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?

  • 1
    $\begingroup$ Did you intend what you wrote , that x is a set of N elements of R$^N$? $\endgroup$ Jan 21, 2019 at 2:51
  • $\begingroup$ @William Thanks for pointing it out, I used the wrong brackets. What I meant is that x has N real components (and so is in R^$N$). All of its components are bounded by 0 and M, and the continuous functions only have one each as argument. $\endgroup$ Jan 21, 2019 at 7:32
  • $\begingroup$ x now is a point in R$^N$ but it is not a subset of R$^N$ as you have written. $\endgroup$ Jan 21, 2019 at 9:46

1 Answer 1


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.

  • $\begingroup$ it seems strange to me to write $f^{-1}$ if there is no inverse... also couldn't the pre-image be not compact? what properties does the pre-image of a closed set (under a continuous function) have? $\endgroup$ Jan 21, 2019 at 13:32
  • $\begingroup$ $f^{-1}[C]$ is denotes the set $\{x ∈ X: f(x) ∈ C\}$. It exists even when $f^{-1}$ as a function does not. Preimage of a closed set under a continuous function is a closed set. In fact, that condition is equivalent to continuity. $\endgroup$
    – user87690
    Jan 21, 2019 at 13:39
  • $\begingroup$ thank you, I was not aware of this theorem proofwiki.org/wiki/… and it works just fine. $\endgroup$ Jan 21, 2019 at 18:08

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