What is the point of countable vs. uncountable sets?

Solution 1:

One important application is this: Consider the set of functions that take an integer argument and return an integer result. It's not hard to show that this set is uncountable. Now consider the set of computer programs. This set is countable. Therefore there are uncomputable functions—in fact most functions cannot be computed.

This is a pretty common type of argument in some contexts.

Solution 2:

One very important (in my opinion) application of countability:

Often associated with a physical system is some function $f:\mathbb R^n\to \mathbb R^m$. It could be the acceleration experienced by a planet, the voltage in a circuit, whatever. We want to study the behavior of these functions, and we find that at some points the behavior is good, but at others the behavior is very bad. In particular, we have a nice mathematical model for the good points, which allows us to perform computations easily, but this model breaks down at the bad ones. To apply this model, we need a "generic" point to be a good point, in the sense that if you choose a random point it will be good with probability $1$. One common way to show this is to show that there are only countably many bad points. Formally, this uses the fact that the Lebesgue measure of any countable set is $0$.

Edit: A concrete example of this arises in my own area of interest, billiards. Consider a billiard ball bouncing around inside a polygon. The ball behaves in a very predictable manner when it strikes an edge (the angle of incidence is equal to the angle of reflection) but it's behavior is ill-defined when it strikes a corner. Luckily, starting from any fixed point there are only countably many different directions the ball could travel in which eventually strike a corner, so for the most part this possibility can be ignored.

Solution 3:

One application is probability and measure theory.

If you have met infinite sums, then you may know that we often want probability to be countably additive. What does that mean? If $\{A_n\mid n\in\Bbb N\}$ is a countable collection of pairwise independent events, then $P(\bigcup A_n)=\sum P(A_n)$. That is to say, the probability that one of these will happen is the sum of probabilities.

The reason this is useful is that despite the fact that in "real world situations" there are only finitely many events we can talk and measure, sometimes we don't know how many exactly, and working with infinitely many makes it easier. Therefore countable additivity is a useful and important property.

But here's a catch. If $X$ is countable then there is no countable additive probability such that $P(X)=1$ and $P(\{x\})=0$ for every $x\in X$. This is because if $X$ is countable we can write it as $X=\{x_n\mid n\in\Bbb N\}$ and then we have $$1=P(X)=P\left(\bigcup_{n\in\Bbb N}\{x_n\}\right)=\sum_{n\in\Bbb N} P(\{x_n\})=\sum_{n\in\Bbb N} 0=0.$$

What all this have to do with your question? Well, easy. We often talk about a "uniform" probability over $[0,1]$ where every singleton has probability zero. If $[0,1]$ is countable, we can't do that.

Solution 4:

One interesting fact that results in topology is that removing a countable set of points from the plane cannot disconnect the plane. This is because you have an uncountable number of slopes to pick from and a countable number of 'obstacles'.

If the plane were disconnected, then we would have an injective map from the set of slopes, $\mathbb{R}$, to the set of natural numbers, $\mathbb{N}$; absurd.