Revision of Use basic theorems to prove that sets are open or closed from Wed, 13/05/2009 - 22:48

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 instead.

Prerequisites

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

This article is incomplete. This article needs examples to illustrate the use of all four of the principles below.

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