To prove that a set is open or closed, use basic theorems rather than direct arguments

Quick description

If you want to prove that a set is open or closed, then it is tempting to argue directly from the definitions of "open" and "closed". But one can often argue much more cleanly by using some basic facts and avoiding epsilons and deltas.

Prerequisites

Basic analysis, definitions of open and closed sets, easy theorems about open and closed sets.

General discussion

Here are some theorems that can be used to shorten proofs that a set is open or closed.

A union of open sets is open, as is an intersection of finitely many open sets.
An intersection of closed sets is closed, as is a union of finitely many closed sets.
If $X\rightarrow Y$ is a continuous function and $Z\subset Y$ is open/closed, then $f^{-1}(Z)$ is open/closed.
A subset $F$ of $\R^n$ (or more generally of a metric space) is closed if and only if whenever $(x_n)$ is a sequence of elements of $F$ and $x_n\rightarrow x$ , then $x$ is also an element of $F$ .

Example 1

Let $X$ be the set of all real numbers $\alpha$ such that there exists a rational number $p/q$ such that $|e^{\alpha}-p/q|<1/2^q$ . What is the neatest way of showing that $X$ is open?

One method that involves nothing more than formal manipulations is to express the definition of $X$ as

$X=\bigcup_{p/q\in\Q}\{\alpha\in\R:e^{\alpha}\in(p/q-1/2^q,p/q+1/2^q)\}=\bigcup_{p/q\in\Q}f^{-1}(I_{p/q}),$

where $f(x)$ is the continuous function $e^x$ and $I_{p/q}$ is the open interval $(p/q-1/2^q,p/q+1/2^q)$ . Since each $I_{p/q}$ is open and $f$ is continuous, so is each $f^{-1}(I_{p/q})$ , and therefore so is their union. And we have shown this without dirtying our hands with epsilons and deltas.

General discussion

Often in analysis it is helpful to bear in mind that "there exists" goes with unions and "for all" goes with intersections. For example, the set of all real numbers $x$ such that there exists a positive integer $n$ with $|x-n|<1/3$ is the union over all $n$ of the set of $x$ with $|x-n|<1/3$ . Perhaps writing this symbolically makes it clearer:

$\exists n\in\N\ |x-n|<1/3\}=\bigcup_{n\in\N}\{x:|x-n|<1/3\}.$

This often makes it possible to show that a set is open by showing that it is a union of sets that are more obviously open. Similarly, one can often express the set of all $x$ that satisfy some condition as the inverse image of another set under a continuous function. Here is an example.

Example 2

Prove that the set of all non-singular $n\times n$ matrices is open (in any reasonable metric that you might like to put on them).

A quick argument is that this set is equal to $\det(A)\ne 0\}$ , which is the inverse image of the open set $\R\setminus\{0\}$ under the continuous map $M_n\rightarrow\R$ .

General discussion

This example differs from the previous one in that the definition of "non-singular" was not in a form where we could immediately apply the basic theorems. Instead, we had to search for an appropriate function (the determinant) that would do the job for us. In spirit, this argument is a bit like proving that a subgroup $H$ of a group $G$ is normal by finding a homomorphism from $G$ to some other group with $H$ as its kernel.

Example 3

Let $\R\rightarrow\R$ be a continuous function. Then the graph of $f$ is closed. The graph of $f$ is the set of all points of the form $(x,f(x))$ .

A direct proof of this would be to take some point $(x,y)$ with $y\ne f(x)$ and argue that there exists $\delta>0$ such that if $(x',y')$ has distance at most $\delta$ from $(x,y)$ then $y'\ne f(x')$ . That can be done, but it is slightly tedious.

A quick proof is to consider the map $(x,y)\rightarrow y-f(x)$ . This is continuous, and the graph of $f$ is $F^{-1}(\{0\})$ . Therefore, the graph is closed.

General discussion

Again, we chose a continuous function in order to solve this problem. Was any ingenuity required? Well, the graph of $f$ is $y=f(x)\}$ . If one is trying to express it as the inverse image of a closed set under a continuous function, then it doesn't take too much ingenuity to rewrite this as $y-f(x)=0\}$ . But if you want a proof that takes no ingenuity at all, then notice that the statement $y=f(x)$ is telling us that the pair $(f(x),y)$ lies on the line $x=y$ (to put it slightly confusingly). So we could have argued first that the line $L$ consisting of all points of the form $(\lambda,\lambda)$ is closed and then that the map $(x,y)\rightarrow(f(x),y)$ is continuous. The graph of $f$ is $G^{-1}(L)$ so we are done.

But how would we prove that $L$ is closed if we wanted to do no work? Probably we'd end up considering the map $(x,y)\mapsto x-y$ and then we would be back with the earlier argument.

Example 4

Is there another work-free way to prove that the graph of a continuous function is closed? How about using the sequence definition of closed sets? So we take a sequence $(x_n,f(x_n))$ of points in $\R^2$ that converges to some other point in $\R^2$ . That implies that $x_n\rightarrow x$ for some $x$ . Since $f$ is continuous, $f(x_n)\rightarrow f(x)$ . But then $(x_n,f(x_n))\rightarrow(x,f(x))$ , which belongs to the graph of $f$ , so we are done.