Density of rationals and real numbers clamped between close rational numbers

Question

I was watching MIPT lection on math in russian

It was stated there that $$\forall \alpha,\beta\in\mathbb{R}, \alpha<\beta, \exists r\in\mathbb{Q}: \alpha<r<\beta$$

(some article on internet claims that this is true for $\mathbb{R}\backslash\mathbb{Q}$ as well rat_dense.pdf)

That is ok

But then it is proven (in video) that

if $$\forall\alpha,\beta\in\mathbb{R},\\ \forall\varepsilon\in\mathbb{Q},\varepsilon>0,\\ \exists S_1,S_2\in\mathbb{Q}:S_1<\alpha<S_2, S_1<\beta<S_2,S_2-S_1<\varepsilon$$

then $\alpha=\beta$

it is proven by assuming $\alpha\neq\beta$ (case of $\alpha<\beta$) which mean that there can be two rational numbers between them $r_1,r_2\in\mathbb{Q}, S_1<\alpha<r_1<r_2<\beta<S_2$, and difference of that rational numbers is rational number and $S_2-S_1>r_2-r_1>0$ and if $\varepsilon=r_2-r_1$ then $S_2-S_1<\varepsilon$ will fail, which proves that $\alpha=\beta$

But now I lost. If I understand correctly $\mathbb{Q}\subset\mathbb{R}$ so it is correct that $S_1,S_2\in\mathbb{R}$ as well, which mean that if there can be 1 number between them then there can any amount of $\mathbb{Q}$ or $\mathbb{R}\backslash\mathbb{Q}$ which contradicts $\alpha=\beta$ and $S_1<S_2$, it seems to me that $S_1=S_2$ but that will cuase whole stetement to fail. I don't understand this.

Answer 1

That's the point - we want it to fail, because we're trying to prove a contradiction. To put it into some more words:

1. Our assertion is that we can always "trap" both $\alpha$ and $\beta$ in an interval whose endpoints are rational numbers but whose width is arbitrarily small (i.e. if you pick a positive value $\epsilon$, you can always find an interval smaller than that).

2. We then assume that $\alpha \neq \beta$ (and without loss of generality, that $\alpha < \beta$).

3. We use that to find two distinct rational numbers between them.

4. Those rational numbers also have to be trapped in the arbitrarily small interval, so we choose $\epsilon$ to be the width of their interval.

5. From our initial assertion, we know that there must be rational $S_1$ and $S_2$ that are closer than $\epsilon$ apart but which sandwich $\alpha$ and $\beta$ between them ...

6. But that means that $r_1$ and $r_2$ are also sandwiched between $S_1$ and $S_2$, so $S_1$ and $S_2$ are forced to be less than $\epsilon$ apart but also further than $\epsilon$ apart at the same time ...

7. So we've reached a contradiction, which means we made a false assumption. And the only assumption was that $\alpha \neq \beta$.