In Sheldon Axler's Linear Algebra Done Right third edition the following is given as an example of a subspace:

The set of differentiable real-valued functions on $\mathbb R$ is a subspace of $\mathbb R^{\mathbb R}$

I'm looking for an intuitive explanation of the statement? Letting $S$ be the set of all differentiable real-valued functions, in order for the statement to be true, $S$ must be a subset of $\mathbb R^{\mathbb R}$(a subspace needs to be a subset).

-How can $S \subset \mathbb R^{\mathbb R}$ when $S$ is a set containing functions and $\mathbb R^{\mathbb R}$ is a set containing real numbers?

-What are the elements in $\mathbb R^{\mathbb R}$? How can we think of $ \mathbb R^{\mathbb R}$ as a tuple?

  • 6
    $\begingroup$ $\mathbb{R}^\mathbb{R}$ is not a set of real numbers, it is a set of $\mathbb{R}$-tuples. An element of $\mathbb{R}^\mathbb{R}$ is of the form $(a_x)_{x \in \mathbb{R}}$; i.e. for every $x \in \mathbb{R}$, there is an associated $a_x \in \mathbb{R}$. So we can equivalently think of $(a_x)_{x \in \mathbb{R}}$ as a function $f: \mathbb{R} \to\mathbb{R}$ with $f(x)=a_x$. $\endgroup$
    – kccu
    Dec 29, 2015 at 1:23
  • $\begingroup$ Perhaps the Axler book has a definition of $\mathbb R^{\mathbb R}$ in it somewhere... $\endgroup$
    – GEdgar
    Dec 29, 2015 at 2:25
  • $\begingroup$ @GEdgar No but it does have a definition that applies to $\mathbb R^{\infty}$. Thanks Ill look into that! $\endgroup$
    – Red
    Dec 29, 2015 at 2:28
  • $\begingroup$ Exercise: First interpret any tuple $(x_1,x_2)\in\mathbb R^2$ as a function $[2]\to\mathbb R$, where $[2]$ is the two-element set $\{1,2\}$. $\endgroup$
    – user856
    Dec 29, 2015 at 3:22
  • $\begingroup$ @Rahul That's the trouble I'm having with this notation. That would mean f(1) and f(2) make up the entire codomain R? $\endgroup$
    – Red
    Dec 29, 2015 at 3:31

4 Answers 4


$\mathbb R^{\mathbb R}$ denotes the set of all maps from $\mathbb R$ to $\mathbb R$.

So $S \subset \mathbb R^{\mathbb R}$ is given.

To check if $S$ is a linear subspace, see if the zero element is in $S$, see if the sum of two members of $S$ is in $S$, and similarly for multiplication by a scalar.


Actually one usually uses the notation $A^B$ to mean the set of all functions from $B$ to $A$, so in your case we have

$$\Bbb{R}^{\Bbb{R}} = \{f\colon\Bbb{R}\to\Bbb{R}\ |\ f\text{ is a function}\}.$$

This is why it makes sense to consider $S$ as a subset of $\Bbb{R}^{\Bbb{R}}$.

  • $\begingroup$ I see. So I remember a theorem that says if set $A$ has m elements and set $B$ has n elements then the total number of functions is $|B|^{|A|}$ or $n^m$. So in other words, there are $\Bbb{R}^{\Bbb{R}}$ many functions from $\mathbb R$ to $\mathbb R$. I still don't see how it can be seen as uncountably infinite vector space $\endgroup$
    – Red
    Dec 29, 2015 at 2:18
  • $\begingroup$ @Red most classical vector spaces are uncountably infinite. All $\Bbb{R}^n$ and $\Bbb{C}^n$ for $n>0$ are uncountable. Maybe you were talking about the dimension of $\Bbb{R}^{\Bbb{R}}$, which is uncountable. $\endgroup$
    – Mankind
    Dec 29, 2015 at 8:36
  • 3
    $\begingroup$ @Red: To see why the dimension of $\mathbb{R^R}$ is uncountably infinite, consider the family of functions $\chi_t:\Bbb R\to\Bbb R$ defined by $$\chi_t(x):=\begin{cases}1 & x=t\\0 & x\ne t.\end{cases}$$ You should be able to show that $\{\chi_t:t\in\Bbb R\}$ is linearly independent, and that for any $f:\Bbb R\to\Bbb R$ we have $$f=\sum_{t\in\Bbb R}f(t)\chi_t.$$ Hence, $\{\chi_t:t\in\Bbb R\}$ is a (clearly uncountable) basis for $\mathbb{R^R}.$ $\endgroup$ Dec 29, 2015 at 12:46
  • $\begingroup$ @CameronBuie: How will it become a basis for the vector space? It is clear that the set of all characteristic functions of singletons forms a linearly independent set which is uncountable. And hence, the space is of uncountable dimension. But, how do we even define that infinite sum if we do not have a topology to talk about convergence? Maybe that can be considered as a formal sum, but anyway, as per the definition of a basis, we must be able to write every vector as a finite linear combination of members of it. $\endgroup$
    – Janson
    Jun 15, 2020 at 12:44
  • $\begingroup$ By the way, the span of the given 'basis' is a very small subspace of the product space, namely the direct sum! $\endgroup$
    – Janson
    Jun 15, 2020 at 12:45

$\Bbb R^2$ is a two dimensional vector space, a point in $\Bbb R^2$ is a pair $(x,y)$. Now $\Bbb R^3$ is three dimensional, a point in $\Bbb R^3$ is a triple $(x,y,z)$. Now $\Bbb R^{\Bbb N}$ is countably many dimensions, indexed by the index set $\Bbb N$. So a point in $\Bbb R^{\Bbb N}$ can be thought of as an infinite sequence $(x_1,x_2,x_3,\dots)$. You can also think of that as a function from $\Bbb N$ to $\Bbb R$. Next $\Bbb R^{\Bbb R}$ is a vector space with uncountably many dimensions, indexed by $\Bbb R$. So a point is an association of a value in $\Bbb R$ to every element of (the index set) $\Bbb R$. You can think of a number of ways to denote it, but if you think about it that's exactly just the same thing as a function from $\Bbb R$ to $\Bbb R$. That way an entire function from $\Bbb R$ to $\Bbb R$ is captured as a single point in $\Bbb R^{\Bbb R}$. Do you see how that works now?


If you go back to page 14 in chapter 1 of the text, at the bottom left corner of the page there is a footnote which mentions how $\mathbb{F}^{\infty}$ and $\mathbb{F}^n$ are special cases of $\mathbb{F}^S$, which Axler defines as the set of functions $g:S\to \mathbb{F}$.

Why is this so?

We can think of the $n$-tuples in $\mathbb{F}^n$ as being the assignment of elements of the set $\{1,2,3,....,n\}$ to elements of $\mathbb{F}$ by $g \in \mathbb{F}^{\{1, 2, 3,...,n\}}$ as $g(1)=x_1$, $g(2)=x_2$,...., $g(n)=x_n$ for the $n$-tuple $(x_1, x_2,..., x_n)$. We can also think of this as indexing a subset of $\mathbb{F}$ with $\{1,2,3,...,n\}$.

So similarly for $\mathbb{F}^{\infty}$, we have that the natural numbers are indexing our set (as Axler defined it), so $\mathbb{F}^{\infty}$ is really just $\mathbb{F}^{\mathbb{N}}$ i.e. functions $g: \mathbb{N} \to \mathbb{F}$ defined as $g(1)=x_1$, $g(2)=x_2$,...... which assign all of the natural numbers to elements of our field $\mathbb{F}$ in the form of countably infinite tuples $(x_1, x_2, ......)$.

Now to your question. From the perspective I just explained, we can see that $\mathbb{R}^{\mathbb{R}}$ are just functions from $\mathbb{R} \to \mathbb{R}$, and that we are indexing elements $\mathbb{R}$ with elements of $\mathbb{R}$. Hence each real valued function is just a single ordering of a subset of $\mathbb{R}$ into an uncountably long tuple!(since $\mathbb{R}$ is uncountable) And so these real valued functions correspond to single points in $\mathbb{R}^{\mathbb{R}}$. Amazing isn't it?

From this perspective we can see how the set of differentiable real valued functions is a subset of $\mathbb{R}^{\mathbb{R}}$, since these functions are just tuples which order the elements of $\mathbb{R}$ in a way such that the graph of the function $f(i)=x_i$ for $i,x\in \mathbb{R}$ is differentiable.

To start you off, note that our additive identity is just $g:\mathbb{R} \to \{0\}$ (since constant functions are differentiable). This is just an uncountably long tuple consisting only of zeroes, and this tuple is the origin of our $\mathbb{R}$-dimensional space! Since $0$ is the additive identity for $\mathbb{R}$, we can see how each component of any tuple we add with our uncountably long tuple of $0$'s will remain unchanged under addition of a $0$, hence the zero function is our additive identity.

Hope that helps :)

  • 1
    $\begingroup$ If you'd like more info on the indexing functions, I can send you some notes via email. $\endgroup$ Dec 30, 2015 at 5:33

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