What is $\mathbb{R}^0$? More specifically, I assume, the Cartesian product: $$\prod_\emptyset \mathbb{R}.$$ Is there a convention here?
I ask because I've stumbled across a definition in topology where the $0$ dimensional closed unit ball and interior of the $0$ dimensional closed unit ball are equal and seem to be a point (although I'm just assuming the last part by context, hence my confusion).
Formally, the Cartesian product can be defined as the set of functions from an index set to the disjoint union of the factors of the product. That is:
Let $J$ be an index set (possibly infinite), and let $\{X_j\}_{j\in J}$ be a family of sets indexed by $J$. Then $$ \prod_{j\in J} X_j := \left\{ f : J \to \bigsqcup_{j\in J} X_j\ \middle| \ f(j) \in X_j \text{ for all $j\in J$}\right\}.$$
In other words, the Cartesian product is the collection of all functions from the index set to the disjoint union of the family of sets such that the value of the function at any particular index is contained in the set of the same index. With respect to this definition $$ \prod_{j\in\emptyset} \mathbb{R} = \left\{ f : \emptyset \to \bigsqcup_{j\in\emptyset} \mathbb{R}\ \middle|\ \ f(j) \in \mathbb{R} \text{ for all $j\in \emptyset$}\right\}. $$ The codomain of such a function $f$ may be a little dubious, and the requirement that $f(j) \in \mathbb{R}$ for all $j\in\emptyset$ is vacuously satisfied for any function with domain $\emptyset$, but this ends up not really mattering—there is only one function $f : \emptyset \to X$ for any $X$: the empty function. (As comments are ephemeral by nature and I think that Fly by Night's comment is perfect and should be preserved: there is only one function $f: \emptyset\to X$ because "'not doing anything' counts as 'doing something.'") This implies that $$ \prod_{j\in\emptyset} \mathbb{R} = \{ \ast \}, $$ where $\{\ast\}$ denotes a one-point set.
Perhaps more intuitively (though much less formally, and with very little rigor), we can think by analogy to sums and products of real numbers. We typically adopt the convention that $$ \sum_{j\in\emptyset} a_j = 0 \qquad\text{and}\qquad \prod_{j\in\emptyset} a_j = 1. $$ In the case of the sum, we can think of starting with nothing, then adding to that the terms of the sum. That is, $$ \sum_{j\in\{1,2,3,\dotsc,N\}} a_j = 0 + a_1 + a_2 + a_3 + \dotsb + a_N. $$ Thus if we add no numbers together, the result we get at the end of the day should be nothing. That is, the empty sum is zero. By a similar kind of argument, the empty product should be one. Essentially, the empty sum is the additive identity, and the empty product is the multiplicative identity.
In the case of Cartesian products, we would expect the empty product to be some kind of identity element. Since $$ \{\ast\} \times X \cong X $$ for any set $X$, the one-point set is a kind of identity element for the Cartesian product. Hence there is some heuristic reason to believe that the empty Cartesian product should be a one-point set.
In order to understand this, one has to go to low level set theoretic concepts.
A function is a set $f$ of ordered pairs such that
for all $x,y,z$, if $(x,y)\in f$ and $(x,z)\in f$, then $y=z$.
This property allows us to write $f(x)$, where $x$ is in the domain of $f$, to denote the unique object such that $(x,f(x))\in f$: one such object exists because $x$ is in the domain of $f$ and it is unique because of the function property. Such object $f(x)$ is the value of $f$ at $x$.
Once we have a function, we say that $\{x:\text{there is $y$ such that $(x,y)\in f$}\}$ is the domain of $f$ and $\{y:\text{there is $x$ such that $(x,y)\in f$}\}$ is the range of $f$. (Axioms of set theory guarantee that both the domain and the range are sets.)
We say that a function $f$ is a function from $A$ into $B$ if $A$ is the domain of $f$ and the range of $f$ is a subset of $B$ and we write $f\colon A\to B$.
There exists a single function with empty domain, namely the empty set, which in such context is often denoted by $0$, but I'll use the other usual symbol $\emptyset$.
A family of sets indexed by a set $I$ is just a function with domain $I$. Note that in set theory everything is a set.
When talking about a family of sets indexed by $I$, the special notation $(X_i)_{i\in I}$ is often used; $X_i$ is just the value of the function at $i$. The union of this family is $$ \bigcup_{i\in I}X_i= \{x:\text{there exists $i$ such that $x\in X_i$}\} $$ An axiom of set theory guarantees this is a set.
Now we're ready to define the product of a family of sets: $$ \prod_{i\in I}X_i= \Bigl\{ f\colon I\to\bigcup_{i\in I}X_i \Bigm| \text{for all $i\in I$, $f(i)\in X_i$} \Bigr\} $$ The axiom of choice guarantees that, if every $X_i$ is not empty, then the product is not empty, but this is not needed here.
In the special case where, for every $i$, $X_i=X$ (a fixed set), the product of the family is simply the set of all functions $f\colon I\to X$, because the union is obviously $X$.
What happens when $I=\emptyset$? Well there is exactly one function $f\colon I\to\bigcup_{i\in I}X_i$, namely the empty function.
What is $\mathbb{R}^n$ or, more generally, $X^{n}$, where $X$ is a set and $n$ is a natural number? It is just the product of the family $(X_i)_{i\in I}$ as above with $$ I=\{1,2,\dots,n\},\qquad X_i=X $$ Oh, well! What about $X^{0}$? In this case the index set contains no element, so it is empty. Therefore $$ X^{0}=\{\emptyset\} $$ Note that this set is not empty!
Under the axiom of choice, the product of a family of nonempty sets is not empty.
Conversely, if one of the sets in the family is empty, say $X_{i_0}=\emptyset$, then the product is empty, because $$ x\in X_{i_0}=\emptyset $$ is false for every $x$, so we cannot find $f(i_0)\in X_{i_0}$.
However, if the index set is empty, no set of the family can be empty, so the product is not empty. Hence we can state
the product $\prod_{i\in I}X_i$ is empty if and only if there exists $i\in I$ such that $X_i=\emptyset$.
But what about $\emptyset^0$? It is the same as above: $\emptyset^0=\{\emptyset\}$ (not empty), because the empty function has empty domain and empty range, so it qualifies as an element of $\emptyset^0$.
In the context of vector space, $\mathbb{R}^{0}$ is the unique vector space with one element, that is $\{0\}$. Not a big difference with the above, because a one element set can be endowed with a single structure of vector space. Note that this is consistent with the standard property that $$ \mathbb{R}^m\times\mathbb{R}^n\cong\mathbb{R}^{m+n} $$
