A subset of a linear space $X$ is called *infinite-dimensional* if it is not contained in a finite-dimensional linear subspace of $X$.



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> **Problem.** Let $L$ be an infinite-dimensional subset of the linear space $\mathbb R^\omega$. Is there an infinite set $I\subseteq\omega$ such that for every infinite set $J\subseteq I$ the projection of $L$ to $\mathbb R^J$ is infinite-dimensional?
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**Remark.** The answer to this problem is affirmative if every function $f\in L$ has finite range.



Infinite-dimensional projections of linearly independent sets
For $n :=\{0,1,\dots,n-1\}\in\omega$ let’s denote $P\_n:\mathbb{R}^\omega\to \mathbb{R} ^n$ the projection, which is the restriction map $f\mapsto f\_{|n}$. The dimensions of the subspaces $P\_n(L)\subset \mathbb{R} ^n$ can’t be stationary, because that would mean that for some $n\_0$, every function $f\in P\_{n\_0}(L)$ has a unique extension to a function $\tilde f\in L$, in which case $ P\_{n\_0}(L)\ni f\mapsto \tilde f\in L$ is a linear bijection, whereas $L$ was assumed infinite-dimensional. So let $(n\_k)\_k$ be a strictly increasing sequence such that $\dim P\_{n\_k+1}(L)> \dim P\_{n\_k}(L)$. Thus $\ker \big(P\_{n\_k}: P\_{n\_k+1}(L)\to P\_{n\_k}(L) \big)\neq(0)$: there is a function $f\_k\in L$ such that $f\_k|{n\_k}=0$ and $f\_k(n\_k+1)=1$. So if we take $ I:=\{n\_k+1\}\_{k\in\omega}$, for any infinite $J\subset I$ The space $P\_J(L)$ contains the functions $P\_Jf\_k={f\_k}\_{|J}$, which are clearly linearly independent.


