Disproving that $\mathbb N$ is a complete Set

Question

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?

Note that $A \not\subseteq \mathbb N$ and that $\mathbb N$ is not a field. But $\mathbb N$ is a complete metric space with respect to $d(x, y) = | x - y |$ in the sense that every Cauchy sequence converges. — ViktorStein, Jan 06 '21 at 10:49
In your definition of completeness using sups and infs, you need to say that if $A$ is bounded above then it has a sup in $S$, and if $A$ is bounded below then it has an inf in $A$. For example, $\mathbb{R}$ is complete, but the set $A=\mathbb{R}$ has no sup or inf in $\mathbb{R}$. (And, as already pointed out, note that the definition quantifies over $A$ a subset of $S$, and in your example $A$ is not a subset of $\mathbb{N}$). — halrankard2, Jan 06 '21 at 11:29

score 3 · Answer 1 · answered Jan 06 '21 at 10:52

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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.

answered Jan 06 '21 at 10:52

mathcounterexamples.net

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Derek Luna · Accepted Answer · 2021-01-06T10:59:16.583

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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.

edited Jan 06 '21 at 10:59

answered Jan 06 '21 at 10:54

Derek Luna

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score 1 · Answer 3 · answered Jan 06 '21 at 10:59

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 $\mathbb R$.

Disproving that $\mathbb N$ is a complete Set

3 Answers3