Show that $\inf(\frac{1}{n})=0$.
We are given the following definition:
If a sequence $(a_n)$ is bounded from below then there is a greatest lower bound for the sequence called the infimum.
$m=\inf(a_n)$ if
i) $(a_n) \geq m \ \forall n \in \mathbb{N}$.
ii) For each $\epsilon >0 \ \exists \ n_\epsilon \ \in \mathbb{N}$ such that $a_{n_\epsilon} < m + \epsilon $.
Show using the definition that, if $a_n=\frac{1}{n}$ then $\inf(a_n)=0$.
Here is what I have:
i) RTP: $\frac{1}{n} \geq 0 \ \forall n\in \mathbb{N}$.
Assume that $\frac{1}{n} < 0$ for some $n \in \mathbb{N}$.
$\implies n < 0$.
But we know that $n \in \mathbb{N}, n \geq 1$, and thus we can say that$\frac{1}{n} > 0\ \forall n \in \mathbb{N}$.
ii) Let $ \epsilon > 0$ be given.
RTP: $\exists n_\epsilon \in \mathbb{N}$ such that $\frac{1}{n_\epsilon}<0+\epsilon = \epsilon$.
Proof:
$\frac{1}{n_\epsilon} < \epsilon \implies n_\epsilon > \frac{1}{\epsilon}$, since both values are positive.
For the given $\epsilon > 0$, choose $n_\epsilon > \frac{1}{\epsilon} \in \mathbb{N}$.
Now from the Archimedes Postulate, we know $\exists n \in \mathbb{N}$ such that $n \geq n_\epsilon$.
$\implies n \geq n_\epsilon > \frac{1}{\epsilon}$
$\implies \frac{1}{n} \leq \frac{1}{n_\epsilon} < \epsilon$, since all values are positive.
It's correct but can be written in fewer words:
- $n>0$ implies $\frac{1}{n}>0$ so $a_n>0$ for all $n>0$;
- given $\epsilon>0$, just pick any integer $n>\frac{1}{\epsilon}$, then $a_n=\frac{1}{n}<\frac{1}{1/\epsilon}=\epsilon$.
By pure substitution,
$0=\inf\left(\frac1n\right)$ if
i) $\left(\frac1n\right) \geq 0 \ \forall n \in \mathbb{N}$.
ii) For each $\epsilon >0 \ \exists \ n_\epsilon \ \in \mathbb{N}$ such that $\frac1{n_\epsilon} < \epsilon $.
Then
i) is obvious.
ii) holds by taking $n_\epsilon=\lceil\frac1\epsilon\rceil$.