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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.


marked as duplicate by Asaf Karagila, Namaste, Xander Henderson, Parcly Taxel, user99914 Jul 18 '18 at 2:32

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  • $\begingroup$ Perhaps an integral? $\endgroup$ – Adrian Keister Jul 17 '18 at 14:38
  • $\begingroup$ I feel like this question is missing an answer addressing Schauder bases. $\endgroup$ – R.. Jul 18 '18 at 1:08

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.

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).


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.

  • 1
    $\begingroup$ Presumably you want the $f \in B$ to take some fixed value, otherwise all multiples of an element of $B$ are also in $B$. $\endgroup$ – Mees de Vries Jul 17 '18 at 14:47
  • $\begingroup$ Thanks for catching that. I did, I meant it to have value one. $\endgroup$ – copper.hat Jul 17 '18 at 14:47
  • $\begingroup$ Thank you for your answer. Now, how do you write a vector as a linear combinaison of uncountable vector ? By the way, in uncountable basis, is it possible to have 2 basis with a different number of elements ? $\endgroup$ – Peter Jul 17 '18 at 15:23
  • $\begingroup$ @Peter - if $f : S \to V$ for some vector space $V$ satisfies that $f^{-1}(V\setminus\{0\})$ is finite, then there is some enumeration $\{s_k\}_{k=1}^n$ of $f^{-1}(V\setminus\{0\})$. It is customary to define $$\sum_{s \in S} f(s) := \sum_{k=1}^n f(s_k)$$So the answer to your first question is simply $$f = \sum_{x\in\Bbb R} f(x)1_{\{x\}}$$. The answer to your second question is "no", but it takes more explaining than there is room for in a comment. $\endgroup$ – Paul Sinclair Jul 17 '18 at 16:55
  • $\begingroup$ @PaulSinclair: I didn't know that $\sum_{x\in\mathbb R}f(x)1_{\{x\}}$ make sense... I always thought that such a sum where an integral (continuous sum). $\endgroup$ – Peter Jul 17 '18 at 17:05

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 $F$.

  • $\begingroup$ I think you mean "which take non-zero values only at finitely many points". $\endgroup$ – Mees de Vries Jul 17 '18 at 14:44
  • $\begingroup$ Indeed. I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Jul 17 '18 at 14:46
  • $\begingroup$ Thank you for your answer. Now, how do you write a vector as a linear combinaison of uncountable vector ? By the way, in uncountable basis, is it possible to have 2 basis with a different number of elements ? $\endgroup$ – Peter Jul 17 '18 at 15:22
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    $\begingroup$ When we write a vector as linear combination of infinitely many vectors, then all but finitely many coefficients are equal to $0$. And any two bases of a vector space have the same cardinal. $\endgroup$ – José Carlos Santos Jul 17 '18 at 15:24

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.


$\ell^p$ 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 $\dim V = \operatorname{card} V$.

  • $\begingroup$ What do you mean by : they have a dimension c ? if so, they are of finite dimension, and thus can't have a uncountable basis... $\endgroup$ – Peter Jul 17 '18 at 19:42
  • $\begingroup$ @Peter $c$ is the cardinality of $\mathbb{R}$. See here. $\endgroup$ – mechanodroid Jul 17 '18 at 19:47
  • $\begingroup$ You should probably clarify that you mean Hamel basis, since that's not what one normally means when talking about the basis for an $\ell^p$ space. $\endgroup$ – R.. Jul 18 '18 at 1:07
  • $\begingroup$ @R.. I believe it's clear I mean Hamel basis because the question is about Hamel bases. $\endgroup$ – mechanodroid Jul 18 '18 at 1:11
  • $\begingroup$ Maybe it's just differences of backgrounds/conventions, but I wouldn't use names for the spaces that imply a topology unless you're treating them with the topology or making it explicit that you're not. $\endgroup$ – R.. Jul 18 '18 at 1:15

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