Specifically, I get confused at "for which it has no convergent subsequence". Am I being stupid, or should you rephrase? $\endgroup$ Dec 27, 2019 at 17:07
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    $\begingroup$ @mathworker21 This sentence means that if you take an ordering of the set of polynomials with rational coefficients to form a dense sequence $(p_n)_{n \geq 1}$ then this sequence will have a cluster point $x$ for which there is no subsequence $p_{n_k}$ such that $p_{n_k} \to x$. Indeed, since $(p_n)_{n \geq 1}$ is dense, every point in $T \setminus \{p_n: n \geq 1\}$ is a cluster point of $p_n$. To see the existence of $x$, then note that the set of subsequences of $p_n$ has cardinality $2^{\aleph_0}$ but $T \setminus \{p_n: n \geq 1\}$ has cardinality $2^{2^{\aleph_0}}$. $\endgroup$ Dec 27, 2019 at 17:28
  • $\begingroup$ Try $f_n=\sin(n)$ $\endgroup$
    – Hyperplane
    Dec 28, 2019 at 7:49

1 Answer 1

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There are distinct irrationals $b_t$ for $t\in I$ such that $\{1\}\cup\{b_t\mid t\in I\}$ is a $\mathbb Q$-basis for $\mathbb R.$ Define

$$f_n(t)=nb_t-\lfloor nb_t\rfloor$$ i.e. it's the fractional part of $nb_t.$

The set $\{f_n\}$ is dense

For any finite subset $J\subset I,$ the set $\{(nb_t)_{t\in J}\mid n\in\mathbb N\}$ is dense in $(\mathbb R/\mathbb Z)^{J}$ by Weyl's equidistribution criterion; see for example https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/ Exercise 5. Equidistribution is stronger than density, and density in $(\mathbb R/\mathbb Z)^{J}$ is the same as density in $[0,1)^{J}.$

No subsequence $f_{n_k}$ is convergent

Fix $n_k.$ Convergence will always mean as $k\to\infty.$

For each positive integer $m,$ the set of reals $r$ such that $\exp(2\pi i n_kmr)$ converges is a null set - see zhw's answer at https://math.stackexchange.com/a/1380389/467147. So the union over $m$ is also a null set. Pick $r$ such that for all positive integers $m,$ the sequence $\exp(2\pi i n_kmr)$ does not converge.

By the $\mathbb Q$-basis property, there are rationals $q_t$ for $t\in I,$ at most finitely many non-zero, such that $r-\sum_tq_tb_t\in\mathbb Q.$ By clearing denominators, there is a positive integer $m$ and integers $m_t$ such that $$mr-\sum_tm_ib_t\in\mathbb Z.$$

Since $\exp(2\pi i n_kmr)$ does not converge but equals $\prod_t\exp(2\pi i n_km_tb_t),$ there is some $t\in I$ such that $\exp(2\pi i n_km_tb_t)=\exp(2\pi i f_{n_k}(t))^{m_t}$ does not converge. Since $x\mapsto \exp(2\pi i x)^{m_t}$ is continuous on $[0,1),$ the sequence $f_{n_k}(t)$ cannot converge.

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