I am watching a lecture about multivariable mathematics (Math 3500 Day 20: Continuity and Preimages) by Prof. Theodore Shifrin.
Let $f:\mathbb{R}^3\to\mathbb{R}$ be a function.
Let $c\in\mathbb{R}$ be a real number.
Suppose that $\{x\in\mathbb{R}^3 | f(x)=c\}\neq\emptyset$.
Intuitively, we expect the set $\{x\in\mathbb{R}^3 | f(x)=c\}$ to be a $2$-dimensional surface.
(1) Why do we expect this set $\{x\in\mathbb{R}^3 | f(x)=c\}$ is $2$-dimensional?
For example, if $f(x) = c$ for all $x\in\mathbb{R}^3$, then the set $\{x\in\mathbb{R}^3 | f(x)=c\}$ is $\mathbb{R}^3$, which is not a $2$-dimensional surface.
But we don't think this is a typical case.
(2) For what $f:\mathbb{R}^3\to\mathbb{R}$, is the set $\{x\in\mathbb{R}^3 | f(x)=c\}$ $2$-dimensional?
(3) I don't know the definition that a subset of $\mathbb{R}^3$ is $2$-dimensional. What book should I read?
This can be understood as follows. Consider the Jacobian matrix of the function $f$ at a point $p$ with $f(p) = c$. This is a linear approximation for $f$ near $p$, and takes the form of a $1 \times 3$ matrix transformation. Every vector in $\mathbb{R}^3$ in the null space of this matrix corresponds to a direction one can move away from $p$ on which $f$ is constant, thus a direction that lies on the set $\{f(x) = c\}$. If the linear transformation is surjective, the null space of the Jacobian is two-dimensional, meaning that the set $\{f(x) = c\}$ is two-dimensional near $p$. This provides the intuitive condition we need for the result to be true: the Jacobian should always be surjective, which when mapping to $\mathbb{R}^1$ just means that it is nonzero.