Monotone-sequences property ⇒ Archimedean property
Today I just started learning this and having trouble understanding parts of the proof. Sorry if these are easy to some, I just couldn't find this answered online.
I am using here Archimedean property $\frac{1}{n}→0$. I also have to use this lemma:Let $(a_n)$ be a sequence and suppose that $a_n \to a$. If $(\forall n)\, a_n \leq x$, then $a\leq x.$
My proof:
We observe that $\frac{1}{n}$ is decreasing and bounded below by $0$. Then by the Monotone Sequences Property, it converges to some limit $\delta$. Noticing $\frac{1}{n} > 0$ for all $n$, applying the lemma, we get $\delta \geq 0$.
Assume for contradiction that $\delta > 0$.
Let's set $\epsilon = \delta$. Since $\frac{1}{n} \to \delta$, by the definition of convergence, there exists $N$ such that for all $n \geq N$, we have: $\left|\frac{1}{n} - \delta\right| < \delta$.
This simplifies to: $0 < \frac{1}{n} < 2\delta$ for all $n \geq N$.
Now, since this holds for all $n \geq N$, consider specifically $n \geq 4N$. For such $n$, we have: $\frac{1}{n} \leq \frac{1}{4N}$.
Since $\frac{1}{N} < 2\delta$, multiplying both sides by $\frac{1}{4}$ gives: $\frac{1}{4N} < \frac{\delta}{2}$.
Combining these results, for $n \geq 4N$, we get: $\frac{1}{n} \leq \frac{1}{4N} < \frac{\delta}{2}$.
This contradicts our earlier conclusion that $0 < \frac{1}{n} < 2\delta$ for all $n \geq N$, because $\frac{1}{n} < \frac{\delta}{2}$ implies that $\delta$ is not the smallest bound. Thus, the assumption $\delta > 0$ must be false.
Therefore, $\delta = 0$.
My question:
How exactly are we getting the contradiction here from $\frac{1}{n} < 2\delta$ and $\frac{1}{n} < \frac{\delta}{2}$? Clearly $\frac{\delta}{2}$ fits into the interval $0 < \dots < 2\delta$, so they both can be true. So where are we getting at?