Does $\mathbb{R}^\mathbb{R}$ have a basis?

If $V$ is vector space over a field $F$ then a subset $B$ is called Hamel basis (or simply a basis) for $V$ if $B$

  • is linearly independent, which means that for any finite subset $\{b_1,\dots,b_k\}\subseteq B$ the implication $$c_1b_1+\dots+c_kb_k=0 \Longrightarrow c_1=\dots=c_k=0$$ holds for any $c_1,\dots,c_k\in F$.
  • generates $V$, which means that any $v\in V$ can be expressed in the form $$v=c_1b_1+\dots+c_kb_k$$ for some $b_1,\dots,b_k\in B$, i.e. as a finite linear combination.

In the case of finitely-dimensional spaces this is precisely the same thing as the usual notion of basis, which is usually introduced in the first course in linear algebra.

Many properties valid for the finitely-dimensional case are true for Hamel basis as well, e.g.:

  • $B$ is a basis if and only if every vector $v\in V$ can be uniquely expressed as $v=c_1b_1+\dots+c_kb_k$, i.e. as the linear combination of the basic vectors.

  • For any linearly independent set $A$, there exists a basis $B\supseteq A$. (The proof of this fact uses Zorn's lemma which is equivalent to axiom of choice.) This result implies that every vector space has a Hamel basis.

  • If you prescribe the values of a map $f: V\to W$ for elements of $B$ (where $W$ is a vector space), this determines uniquely a linear map $f:V\to W$. (More formally: For any $g:B\to W$ there is there exists exactly one linear map $f:V\to W$ such that $f|_B=g$.)

  • If $B_1$, $B_2$ are Hamel bases of a vector space $V$, then $|B_1|=|B_2|$, i.e. any two bases of the same space have the same cardinality. Hence it is possible to define Hamel dimension of a vector space.

As I have mentioned above, the proof that every vector space $V$ has a basis uses Axiom of Choice. In fact, this claim is equivalent to Axiom of Choice. This means that we are not able (at least not for all spaces) to write down explicitly a basis. In some cases it might be possible.

EXAMPLE: Let $c_{00}$ be the space of all real sequences which have only finitely many non-zero terms. Then $\{e^{(i)}; i\in\mathbb N\}$, where the sequence $e^{(i)}$ is given by $e^{(i)}_n=\delta_{in}$, is a Hamel basis of this space. (Note that $c_{00}\subseteq\mathbb R^{\mathbb N}$. Thus we have just found infinite linearly independent subset of $\mathbb R^{\mathbb N}$, namely the set $\{e^{(i)}; i\in\mathbb N\}$. This implies $\mathbb R^{\mathbb N}$ cannot have finite basis. The argument that $\mathbb R^{\mathbb R}$ cannot have a finite Hamel basis is very similar.)

In fact, we can see quite easily that Hamel dimension of $\mathbb R^{\mathbb R}$ (as a vector space over $\mathbb R$ ) is $2^{\mathfrak c}$. Since the functions $\delta_x$ given by $\delta_x(x)=1$ and $\delta_x(y)=0$ for $y\ne x$ form a linearly independent set, we get that $|B|\ge|\mathbb R|$.

Every element of $V=\mathbb R^{\mathbb R}$ is determined by a finite sequence of pairs $(c,b)\in\mathbb R\times B$. The cardinality of all such finite sequences is $|\bigcup_{n\in\mathbb N} (B\times \mathbb R)^n| = \aleph_0 \cdot |B\times \mathbb R| = |B|$. Hence we get $|B|=|V|=|\mathbb R^{\mathbb R}|=2^{\mathfrak c}$.

You might notice that I stressed the word finite several times. In fact, if we only work with the structure of vector space, we are only able to do finite linear combinations. But if there is an additional structure on the space $v$, which enables us to define convergence (e.g. if we have topology, norm, metric) then we can also consider something like "infinite linear combinations" and ask whether there is a set $B$ such that every element $v\in V$ is expressible precisely in one way as $\sum_{b\in B} c_b\cdot b$. Such bases are indeed studied, example of this type of base is Schauder basis.

Let me mention a few references where you can learn more:

  • Christopher Heil. A Basis Theory Primer, New York, 2011. Draft version of that book is available at author's website.

  • Marek Kuczma. An introduction to the theory of functional equations and inequalities, Birkhäuser, Basel, 2009.

  • Or you can simply try to google for Hamel basis.

Typically, Hamel basis is often mentioned in books about set theory (since it is connected to axiom of choice), about functional analysis (but here other types of bases, like Schauder basis, are more interesting) or about functional equations (since it is related to Cauchy equation).

See also this question: basis of a vector space


Even though $\mathbb{R}^\mathbb{R}$ has a basis, it is not finite, because $\mathbb{R}^\mathbb{R}$ is not finitely spanned (there exists no finite set of vectors $\{v_1, ...,v_n\}$ so that $S[v_1,...,v_n] = \mathbb{R}^\mathbb{R}$).

A basis for $\mathbb{R}^\mathbb{R}$ has not been found yet, I believe, so unfortunately, we won't be able to give you one.


The dimension of $\mathbb R^\mathbb R$ over $\mathbb R$ is $2^{\frak c}$. It is not even the size of the continuum. As Jeroen says, this space is not finitely generated. Not even as an algebra.

Even as an algebra it is not finitely generated. What does that mean? Algebra is a vector space which has a multiplication operator. In $\mathbb R[x]$, the vector space of polynomials, we can write any polynomial as a finite sum of scalars and $x^n$'s. This is an example for a vector space which is not finitely generated, but as an algebra it is finitely generated.

In introductory courses it is customary to deal with well understood spaces. In the early beginning it is even better to use only finitely generated spaces, which are even better understood.

The axiom of choice is an axiom which allows us to "control" infinitary processes. Assuming this axiom we can prove that every vector space has a basis, but we cannot necessarily construct such space.

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Linear Algebra

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