Functional Analysis – Cardinality of Projections of Subsets of the Hilbert Cube by Inner Products

Question

I have three related questions.

Question 1: Is there a subset $X$ of the Hilbert cube $[0,1]^{\Bbb N}$ of cardinality continuum, such that for each sequence $a\in [0,1]^{\Bbb N}$ with $\sum a_n$ finite, the cardinality of the projection
$$\Big\{\sum_{n=1}^\infty x_na_n : x\in X\Big\}$$
is smaller than the continuum?

Question 2: Is there consistently such a set?

It would have been awesome if there is, provably, no such set $X$. Too good to be true?

Question 3: What if the Continuum Hypothesis holds?

It is known that if the cardinality of $X$ is smaller than the continuum, then there is a bijective projection as above. This is proved in Carlson's paper Strong measure zero and strongly meager sets, using analytic functions. The argument does not extend to sets of cardinality continuum.

Answer 1

Consistently, there is no such set.

Given $X\subset [0,1]^{\mathbb N}$ of cardinality continuum it suffices to find $z\in(0,1)$ such that $x\mapsto \sum_n x_n z^n$ is injective on some subset $X'\subseteq X$ of cardinality continuum. This is then more or less the same setting as [1, Theorem 2.1] where it is shown that $z$ exists in certain Cohen forcing extensions. The differences are that we want a real $z\in(0,1)$ instead of $z\in\mathbb C,$ and the functions are not entire, they're just holomorphic in the unit disc. But the proof goes through with those changes.

[1] Kumar, Ashutosh; Shelah, Saharon, On a question about families of entire functions, Fundam. Math. 239, No. 3, 279-288 (2017). ZBL1390.03044. Also at Shelah's archive.

Under CH, there is such a set.

Assume CH. Take a Hamel basis $\{b_\beta: \beta\in\omega_1\}$ of $\ell^1(\mathbb N),$ the space of absolutely summable sequences. Define $\phi_\beta$ to be the projection by $b_\beta,$ so $\phi_\beta(x)=\sum_n b_\beta(n) x(n)$ for all $x$ in the space of bounded sequences $\ell^\infty(\mathbb N).$

Lemma. For each $\alpha\in\omega_1\setminus\omega$ there exists $x_\alpha\in[0,1]^{\mathbb N}$ such that $\phi_\beta(x_\alpha)$ is rational for $\beta<\alpha$ and irrational for $\beta=\alpha.$ (The restriction to infinite $\alpha$ is just to avoid having to treat the finite case.)

Proof. Pick a bijection $\beta:\omega\to\alpha+1.$ The functions $\phi_{\beta(n)}$ are linearly independent (as functionals on $\ell^\infty(\mathbb N)$). By linear algebra, for each $n\in\omega$ we can pick $z_n\in \ell^\infty(\mathbb N)$ such that $\phi_{\beta(m)}(z_n)=0$ for $m<n$ and $\phi_{\beta(n)}(z_n)\neq 0.$ Let $c=(\tfrac12, \tfrac12, \dots).$ Let's try a solution of the form $$x_\alpha=c+\sum_{n=0}^\infty t_nz_n$$ with $$\max_{i}|t_nz_n(i)|\leq 2^{-n-2}\tag{1}$$ for each $n\in\omega.$ This ensures that $x_\alpha\in[0,1]^{\mathbb N}.$ Using absolute convergence to justify swapping sums, for all $n$ we have \begin{align*} \phi_{\beta(m)}(x_\alpha) &=\phi_{\beta(m)}(c)+\sum_{n=0}^{\infty} t_n\phi_{\beta(m)}(z_n)\\ &=\phi_{\beta(m)}(c)+\sum_{n=0}^{m} t_n\phi_{\beta(m)}(z_n).\tag{2} \end{align*} In the last equality, the $t_n$ terms have vanished for $n>m$ because then $\phi_{\beta(m)}(z_n)=0.$ The coefficient $\phi_{\beta(m)}(z_m)$ of $t_m$ is non-zero. So it's easy to construct $t_m$ by induction on $m$ satisfying (1) (with $n=m$) and such that (2) is rational iff $\beta(m)\neq \alpha.$ $\square$

Pick such an $x_\alpha$ for each $\alpha\in\omega_1\setminus\omega$ and take $X=\{x_\alpha:\alpha\in\omega_1\setminus\omega\}.$ It's uncountable because the $x_\alpha$ are distinct: $\phi_\beta(x_\alpha)\neq \phi_\beta(x_\beta)$ for $\beta<\alpha.$ All the projections by basis vectors are countable: $\phi_\beta[X]\subseteq\mathbb Q\cup \{\phi_\beta(x_\alpha):\alpha\in(\beta+1)\setminus\omega\}.$ We can write an arbitrary absolutely summable sequence as $a=\sum_{\beta} \lambda_\beta b_\beta$ with reals $\lambda_\beta,$ at most finitely many non-zero. The projection by $a$ is then contained in the countable set $\{\sum_{\beta:\lambda_\beta\neq 0}\lambda_\beta q_\beta : q\in \prod_{\beta:\lambda_\beta\neq 0}\phi_\beta[X]\}.$