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First please see this very simple fact:

Fact: $\ $ Any linear map $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ has a proper invariant linear subspace.

By an invariant subspace we mean a space $M$ satisfying $Tx \in M$ for all $x \in M$ and by a proper space we mean a space that is not $\{0\}$ or the whole space. To show the fact, it suffices to take an eigenvalue $\lambda =a+ib$ and an eigenvector $v=x+iy$ of $\lambda$. Since $Tv=\lambda v$, we have $Tx=ax-by$ and $Ty=ay+bx$. Thus $span \{x, y\}$ is an invariant subspace of dimension $1$ or $2$.

Now I wish to change the linear map $T$ into a nonlinear one. I have the following question:

Question:$\ $ Given any surjective continuous map $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ satisfying $f(0)=0$, does there always exist a proper topological subspace $M$ of $\mathbb{R}^3 $that is invariant under $f$ ? For example, a curve, or a surface?

That is, I am looking for a link between linear maps and nonlinear ones.

(I know very little about nonlinear maps or geometry but I am willing to learn them. No matter if you know the answer to the question or not, if you could show me some references to related topics, I would appreciate it a lot!)

EDIT: I am looking for an invariant curve or a surface, not a set of discrete points.

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    $\begingroup$ Take a point $p_0\in\mathbb{R}^3$ and let $S$ be the set of points $p_0,f(p_0),f(f(p_0)), \dots,$. Clearly $S$ is invariant under $f$, and any set is a topological subspace, with the induced topology. So I think you want to require an invariant topological curve or topological surface. $\endgroup$
    – Ben McKay
    Commented Jun 5, 2020 at 13:26
  • $\begingroup$ @BenMcKay Oh yes I am looking for an invariant curve or surface. I have added the newest edit. Thank you for reminding me of this. $\endgroup$
    – user444628
    Commented Jun 5, 2020 at 13:40
  • $\begingroup$ The analogoues of eigenvectors in nonlinear case are stable/unstable/center manifolds. Check out stable manifold theorem for instance. When they exist, they are invariant manifolds that you seek. $\endgroup$
    – Piyush Grover
    Commented Jun 5, 2020 at 14:41

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