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Define $D^n = \{x \in \mathbb{R}^n : |x| \le 1\}$. Is $D^n$ defined when $n = 0$? I would say no, since I don't think $\mathbb{R}^0$ is defined.

user77429
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    $\mathbb R^0$ is defined in most contexts: it is just the one point set. If you want, it is the underlying topological space of the real vector space of dimension zero. – Mariano Suárez-Álvarez May 12 '13 at 05:37
  • Even if you look at $\mathbb R^n$ as the direct product of $n$ copies of $\mathbb R$, then $\mathbb R^0$ should be the direct product of zero copies of $\mathbb R$, and that is in prettymuch every context defined to be a one-point set. This is a standard convention, and the only one which makes sense in the grand scheme of things. – Mariano Suárez-Álvarez May 12 '13 at 05:38
  • I'd vote against closing this question. – Zev Chonoles May 12 '13 at 05:56

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Using a definition of $\mathbb{R}^n$ like $$\{(a_1,\ldots,a_n)\mid a_i\in\mathbb{R}\},$$ there would appear to be no definition of $\mathbb{R}^0$ (or it may even seem like $\mathbb{R}^0$ should be the empty set; but this would be terrible, since the rule $\mathbb{R}^{n+m}\cong\mathbb{R}^n\times\mathbb{R}^n$ would no longer be true).

For the purposes of generalization, the above definition is not ideal. A good definition to use is that $\mathbb{R}^n$ is $\color{red}{\textbf{the}}$ product (in an appropriate category) of $n$ copies of $\mathbb{R}$. $\color{red}{\textbf{The}}$ product of an empty collection of objects, when it exists, is $\color{red}{\textbf{the}}$ terminal object of the category; thus, thinking of $\mathbb{R}$ as a topological space, we would say that $\mathbb{R}^0$ is $\color{red}{\textbf{the}}$ topological space with a single point, while thinking of $\mathbb{R}$ as an abelian group, we would say that $\mathbb{R}^0$ is $\color{red}{\textbf{the}}$ trivial group.

You are thinking of $\mathbb{R}^n$ as a normed real vector space (which is a special case of both of the above examples), and as a result, we would say that $\mathbb{R}^0$ is $\color{red}{\textbf{the}}$ zero-dimensional real vector space (with its only possible norm), because such an object is terminal in the category of normed real vector spaces. Because the only element of $\mathbb{R}^0$ is its zero vector, we would therefore have that $$D^0=\{x\in\mathbb{R}^0:|x|\leq 1\}=\mathbb{R}^0.$$ Thus, $D^0$ is $\color{red}{\textbf{the}}$ one-point topological space.

Why have I marked my uses of the word $\color{red}{\textbf{the}}$ this way? Because it really doesn't mean "the", in the sense you may be expecting. Really, any isomorphic object will do equally well, and there is no single thing to point to as being "the" product. For example, the one-point topological space $$\{a\} \quad\text{with the topology}\quad\mathcal{T}=\{\varnothing,\{a\}\}$$ is no better or worse than the one-point topological space $$\{b\} \quad\text{with the topology}\quad\mathcal{T}=\{\varnothing,\{b\}\}$$ so which one should we say is $\mathbb{R}^0$? The correct action is not to choose. See this article on the nLab.

Zev Chonoles
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