# Can a linear subspace in Banach space be the union of several other subspaces?

It's a well known exercise for students to prove that a linear subspace of $\mathbb{R}^n$ can't be expressed as the countable union of other subspaces. The proof is quite simple, including only the comparison of Lebesgue measure.

However, it is not true for the linear space of polynomials. It can be the union of the subspace of degree 1 polynomials, polynomials of degree $\leq 2$, degree $\le 3$, ....

By Baire's theorem, a Banach space cannot be covered by countable number of closed subspaces. So my question is whether a (Banach) vector space of uncountable dimension can be written the union of countably many proper subspaces.

PS: I know it is rude to consider union of linear subspaces--it may be more appropriate to consider sums.... Just like Goldbach, one should not try to add primes, but multiply them:)

• If you drop the countability, then it is easy: Every Banach space is the union of its one-dimensional subspaces.
– gerw
Jan 31, 2018 at 12:19
• it's of course confusing if you highlight a question and then say that it's easy... please make some effort to emphasize the question you want to ask. If you ask about non-closed subspaces, then the Banach structure has nothing to do, and the question becomes whether a vector space of uncountable dimension is the union of countably many proper subspaces.
– YCor
Feb 2, 2018 at 14:11
• @YCor You are right Feb 2, 2018 at 14:12

1) By the Baire theorem, a Banach space can never be expressed as a countable union of closed proper subspaces.

2) Over an arbitrary field, any vector space $V$ of infinite dimension is union of countably many proper subspaces. Indeed, let $I$ be a basis. Write $I=J\sqcup K$ with $J$ infinite countable, $J=\{j_i:i\ge 1\}$. Write $I_n=K\cup\{j_i:i\le n\}$. Let $V_n$ be the subspace generated by $I_n$. Then $V_n$ is a proper subspace and $V=\bigcup V_n$. (If you rather like hyperplanes, rather define $I_n=I\smallsetminus\{j_n\}$).

3) Just about the "exercise for students", here's an approach without Lebesgue measure: a finite-dimensional vector space $V$ over an uncountable field is not the union of countably many proper subspaces.

More generally, for any infinite cardinal $\kappa$, a finite-dimensional vector space $V$ cannot be written as union of $<\kappa$ proper subspaces. Proof by induction on $\dim(V)$:

• $\dim V=0$: there is no proper subspace, so one can't cover 0!
• $\dim V=1$: $\{0\}$ is the only proper subspace, so one can't cover nonzero elements;

• $\dim V\ge 2$: Indeed, assume we have such a covering. One can suppose these are all hyperplanes. Each hyperplane is defined by an element of the projective space $V^*$, which has cardinal $\kappa$. So there's a hyperplane $H$ that is distinct to all others of the given family; hence it is not contained in any of the other ones. Then by induction, $H$ is not covered by the union of other hyperplanes. This is a contradiction.

Let me complement YCor's answer in the case where the Continuum Hypothesis fails.

Suppose that $$X$$ is a Banach space with a long Schauder basis $$(e_\alpha)_{\alpha < \mathfrak{c}}$$ of cardinality continuum, for example, $$X = \ell_p([0,1])$$ for $$p\in [1,\infty)$$. Then every $$x\in X$$ is countably supported with respect to the basis. Denote by $$(e^*_\alpha)_{\alpha < \mathfrak{c}}$$ the coordinate functionals associated to the basis.

Bearing in mind that $$\aleph_1$$ is less than the continuum (we assume that CH fails), we pick $$\aleph_1$$ among the basis vectors, say, $$e_\alpha$$ for $$\alpha \in A$$, where $$A$$ has cardinality $$\aleph_1$$. Then,

$$X = \bigcup_{\alpha\in A} \ker e_\alpha^*,$$

because for $$x\in X$$ the set $$\{\beta< \mathfrak{c}\colon \langle e_\beta^*, x\rangle \neq 0\}$$ is countable.