Show following assertion: if $(a_n)$ is a Cauchy sequence in $\mathbb{N}$, then $a_n$ is constant starting from a certain index.

Question

I need to show the following assertion: if $(a_n)$ is a Cauchy sequence in $\mathbb{N}$, then $a_n$ is constant starting from a certain index.

In other words: $\exists (n_o, N \in \mathbb{N)}, \forall n \geq n_o: a_n = N $

I don't understand how this can be true. This implies that a Cauchy sequence will eventually become constant, which is not true. It converges to a certain value, but never actually reach it. So there clearly is something else I'm not taking into account but I can't figure out what.

Thank you very much. Any help is highly appreciated

Answer 1

You may be slightly mixed up on the difference between being Cauchy and convergent. (Though maybe not; it's hard to tell.) Whether or not a Cauchy sequence must converge depends on the space you're working in. In $\mathbb{R}$, every Cauchy sequence converges, and this is a very special property of the reals. In $\mathbb{Q}$, there are Cauchy sequences that fail to converge, because what they "should" converge to is not rational. (They will still converge to a real number, though, but that's "outside" your space.) The good example is the increasingly-long decimal approximations to $\sqrt{2}$.

In your problem, you have a Cauchy sequence of natural numbers. And yes, in your case, this will necessarily converge (to the eventual constant). Let $\{a_n\}$ be your Cauchy sequence of naturals. Let $\epsilon = 0.5$. Then there is an index $k$ such that $m,n \geq k$ implies that $|a_m-a_n| < 0.5$. But the closest any two distinct natural numbers can ever be is $1$: this occurs only when they are consecutive, of course. Since $a_m$ and $a_n$ are natural, this implies that we must have $a_m=a_n$ for all indices $m,n \geq k$. In particular, the sequence must be constant starting at index $k$ (maybe earlier) and beyond. Hence the sequence is eventually constant with value $a_k$, and that will be what it converges to.

It is good to work out what goes wrong with this argument in $\mathbb{Q}$ or $\mathbb{R}$, because as you pointed out, it is definitely not true there.

Answer 2

Suppose $a_n$ is a cauchy sequence in $\mathbb N$. So, for $0<\epsilon <1$ you have a $k\in \mathbb N$ so that, $|a_n-a_m|< \epsilon $ for all $n,m\geq k$ choose $ m$ to be $k$ and you get, $a_n= a_k$ for all $n\geq k$.

So you obtain that the sequence is eventually constant.

Note that, we were totally inside $\mathbb N$ while discussing this. So, we know distance between two distinct points is at least 1. This is not the case in $\mathbb R$. So, this is not true that a real cauchy sequence is eventually constant.

If you know little bit about metric spaces, then all of this discussion holds true for discrete metric space also.

Note that, if you have a subset of $\mathbb R $ (or more generally a metric space) having $inf \{d(x, y) : x\neq y\}>0 $ the same result holds there. And it is essentially proved as we did here. Just modify the upper bound 1 with some suitable number. Here $d(x, y) =|x-y|$.

Answer 3

Because $(a_n)$ is Cauchy, there exists $N$ such that $m,n\ge N$ implies $|a_n-a_m|<1/2.$ Now if $a,b\in \mathbb N$ are distinct, then $|a-b|\ge 1.$ It follows that among the integers $a_N,a_{N+1},\dots,$ no two of them can be distinct. Therefore $a_n=A_N$ for $n\ge N.$