# Does adding a continuous inequality constraint over a compact set lead to another compact set?

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?

• Did you intend what you wrote , that x is a set of N elements of R$^N$? Jan 21, 2019 at 2:51
• @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. Jan 21, 2019 at 7:32
• x now is a point in R$^N$ but it is not a subset of R$^N$ as you have written. Jan 21, 2019 at 9:46

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.
• 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? Jan 21, 2019 at 13:32
• $f^{-1}[C]$ 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. Jan 21, 2019 at 13:39