If $a\in \mathbb{R}$ and $0\leq a<\frac{1}{n}, ~\forall ~n\in \mathbb{N}$, show that $a=0$.
I can prove it by sandwich rule (squeeze lemma) of limits. But I want to use the concept of Real Number system only rather than the concept of limit. Please help.
Solution 1: Let $\epsilon>0$. Then by the Archimedean Property, there exists $n\in\Bbb N$ such that $\frac{1}{n}<\epsilon$. Thus, $0\leq a<\frac{1}{n}<\epsilon.$ Thus, $0\leq a<\epsilon \quad \forall \epsilon >0$. Hence, $0\leq a\leq 0$. Thus, $a=0$.
Solution 2: We know that $$0\leq a\tag 1$$ and $$a<\frac{1}{n}\quad\forall n\in\Bbb N\tag 2.$$ Suppose that $0<a$. Then, Archimedean Property says that $$\frac{1}{n}<a\quad\text{for some }n\in\Bbb N\tag 3.$$ Statement $(2)$ contradicts statement $(3)$. Therefore, we cannot assume that $0<a$. Because statement $(1)$ is true and $0<a$ is false, it follows that $0=a$.
Suppose $a \ne 0$. Because $a \ge 0$ from hypothesis it follows $a \gt 0$. Now let's take $n = [\frac 1a] + 1$ where [] is the integer part. We have $n \in \mathbb{N}$ and it's easy to show that $\frac 1n \lt a$, which contradicts the hypothesis.