Proving that there are infinitely many real numbers between two real numbers

Question

I want to prove that: $$\forall\alpha,\beta\in\mathbb{R}\text{ exactly one of the following holds:}$$

1. $\alpha=\beta$
2. There are infinitely many numbers that are strictly between $\alpha$ and $\beta$.

My proof:

If $\alpha=\beta$, case (1) is satisfied and we don't need any more proof. Let us now assume that $\alpha\neq\beta$ and prove (2).

Let us consider the interval $[\alpha,\beta]$. Since $\alpha\neq\beta$, this interval isn't empty and contains at least $1$ element other than $\alpha$ and $\beta$. Let us assume that there is a finite number of other elements in this interval, let's say $n$ elements, with $n \in\mathbb{N}$. We'll call this statement $S(n)$.

Let us now consider the following interval: $$\left[\alpha, \alpha+\frac{|\alpha-\beta|}{2}\right]$$

This interval isn't empty because $\frac{|\alpha-\beta|}{2} > 0$, due to our hypothesis $\alpha\neq\beta$.

This interval is half as long as the original one, and will have at most $n-1$ elements.

If we repeat the procedure, we'll get another interval that is half the distance of the second one, but we are still guaranteed that it won't be empty because of our initial hypothesis of the two extrema points not being equal.

Iterating this process $n+1$ times will yield an interval that still has at least one element other than the extrema. This is a contradiction to the statement $S(n)$, and quickly shows that there is no finite $n$ for which the statement holds true.

Therefore, the number of elements in our interval is infinite, for all $\alpha,\beta$ reals.

Is this proof complete or am I missing anything?

Answer 1

Yes, it is correct, but you didn't have to make the whole $S(n)$ proof by contradiction argument. You could have just said that each time we halve the interval, we get a new element. Since we can halve the interval a countable number of times, we have at least a countable number of elements in the original interval.

Answer 2

if $\alpha =\beta$ then we are done.

When $\alpha \neq\beta$, without loss of generality suppose $\alpha \gt\beta$.

Then consider $x_n=\beta+\frac{\alpha-\beta}{n}$.

Clearly $x_n\gt \beta$ for each $n\in\mathbb Z^+$. And for $n\geq2$ if $x_n\geq \alpha \implies n\leq1$ which is a contradiction. So $x_n\lt \alpha$ for $n\geq2$.

And for $n\neq m$ clearly $x_n\neq x_m$ thus there are infinitely many reals in between $\alpha,\beta$.