Is it always true that the level set of a three variable function is a surface?

Question

I know that it is usually a surface, but I am curious if there exists an exception. Thank you

Answer 1

No, it can go very wrong, even with nice functions. The level set $x^2+y^2+z^2=0$ is just a point. The level set $x^2+y^2=0$ is a curve (the $z$-axis). There more complicated variants of this same phenomenon.

The appropriate notion is that of a regular value. When we look at the level set $f(x,y,z)=c$, it will be a (nice, smooth) surface whenever $\nabla f(x,y,z)\ne 0$ at every point $(x,y,z)$ of the level set.

Answer 2

Sure it is... the edge of a three-dimensional region!

Or just think about $x^2 + y^2 + z^2 = 4$, i.e., a sphere.

There are indeed "non-nice" or "corner cases" (e.g., $f(x,y,z) = \delta(x,y,z)$), and if that is really what you're asking, then sure: $f(x,y,z) = c$ is constant throughout the whole space and has no well-defined level-set contours.

If such an example is all the OP is seeking, I think we're done here.

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More fully (and to placate some of the commenters, and in the context of a question from a newbie OP with reputation $1$):

In general (defined below), a function of three variables has level sets, i.e., solutions to $f(x,y,z)$ (a constant), that have topological dimension $2$... that is, they can be described as two-dimensional surfaces.

@Steven Stadnicki points out that there are of course an infinite number of functions that are discontinuous such that its level sets are dense in space. That alone does not necessarily mean that such a function lacks a topologically two-dimensional level set. Regardless, here we should craft a solution given our best inference of what the OP knows and is asking. Is the OP really asking about functions that approximate a three-dimensional Serpinski gasket, for instance... for which a derivative is not well-defined? Well, if so, then I guess we should point this out. But my reading of his question is that it is lower-level.

Consider a function inspired by @nicomezi's example: $f(x,y,z) = x y z$. The contours are perfectly well defined for all non-zero points... this is what I meant by a "corner case." It is true that for @nicomezi's actual function, $x y z = 0$ (or indeed any constant), means the entire space is constant and there are no well-defined level-set contours. Fine. If that satisfies the OP, I'd love to hear it from @FastCreepyBaby.

To all the commenters: If you're so sure your comments and answers answered the OP's question, WHY was the question closed? The stated reason is "lacks context." Exactly. We should really understand better what @FastCreepyBaby understands and seeks.