How do I prove that $\forall x\in\mathbb{R}:(\forall y\in\mathbb{R}: y>0 \implies 0\leq x\lt y)\implies x=0$

Question

When reading this proof of the uniqueness of limits of sequences I stumbled across this argument (in the last few lines):

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We have a $y\in\mathbb{R}$ with $y>0$. (Originally called $\epsilon$)

We have a $x\in\mathbb{R}$ with $x\geq0$. (Originally called $|x-y|$)

We have $x\leq(\text{some value})\lt y$.

Since $y>0$ is arbitrary, we conclude that $$x=0$$

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I think this arguments validity would follow from a proof of this Proposition:

$$\forall x\in\mathbb{R}:(\forall y\in\mathbb{R}: y>0 \implies 0\leq x\lt y)\implies x=0$$

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Does this arguments validity follow from this proposition and if so, how do I prove it?

(Also this feels awfully close to the squeeze theorem. Are they related?)

Answer 1

Yes, you could say it follows from that proposition, or rather, they are actually equivalent statements. This is clear if you write them side by side.

In the original one that you referred to, we are forced to conclude that

$$x = y \text{ if } 0 \leq |x - y| < \varepsilon \text{ for all } \varepsilon > 0.$$

In this one, we are forced to conclude

$$x = 0 \text{ if } 0 \leq x < \varepsilon \text{ for all } \varepsilon > 0.$$

The idea is that if a non-negative number $x$ is smaller than any number $\varepsilon$ that can be made really-really small (so it must be smaller than $\varepsilon_0 = 10^{-9999999}$ and $\varepsilon_1 = 10^{-2020201920182017}$, etc.), then it must be $0$ itself.

To prove the statement, you could do it by contradiction:

Proof. Suppose for all $y > 0$, we have $0 \leq x < y$. Suppose for contradiction that $x \neq 0$. Then by considering $y_0 = x > 0$, we get a contradiction since we have both $x < y_0$ and $x = y_0$ (where the first is by hypothesis and the second is by our assumption for contradiction). So $x$ has no choice but to be $0$.

Answer 2

The proof indeed follows from this proposition; the last lines of the proof are more or less a rephrasing of it.

The proposition can be proved by contradiction. Suppose that $x \geq 0$ has the property that for any $y \in \mathbb{R}$, $$ y > 0 \implies x < y. \tag{1} $$ For a contradiction, suppose that $x > 0$. Since $x > 0$ then it follows directly from $\frac{1}{2} < 1$ that $\frac{x}{2} < x$. However, since $x > 0$, then $\frac{x}{2} > 0$. Hence, by property $(1)$, then $x < \frac{x}{2}$, which contradicts the inverse inequality. So $x \not> 0$, that is $x \leq 0$. Since $x \geq 0$, then $x = 0$.

So any $x \geq 0$ that has property $(1)$ must be zero.

Answer 3

Arbitrary $\varepsilon>0$ means you can consider $\varepsilon$ to be arbitrarily small.

Hence, the argument $\left|x-y\right|<\varepsilon$ for any $\varepsilon>0$ is basically saying that the distance between $x$ and $y$ is as small as you want it to be, and therefore they must be the same number.

You are correct, it's kind of a squeeze argument since you are considering $0\leq\left|x-y\right|<\varepsilon$.