What is the Euclidean topology on $\mathbb{R}^0$ like?

Question

I am trying to prove that a topological space $(X,\mathscr{T})$ is a $0$-manifold if and only if it is a countable discrete space. In the process I have to show that there exist a homeomorphism from a singleton in $\mathscr{T}$ to an open set in $\mathbb{R}^0$.

This lead me to think. What is $\mathbb{R}^0$ really? According to my textbook, there is a point in the space. I have some questions:

1. What does the Cartesian product of power $0$ look like? I wasn't able to find that the Cartesian product is defined for powers less or equal to zero, but it should definitely be something I want to learn, if you can explain how one should view it.
2. According to my textbook, $\mathbb{R}^0$ is a single point. What is that point? Is $\mathbb{R}^0 = \{0\}$?
3. How can we know that the Euclidean topology exist on $\mathbb{R}^0$. Does it contain an open ball?

I hope you are able to help me understand how to view this space.

Answer 1

$\mathbb R^0$ is not really the zero times cartesian product of $\mathbb R$, it is just a way to write a zero dimensional space which fits in the pattern of all the other $\mathbb R^n$ spaces. It consists of only one point. It doesn't really matter what the name of that point is. It could be $\{0\}$ if you like, but you could also call it $\{\text{bob}\}$. You know that a topology on a space must contain the empty set and the whole set, thus the only open sets here are $\emptyset$ and $\mathbb R^0$. Thus, $\mathbb R^0$ does contain an open ball, and it is $B_r(x)$ for all $x\in\mathbb R^0$ and all $r\in\mathbb R$. Baisically, this is the simplest kind of topological space imaginable.

Answer 2

$0$ is defined in most set theory settings to be the empty set, and the notation $X^A$ denotes the set of all functions $A\to X$. For any set $X$ there is a unique function from the empty set to $X$, which is the empty function. Thus the zeroeth power Cartesian product of any set is a single point. That point is the empty function, which is the empty set, which is $0$.

A set with a single point has a unique topology consisting of the empty set and the whole space, so there is no need to go out of the way trying to determine what the topology is. It is a metric space, and an open ball is the whole space.

Answer 3

Answer: $\mathbb R^0=\{0\}$, hence its only topology equals $\{\emptyset, \{0\}\}$.

"Proof"/explanation (either A+B+C+E or A+D+E; A+B+C+E is a more exact proof):

A. $0=\emptyset$, $1=\{0\}$, $2=\{0,1\}$, $3=\{0,1,2\}$ etc. (this is the "standard" definition of natural numbers). https://en.wikipedia.org/wiki/Natural_number#von_Neumann_construction

B. $X^A$ "always" means the set of functions $A\rightarrow X$, hence $\mathbb R^0$ denotes the set of functions $0\rightarrow \mathbb R$. REFERENCE: https://en.wikipedia.org/wiki/Function_(mathematics)#Function_spaces "The set of all functions from a set X to a set Y is denoted by X → Y, by [X → Y], or by $Y^X$. The latter notation is motivated by the fact that, when X and Y are finite and of size |X| and |Y|, then the number of functions X → Y is |Y^X| = |Y|^|X|. "

C. A function $f:A\rightarrow X$ is a set $f\subset A\times X$ defined by $f:=\{(a,f(a))\,:\, a\in A\}$. But $\emptyset\times X=\emptyset$, hence $0$ (i.e., $\emptyset$) is the only function $\emptyset\to X$ for any set $X$ (proof: $0\subset\emptyset\times X$, and for each $a\in\emptyset$ there exists a unique $x\in X$ such that $(a,x)\in 0$, QED.) A detailed explanation of this: https://en.wikipedia.org/wiki/Empty_function https://en.wikipedia.org/wiki/Function_(mathematics)#Definition

The above says that $X^0=\{0\}$ for every set $X$. So $\mathbb R^0=\{0\}$:

D. Alternative (to B.+C.) explanation on why $\mathbb R^0=\{0\}$: Using that "standard" definition of natural numbers ("A." above), for every $n=\{1,2,3,\ldots\}$ one can identify $X^n$ with the Cartesian product $X$'s. If $n=0$, you can think $0=\emptyset$ as the neutral element of Cartesian products, the only element of a Cartesian product of $0$ sets. So also that way of thinking works here.

E. The only topology for a set $X$ of one (or zero) elements equals $\{\emptyset,X\}$. https://en.wikipedia.org/wiki/Topological_space#Open_set_definition

(Sorry, SE does not accept long comments, so I had to post this as an answer instead, even though Matt's post says the same with fewer explanations, .)