How to prove real numbers are uncountable using turing machine?
Answers
Step-by-step explanation:
Theorem: There is no bijection from the natural numbers \mathbb{N} to the real numbers \mathbb{R}.
Proof. Suppose to the contrary (i.e., we’re about to do proof by contradiction) that there is a bijection f: \mathbb{N} \to \mathbb{R}. That is, you give me a positive integer k and I will spit out f(k), with the property that different k give different f(k), and every real number is hit by some natural number k (this is just what it means to be a one-to-one mapping).
First let me just do some setup. I claim that all we need to do is show that there is no bijection between \mathbb{N} and the real numbers between 0 and 1. In particular, I claim there is a bijection from (0,1) to all real numbers, so if there is a bijection from \mathbb{N} \to (0,1) then we could combine the two bijections. To show there is a bijection from (0,1) \to \mathbb{R}, I can first make a bijection from the open interval (0,1) to the interval (-\infty, 0) \cup (1, \infty) by mapping x to 1/x. With a little bit of extra work (read, messy details) you can extend this to all real numbers. Here’s a sketch: make a bijection from (0,1) to (0,2) by doubling; then make a bijection from (0,2) to all real numbers by using the (0,1) part to get (-\infty, 0) \cup (1, \infty), and use the [1,2) part to get [0,1] by subtracting 1 (almost! To be super rigorous you also have to argue that the missing number 1 doesn’t change the cardinality, or else write down a more complicated bijection; still, the idea should be clear).
Okay, setup is done. We just have to show there is no bijection between (0,1) and the natural numbers.