Why is a function of $n$ independent variables equivalent to one defined on an open set? In a class I took on Analysis in Several Real Variables, on the first day the lecturer wrote on the blackboard > **Definition** A function of the form > > $\begin{align*} \mathbb{R}^n \supset V &\overset{f}{\to} \mathbb{R} \\\\(x_1,\ldots,x_n) = x &\mapsto f(x) = f(x_1,\ldots,x_n) \end{align*}$ > > is called a **real valued function of n independent real variables** if $V$ is open in $\mathbb{R}^n$ He then stated to the class "Note: independent is equivalent to on open set". I duly noted this in my book suffixed with a sarcastic "(apparently)" since at the time it was one of those things which made absolutely no intuitive sense to me. Looking over my notes now, at the end of the year, this still makes little sense, can someone help clarify it to me?

Presumably you mean $V \subset \mathbb{R}^{n}$? If $V$ is an open subset of $\mathbb{R}^{n}$, then for each point $x=(x_1,...,x_n)\in V$ there is an open ball $B_\Delta(x) \subset V$ that contains all points $(x_1 + \delta_1,..., x_n + \delta_n)$ such that $|\delta_i| < \Delta$ for all $i$ and some $\Delta > 0$. In other words, each component of $x$ can be adjusted _independently_ of all the other components while remaining in $V$.