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?

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It is unclear what your lecturer means by "independent" so who can say? I think you should interpret this as saying that, for example, $\mathbb{R}^{n-1}$ is not open in $\mathbb{R}^n$ and not bother trying to interpret this statement beyond that. – Qiaochu Yuan Jul 6 '12 at 0:02
I would say it was meaningless ramble on the lecturer's part. – copper.hat Jul 6 '12 at 0:04
Are you sure you're saying exactly what your lecturer said? – Mercy King Jul 6 '12 at 0:10
From the responses, it seems I managed to garble something terribly between the blackboard and my notes. You politely ascribe the error to my lecturer, but I think that's unlikely :) So, in hope that something can be retrieved, he was referring to the 'independent' in the function definition, and saying that saying it's being a function of n independent variables was somehow equivalent to its being defined on an open set or something. If this again is insufficiently close to any meaningful sentence then I will admit defeat that my transcription was irretrievably inaccurate... – Mark Allen Jul 6 '12 at 0:52

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 $V$.
Yes, I meant $\mathbb{R}^n$! sorry, I should have looked over my post more carefully before submitting :( – Mark Allen Jul 6 '12 at 0:57
And to respond to your answer, if I understand correctly, this isn't necessarily the case in a closed set since if we take a boundary point, then there will be some direction in which we can't adjust $x$ such that it is still in $V$? – Mark Allen Jul 6 '12 at 1:00
@MarkAllen: Right, or in a closed set you might have to adjust coordinates in some particular ("dependent") way to stay in $V$, e.g., if $V$ were a line in $\mathbb{R}^2$. – mjqxxxx Jul 6 '12 at 2:07
@Mark, warning: "closed" is not the same as "not open". There are sets that are neither open nor closed (e.g. half-open intervals), and there are set that are both (e.g. $\mathbb R$ itself)! – Henning Makholm Jul 6 '12 at 11:33