A sequence with no converging subsequence that clusters everywhere

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?

Solution 1:

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.