Why are subsets of $\mathbb{R}^n$ with soft inequalities closed?

Question

I know that subsets of $\mathbb{R}^n$, e.g. of the form $\{(x,y)\in\mathbb{R}^2|0\leq x\leq 20,y\leq x^2\}$ are closed because the inequalities are not sharp, however, is there an alternative def. of closed sets or a theorem that will allow me to call sets of this form closed? I can't find a formal way to say this.

Answer 1

$\def\R{\Bbb R}$ If you’re happy with the fact that preimages of closed sets under continuous functions are closed then I like the following approach.

The functions $\pi(x,y)=x$ and $f(x,y)=x^2-y$ mapping $\R^2\to \R$ are both continuous by the standard construction theorems. The set you give as an example is then $$ \pi^{-1}([0,20]) \cap f^{-1}([0,\infty)). $$ Since $[0,20]$ and $[0,\infty)$ are closed in $\R$, both factors in the intersection are closed, so their intersection is closed.

Answer 2

I think the best way to see this is working with the idea of a boundary. A boundary point for a set $S$ in $\mathbb{R}^n$ is a point $x$ such that any ball with center $x$ intersects both $S$ and $\mathbb{R}^n\setminus S$.

A set is closed if it contains all its boundary points, and it is open if it contains no boundary points.

Now it is not very hard to see that sets such as yours with "soft" inequalities allow all boundary points to be contained, thus it is closed, while "sharp" inequalities would exclude all boundary points, thus such sets are open.

Answer 3

You can use continous functions.

If you have a continous function $f: \mathbb R^n \to \mathbb R$ and a close set $C \subseteq \mathbb R$ then $f^{-1}(C)$ is closed.

For example you looked at the set $\{(x,y)\in\mathbb{R}^2|0\leq x\leq 20,y\leq x^2\}$.

You can define the continous functions $f_1(x,y) = x$ and $f_2(x)= x^2 - y$. Thus from what we stated above $f_1^{-1}([0,20])$ and $f_2^{-1}([0,\infty))$ are closed.

And now we get that $\{(x,y)\in\mathbb{R}^2|0\leq x\leq 20,y\leq x^2\}$ = $f_1^{-1}([0,20]) \cap f_2^{-1}([0,\infty))$ is closed as an ontersection of closed sets.