Describing the preimage of a continuous function at a point.

838 Views Asked by At

Suppose I have a given continuous function; $f:\mathbb{R}^3\supset (0,\infty)\times[0,2\pi)\times\mathbb{R}\longrightarrow \mathbb{C}$, are there any theorems or techniques I can use in order to describe the preimage: $f^{-1}(\{0\})$?

Specifically, I would like to be able to determine if the preimage as a subset is connected or disconnected and determine if it describes a line, a plane, or a set of points. In effect I suppose I would like to describe the zero-set of a continuous function with as much abstract nonsense as possible.

If needed, are there additional assumptions I would require on $f$ in order to further study the set $f^{-1}(\{0\})$?

1

There are 1 best solutions below

0
On

Since single point sets are closed in the standard topology on $\mathbb{R}^n$ and the preimage of any closed set under a continuous map is closed, you can say that $f^{-1}(\{0\})$ is closed. This is enough to rule out some pathologies, but not the most strong.

You can't manage much more in general. It can be disconnected (consider $sin(x)$), empty (define $f$ such that $f(x) = 1$ for all points in your domain), or even the entire space (define $f$ such that $f(x) = 0$ for all points in your domain). Assuming that $f$ is injective can show that the preimage of a point contains at most one point, but that's just the definition of injectivity.

The sorts of assumptions that make this go better in an interesting way are demanding locally "nice" behavior from a function. There is, for instance, a nice result for complex-differentiable functions on connected open subsets of $\mathbb{C}$: if $\Omega \subset \mathbb{C}$ is open and connected and $f: \Omega \rightarrow \mathbb{C}$ is complex differentiable on $\Omega$, then for any $p \in \mathbb{C}$, $f^{-1}(\{p\})$ either contains no limit point, or $f$ is constantly $p$.