# What if the basis not countable, then what?

I'm second year physics & mathematics student, and self-studying Abstract linear algebra from the book Linear Algebra by Werner Greub.

In the mean time I have come across several times to the notion of countable basis.I know what can or can not do if a some set is countable/uncountable, but while studying linear algebra I do not exactly know what I couldn't do if the basis of space is not countable ?

I generally worked either on abstract (finite / infinite) spaces or finite abstract spaces, and while in the abstract case, we have never assumed that the basis is countable, so I'm not sure what I would lose if I have a space of infinite dimension whose basis is uncountable.

tl;dr

If our vector space has a basis that is not countable, then which properties would be lost or gained compared to the case where we have space having countable basis (finite or infinite).

• One example would be that if an infinite-dimensional normed vector space has a countable basis, then it cannot be complete: see here. – mechanodroid Oct 1 '17 at 10:15
• @mechanodroid The question is about an uncountable bases though. – Wojowu Oct 1 '17 at 10:23
• @ToanPham The infinite dimensional vector space of polynomials of arbitrary degree have infinite, but countable basis. – Arthur Oct 1 '17 at 10:39
• @Peter, yes. For example, the vector space of all functions from the reals to the reals. – Gerry Myerson Oct 1 '17 at 12:11
• Most of the time (not always) the useful infinite dimensional spaces occur in analysis. There you often have uncountable dimension in the strict sense (finite linear combinations of basis elements) but the useful bases are those that span when you allow infinite sums, with a suitable metric or topology. As an example, think of the space of sequences where the sum of the squares converges. – Ethan Bolker Oct 1 '17 at 12:15

For an infinite dimensional vector space we must distinguish form an Hamel (algebraic) basis and a Schauder basis.

The first concept is defined for any vector space and the axiom of choice guarantees that any vector space have a Hamel basis, but this basis might be uncountable.

In such a basis any vector can be expressed as finite linear combination of elements of the basis.

The simpler case is the space of real numbers $\mathbb {R}$ considered as a vector space over the rationals $\mathbb{Q}$ that has an infinite non-countable basis. In this case we cannot have a '' construction'' that enumerate all the elements of the basis.

A Schauder basis can be defined only for a topological vector space, where we can define the convergence of a series. In this case a vector can be expressed as a series ( an ''infinite'' linear combination) of elements of the Schauder basis.

A simple example of a space that has a non countable Hamel basis and a countable Schauder basis is the space of the functions $L^p[0,\pi]$ over $\mathbb{C}$ where the set of functions $\{e^{nix}\: n \in \mathbb{Z}\}$ is a Schauder basis.

A vector space $W$ of countably infinite dimension is never isomorphic to the dual $V^\star$ of some other vector space $V$.

To see this take a basis $B$ of such a hypothetical $V$. Clearly $B$ must be infinite. But then we can find an uncountable subset $Q\subset \mathcal P(B)$ of the powerset of $B$ such that for each $p,q\in Q$ the intersection $p\cap q$ is finite (see here). Then the uncountable family $\{\chi_q\mid q\in Q\}\subset V^\star$ (where $\chi_q(b)=1$ iff $b\in q$ and $\chi_q(b)=0$ otherwise) is linearly independent (see here for more details). Hence $V^\star$ ist not of countable dimension.

• what is $\chi_q$ ? – onurcanbektas Jul 13 '18 at 3:42
• Sorry, I had a typo. $\chi_q$ is the element in $V^\star$ defined by the map which sends $b\in B$ to $1$ if $b\in q$ and to $0$ otherwise. – Tashi Walde Jul 13 '18 at 7:30