Disproving that $\mathbb N$ is a complete Set I want to disprove that $\mathbb N$ is a complete Set.
It makes sense to me that there are Cauchy Sequences in $\mathbb Q$ that converge to a number that is irrational and therefore not in $\mathbb Q$ (e.g. to $\sqrt{2}$), which makes the Set $\mathbb Q$ incomplete.
Is there a way to use this property when working with natural numbers?
Also, there is the Definition, that if every nonempty subset $A \subseteq S$, where $S$ is an ordered field has a $\sup(A) \in S$ and an $\inf(A) \in S$, the Set $S$ is complete (sorry if I understand this incorrectly).
Using this I say:
The set $A=\{5-x:x\in \mathbb N\}$ proves that $\mathbb N$ is incomplete, because $\inf(A) = -\infty \notin \mathbb N$
Is this a valid claim?
 A: In fact $\mathbb N$ is complete (when endowed with the metric induced by the usual one of $\mathbb R$). A Cauchy sequence taking values in $\mathbb N$ is eventually constant and therefore converges.
A: The only way to get a cauchy sequence of natural numbers is to eventually have $|x_{m}-x_{n}| = 0 < \epsilon$ for $m \geq n \geq N$.
Since otherwise there exist $m_{1} \geq n_{1} \geq N$ such that $|x_{m_1} - x_{n_1}| \geq 1 > \epsilon = \frac{1}{2}$ and then the sequence is not a cauchy sequence. This implies that every cauchy sequence in $\mathbb{N}$ converges and so $\mathbb{N}$ is a complete set.
A: If you have a complete metric space $X$ (such as $X=\mathbb R$) and a closed subset $Y$ in it (say $Y=\mathbb N$), then $Y$ will be a complete subspace of $X$.
Proof: Let $(x_n)$ be a Cauchy sequence in $Y$. It follows that it is also a Cauchy sequence in $X$. From the completeness of $X$, it follows that $(x_n)$ converges to some element $x=\lim_{n\to\infty}x_n\in X$. However, as all $x_n\in Y$ and $Y$ is closed, then $x=\lim_{n\to\infty}x_n\in Y$ - in other words, $(x_n)$ converges in $Y$.
Thus, $Y$ is complete. $\blacksquare$
In particular, $\mathbb N$ is complete as it is a closed subset of $\mathbb R$.
