so the problem is as follows. I have the vector space $x=[x_1,x_2,...x_N] \subseteq R^N, 0 \leq x_{1,2,...,N} \leq M$ and I extract from it a subset by adding a constraint of this kind:
$X^1=\{x \in R^N | 0\leq a(x_1)-b(x_2)\leq\epsilon, \epsilon \in R\}$
Where both $a(x_1),b(x_2)$ are invertible functions in their argument (which is a component of $x$).
$c(x)=a(x_1)-b(x_2)$ is the difference of two invertible functions (even if they were invertible on a single component of the whole vector $x$, so $c(x)$ is not invertible, but I'd say it is continuous).
Now, I'd dare to say that $X^1$ is a compact set, since, the set $\{k \in R |0\leq k\leq\epsilon\}$ is closed and bounded (and hence compact for Heine–Borel) and the starting subsets $0 \leq x_{1,2,...,N} \leq M$ were compact. I still miss the criteria to apply here, since $c^{-1}(x)$ does not exist, but this is where I got for now.
Now I wonder, if I apply another constraint to $X^1$ and arrive to $X^2$
$X^2=\{x \in R^N | 0\leq a(x_1)-b(x_2)\leq\epsilon, 0\leq a(x_2)-b(x_3)\leq\epsilon, \epsilon \in R\}$
would $X^2$ still be compact?
In linear cases, this seems to be intuitive since a linear constraint would lead to a convex polytope (a halfplane + the bounds, so a bounded convex polytope), and the intersection of convex polytopes is still a convex polytope (or I could think of it as products of compact sets, maybe).
Since the various functions are invertible, I tough of this simple example: let $a(x_1)=x_1$ and $b(x)=x_2^3$ substituting $x_2$ with $b(x_2)$ would mean transforming the $x_2$ axis by stretching it. In the $x_1 - b(x_2)$ plane I obtain that $X^1$ is a halfplane, and is hence compact. Returning to the $x_1 - x_2$ plane means transforming back the $b(x_2)$ to the $x_2$ axis, "squeezing" it, hence preserving the compactness.
This intuitively is true for any invertible function, but it is not a valid proof of course.
What theorem am I missing? is anything that I wrote incorrect?
Adding such a constraint preserves compactness since it just the intersection with the preimage of a closed set with respect to a continuous function, i.e. $X^2 = X^1 ∩ f^{-1}[C]$ where $f$ is continuous and $C$ is closed. No invertibility of $f$ is needed.