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

I'm studying linear algebra on my own and saw basis for $\mathbb{R}, \mathbb{R}[X], ...$ but there is no example of $\mathbb{R}^\mathbb{R}$ (even though it is used for many examples). What is a basis for it? Thank you

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Cue the AC guys... – Mariano Suárez-Alvarez Dec 3 '11 at 15:51
Conveniently, there is nothing much useful that we could do if we got a basis for $\mathbb{R}^\mathbb{R}$ as a Christmas present. – André Nicolas Dec 3 '11 at 17:47
@MarianoSuárez-Alvarez, what do you mean? – user12205 Dec 3 '11 at 22:11
@Jos: He is referring to the fact that the answer to your question depends very strongly on the axiom of choice (AC). – Zhen Lin Dec 3 '11 at 22:17

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 bases. 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.

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).

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Isn't it time you hit that 10,000 mark? Congratulations! :-) – Asaf Karagila Feb 25 '12 at 10:48

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.

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Can you give some references for the last sentence you wrote? – Damian Sobota Dec 3 '11 at 16:00
@Damian, I recall it from my linear algebra course (2 years ago). I have no formal reference to this. I should perhaps change the sentence to something lighter. – user12205 Dec 3 '11 at 16:03
@DamianSobota : Here's a thought : If $(e_i)_{i \in I}$ is a basis of $\mathbb{R}^{\mathbb{R}}$ then $I$ is strictly bigger than $\mathbb{R}$ (otherwise $\text{span}(e_i)$ is in bijection with finite subsets of $\mathbb{R}^2$, which is in bijection with $\mathbb{R}$). So (assuming the continuum hypothesis) $I$ is in bijection with $\mathbb{R}^{\mathbb{R}}$. That means a basis of $\mathbb{R}^{\mathbb{R}}$ is simply huge. So in my opinion it seems very unlikely we can find an explicit example (or at least very difficult). – Joel Cohen Dec 3 '11 at 17:10
@Joel: What does the Continuum Hypothesis have to do with that? – Asaf Karagila Dec 3 '11 at 19:13
@Joel: $|2^\mathbb R|=|\mathbb R^\mathbb R|$. Therefore the cardinality is much bigger than $\mathbb R$. If $I$ was a spanning set of cardinality continuum, then $\mathbb R$ would have the same cardinality as $\mathbb R^\mathbb R$. - All that, and perhaps more, without even using the axiom of choice; let alone CH, GCH and so on. I did, however, assume that there is a basis to $\mathbb R^\mathbb R$. – Asaf Karagila Dec 3 '11 at 20:51

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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For this space (or simpler the vector space of infinite sequences of reals) one cannot actually write down a basis, even though one can prove using the axiom of choice that every vector space has a basis. Proofs of existence that essentially need the axiom of choice provide no means to actually construct the objects whose existence they establish.

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