# Any Cauchy sequence in the metric space $(\Bbb N, d)$ is either “converging” to infinity or ultimately constant.

Given a metric space $(\Bbb N,d)$ where $d(x,y)= |\frac{1}{x}-\frac{1}{y}|.$

I need to prove that for any Cauchy sequence $(n_j)_{j \in \Bbb N}$ in this metric space, it either satisfies the property that as $j \rightarrow \infty$, $n_j \rightarrow \infty$ or ultimately constant.

Here is the definition of Cauchy seuence:

Let $(x_n)^∞ _{n=1} ⊂ X$ where $(X, d)$ is a metric space. Then $(x_n)$ is a Cauchy sequence if for every $\varepsilon > 0$ there exists an integer $N$ such that $m, n ≥ N ⇒ d(x_m, x_n) < \varepsilon$ .

If the Cauchy sequence satisfies the property that as $j \rightarrow \infty$, $n_j \rightarrow \infty$, then we are done. I need to show that any Cauchy sequence that does not satisfy the property must be ultimately constant.

• Just to confirm, your sequence is a subset of the natural numbers. Right? – codetalker Apr 4 '17 at 5:56
• @Siddhant Yes the sequence is in the metric space $(\Bbb N,d)$. – PropositionX Apr 4 '17 at 5:58

This will be an informal argument, so you can use it to build a proof.

What does it mean that $n_j \to \infty$ as $j \to \infty$?

This means that, given any positive number $\varepsilon$, there is a tail of the sequence above $\varepsilon$. In other words there are, at most, a finite number of elements of the sequence below $\varepsilon$.

What is the negation of this?

Well, this means that there is a particular positive number $\varepsilon_*$ such that, at most, a finite number of elements of the sequence are bigger than $\varepsilon_*$. So the sequence is bounded above (not necessarily by $\varepsilon_*$, can you see this?). But this is a sequence of natural numbers, hence bounded below by $0,$ so we have a bounded sequence of natural numbers.

What can we deduce from this?

Well, a bounded sequence of natural numbers means that there are only a finite number of possible values that the elements of the sequence can take, say $V=\{v_1, v_2, \ldots, v_m \} \subset \mathbb{N}.$

Let $$W = \Big\{|\frac{1}{v_i} - \frac{1}{v_j}| : v_i,v_j \in V, v_i \neq v_j \Big\},$$

the set of distances between elements of the sequence with distinct values. Since there are only a finite number of elements in $V$, the set $W$ is also finite, so take the minimum of these differences, $w_* = \min W.$

What to do with this? The Cauchy hypothesis

Now, let $\epsilon > 0$ such that $\epsilon < w_*$. Since $(n_j)_{j \in \mathbb{N}}$ is a Cauchy sequence, there is a tail of the sequence such that, taking any 2 elements of the tail, their distance is less than (or equal to) $\varepsilon$, which is less than $w_*.$

Now the concluding insight

Suppose that $n_i, n_j$ are on the tail given by the Cauchy assumption, can $n_i \neq n_j?$

Edit:

$$n_j \to \infty \iff \forall \varepsilon \, \exists N : j\geq N \implies n_j > \varepsilon$$

so, the negation would be

$$n_j \not \to\infty \iff \exists \varepsilon \, \forall N :j\geq N \, \land n_j \leq \varepsilon.$$

Note that the statement after $:$ is a statement of the form $p \to q$ which is true if $\, \lnot (p \land \lnot q$), so negating it gets you $p \land \lnot q$.

Hope this helps :)

• For the "boundedness" part of your answer, could you please tell me how can I see that at most, a finite number of elements of the sequence are bigger than ε∗. And how can I then conclude that the sequence is bounded above? – PropositionX Apr 5 '17 at 6:15
• From $(\exists \epsilon>0)(\forall N\in \Bbb N)(\exists j>N)(n_j\le \epsilon)$ I can only see the fact that there are infinitely many terms which is less or equal to $\epsilon$, instead of there are finitely terms which is greater then $\epsilon$. – PropositionX Apr 5 '17 at 6:34
• @PropositionX Edited my answer, hope is clearer – Luis Vera Apr 5 '17 at 9:26
• I think the original statement is $n_j→∞⟺(∀ε)(∃N)(\forall j)(j≥N⟹n_j>ε)$ So the negation would be $nj↛∞⟺(∃ε)(∀N)(\exists j)(j≥N∧_nj≤ε)$. instead of $nj↛∞⟺(∃ε)(∀N)(\forall j)(j≥N∧n_j≤ε)$. Where the latter one is exactly what I need to conclude boundedness. – PropositionX Apr 5 '17 at 9:45
• @PropositionX I was using "Syntactic sugar" but your statement also implies "boundedness". if ($\forall N)(\exists j)$ in particular, for $N=1$ there is a $j \in \mathbb{N}$ such that $(j\geq 1 \land n_j \leq \varepsilon).$ So you see, only those that are less than $j$ can be (not necessarily) above $\varepsilon$ – Luis Vera Apr 5 '17 at 9:52

Your space is isometric to the set $$S = \{1/n\,\vert\,n\in\Bbb N\}\subset\Bbb R$$ with the usual distance.Let be $(x_n)_{n\in\Bbb N}$ a Cauchy sequence in this space. If the sequence takes a finite number of values, is eventually constant. If the sequence takes an infinite number of values, $0$ is an accumulation point of it and exists a convergent subsequence $(x_{n_k})_{k\in\Bbb N}\to 0$, but as $(x_n)_{n\in\Bbb N}$ is a Cauchy sequence, is convergent to the same limit. Now, you can translate this to your space using the isometry.

If $x_n$ is bounded, exists $M\in\mathbb N$ such that, for all $n$, $x_n<M$ . We translate "$\{x_n\}$ is ultimately constant" as "exists $N$ such that for all $n,m>N$, $\vert x_n-x_m\vert=0$". Suppose that $\{x_n\}$ it's not "ultimately constant" so is, for all $N'$ exist $m',n'>N'$ such that $\vert x_{n'}-x_{m'}\vert\gt 0$.

$$d(x_{n'},x_{m'})= \left|\frac{1}{x_{n'}}-\frac{1}{x_{m'}}\right|= \frac{\vert x_{m'}-x_{n'}\vert}{x_{n'}x_{m'}}\ge\frac{1}{x_{n'}x_{m'}}\gt\left|\frac{1}{M^2}\right|=\epsilon$$

So, the sequence cannot be a Cauchy sequence. The "ultimately constant" sequences are obviously Cauchy sequences and it's proved.

• Thanks! An explicit $\epsilon$ is exactly what I expect. But could you please tell me how to show that the sequence is bounded? – PropositionX Apr 5 '17 at 0:22
• By your start of the proof! You said that we have to consider the contrary of diverging: "if the Cauchy sequence satisfies the property that as $j\to\infty, n_j\to\infty$, then we are done." so, we only have to prove that bounded and Cauchy implies "ultimately constant", and I did by proving that if it isn't "ultimately constant" then it cannot be Cauchy – Rafa Budría Apr 5 '17 at 5:02
• In fact diverging sequences exist that aren't Cauchy, but, as you stated correctly, they don't contradict the affirmation we've proved. – Rafa Budría Apr 5 '17 at 5:13
• I do know that you mean. But now what I am asking is that how to do a step-by-step deduction from the negation of $j→∞,n_j→∞$ to for all $n\in\Bbb N, x_n<M$. To be more explicit, how to conclude $(\forall n\in\Bbb N)( x_n<M)$ from the negation of $(\forall \epsilon\in \Bbb N)(\exists N\in \Bbb N)(\forall j\in\Bbb N)(n_j>\epsilon)$, that is $(\exists \epsilon\in \Bbb N)(\forall N\in \Bbb N)(\exists j\in\Bbb N)(n_j\le \epsilon)$ – PropositionX Apr 5 '17 at 5:14
• Oh, no, no, diverging is directly affirm that $\forall M\in\mathbb N,\;\exists n, x_n>M$ – Rafa Budría Apr 5 '17 at 5:21

Note that if $n_j$ is a cauchy sequence in that space $1/n_j$ is a cauchy-sequence in $\mathbb R$. This means that $1/n_j$ converges.

Since $n_j> 0$ we have either $1/n_j\to 0$ in which case $n_j\to infty$ or $1/n_j\to L>0$ which means that $n_j\to 1/L$ (also since $n_j$ is integer this means that $n_j=1/L$ ultimately).