Are there vector spaces with uncountable basis?

Question

Are there vector spaces with uncountable basis ? I was thinking about something as $L^1(\mathbb R)$. A could imagine that $\varphi_x:\mathbb R\to \mathbb R$ defined as $\delta_x(y)=1$ if $y=0$ and $0$ otherwise can generate all function and is uncountable. Moreover there are linear independent (but I'm not sure).

But for an uncountable basis, how we would write for example $\sum_{x\in\mathbb R}f(x)\delta_x$ ? It looks weird, no ?

In general if $V$ has an uncoutable basis $\{e_t\}_{t\geq 0}$, and if $v\in V$, how write $$v=\sum_{t\geq 0}v_te_t,\ \ ?$$ I guess that the previous notation has no sense.

Answer 1

For any set $X$, consider maps $f:X \rightarrow \mathbb R$ such that $f(x)=0$ for all but a finite number of $x$. These form a vector space, with basis $\{\delta_x\}$, where $\delta_x(x)=1$ and $\delta_x(y)=0$ when $x \neq y$. So, the number of basis elements is the same as the cardinality of $X$. Take $X$ uncountable, and this space will have uncountable basis.

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Some confusion may arise from trying to sum an infinite (e.g. uncountable) number of vectors. However, we don't do that! A basis allows any vector be decomposed into a linear combination of basis vectors, and linear combinations are finite by definition (or, equivalently, infinite, but having only a finite number of non-zero coefficients).

Answer 2

Might be worth noting that there are two sorts of bases that are commonly used.

A Hamel basis is one in which any element can be written as a finite linear combination of elements of the basis. A Schauder basis is similar, but one takes the closure of the linear span and hence a topology is needed.

I presume you are looking for an uncountable Hamel basis.

Let $V = \{ f: \mathbb{R} \to \mathbb{R} | f^{-1} (\{ 0\}^c) \text { is finite}\}$ with the usual function addition and scalar multiplication.

It is not hard to see that $B = \{ 1_{\{x\}} \}_{x \in \mathbb{R}}$ is a basis and is uncountable.

Answer 3

Take the space $F$ of all functions from $\mathbb R$ into itself which take non-zero values at finitely many points only. For each $x\in\mathbb R$, let$$e_x(y)=\begin{cases}1&\text{ if }y=x\\0&\text{ otherwise.}\end{cases}$$Then the $e_x$'s form an uncountable basis of $F$.

Answer 4

Yes, there are. In the definition of a vector space you can have as many basis vectors as you want as long as they are linearly independent. We require that the sum for expressing a vector in terms of the basis have a finite number of nonzero terms, so your sum notation makes sense. You can regard the reals as a vector space over the rationals. The basis can have one rational number in it, but you need uncountably many reals so that any real can be expressed as a finite linear combination. The requirement that the sums be finite avoids any complication of convergence of infinite sums.

Answer 5

$\ell^p$ spaces also have uncountable bases.

Namely, the set of these geometric sequences

$$\{(1,t,t^2, t^3, \ldots) \in \ell^p: t \in \langle 0, 1\rangle\}$$

is linearly independent.

$C^k(\mathbb{R})$ spaces also have uncountable bases. The set

$$\{e^{\lambda x} : \lambda \in \mathbb{R}\}$$

is linearly independent.

The above spaces actually have dimension $c$.

On the other hand, the space $\mathbb{R}^\mathbb{R}$ of functions $\mathbb{R} \to \mathbb{R}$ has dimension $2^c$. This is because every vector space $V$ over $\mathbb{R}$ or $\mathbb{C}$ with $\dim V \ge c$ has in fact $\dim V = \operatorname{card} V$.