For E a Normed Vector Space, What Is $\mathbb{R}^{E}$? (Brezis, Theorem 3.16)

Question

I am currently reading Theorem 3.16 in Brezis's "Functional Analysis, Sobolev Spaces, and Partial Differential Equations".

The proof starts of: "Consider the Cartesian product $Y = \mathbb{R}^{E}$, which consists of all maps from $E$ into $\mathbb{R}$".

We denote the elements of $Y$ by $\omega = (\omega_{x})_{x \in E}$, with $\omega_{x} \in \mathbb{R}$. We equip the stanadard product topology to this space, which Brezis notes is the coarsest topology on $Y$ associated with the collection of maps $\omega \rightarrow \omega_{x}$. He then claims that this is "of course" the same as the topology of pointwise convergence.

I have found it hard to find a real definition of this set $Y = \mathbb{R}^{E}$, and I find this notation ambiguous. Each element of $Y$ is a function from $E$ into $\mathbb{R}$, but how exactly are we able to express this as a product of $E$ and $\mathbb{R}$? Could someone please point me to a fully fleshed out definition of $Y$ here?

Answer 1

It's not really a product, it's an "exponential object". In set theory (where this notation comes from), for sets $A$ and $B$, $A^B$ is by definition the set of functions from $B$ to $A$. The Cartesian product however of $A$ and $B$, $A \times B$, is the set of ordered pairs $\{(a,b): a \in A, b \in B\}$ which is a different object altogether.

As $2=\{0,1\}$ (in cardinals and ordinals, this is how the set $2$ is defined!) we can identify $A^2$ with $A \times A$ in an obvious way via a bijection that sends $(a,b) \in A \times B$ to the function $f: 2 \to A$ defined by $f(0)=a, f(1)=b$. So for finite products we usually identify the exponential $A^n$ with an $n$-fold product of copies of $A$. But note that they're not really the same set...

I'd better get used to thinking about these powers as exponential objects, i.e. sets of functions, not "tuples", except in the finite case, maybe.

And the product topology and "the topology of pointwise convergence" are common synomyms for the unique coarsest (smallest) topology that makes all "projections" (or evaluations, sending $f$ to $f(x)$ for some fixed $x$) continuous.

Answer 2

The definition as it is stated is a full definition. A function $\omega: E \to \Bbb R$ can also be viewed as a very long tuple of elements in $\Bbb R$, with an entry $\omega_x$ for every $x \in E$. So you get $\omega(x) = \omega_x$, where the left side is viewing $\omega$ as a function, and the right side is viewing $\omega$ as a tuple.

Now where do tuples live? In a product, so we get $$ \Bbb R^E = \prod_{x \in E} \Bbb R. $$

Just a smaller example, if $E =\{0,1,2\}$ (so it is not a vector space now, but just a set). Then a function $\omega: E \to \Bbb R$ is just a tuple $(\omega_0, \omega_1, \omega_2)$ of three elements in $\Bbb R$. And we have in this case $$ \Bbb R^E = \prod_{x \in E} \Bbb R = \Bbb R \times \Bbb R \times \Bbb R. $$

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Since you also mention the product topology, you may be interested in this Wikipedia page on that subject.