Revision of I have a problem about open or closed sets from Wed, 13/05/2009 - 22:28

Quick description

This page is designed to help if you have a problem concerning open and/or closed sets, particularly in $\R^n$ . Clicking on answers to the questions below will lead to suggestions or further questions.

Prerequisites

Basic real analysis, the definitions of open and closed.

A piece of general advice

When thinking about open or closed sets, it is a good idea to bear in mind a few basic facts.

First, a subset $X$ of $\R^n$ (or any metric space, but this does not apply to all topological spaces) is closed if and only if whenever $(x_n)$ is a sequence of elements of $X$ that converges to a limit $x$ , then that limit $x$ belongs to $X$ as well. In other words a set is closed (in the sense of having a complement that is open) if and only if it is closed under taking limits.
Second, if $\R^n\rightarrow\R^m$ , then $f$ is continuous if and only if $f^{-1}(U)$ is an open subset of $\R^n$ whenever $U$ is an open subset of $\R^m$ . (Again, this holds for arbitrary metric spaces. It also holds for topological spaces, but then it is the definition of continuity.)
Third, a closed bounded subset of $\R^n$ is compact (but a closed bounded subset of an arbitrary metric space does not have to be compact).
Fourth, a finite intersection of open sets is open and any union of open sets is open; and similarly a finite union of closed sets is closed and any intersection of closed sets is closed.

What is your problem?

Which of the following descriptions best fits your problem?

I am trying to prove that a certain set $X$ is open.
- In that case, an obvious approach is to begin your proof by saying "Let $x\in X$ ," and going on to try to prove that there must be some $\delta>0$ such that $y\in X$ whenever the distance between $x$ and $y$ is less than $\delta$ . But often it is much cleaner to use the basic facts above. For example, suppose we define $X$ to be the set of all real numbers $\alpha$ such that there exists a rational number $p/q$ such that $|e^{\alpha}-p/q|<1/2^q$ . We can express this definition 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.

I am trying to prove that a certain set $X$ is closed.

Not yet finished.