1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Just to confirm, your sequence is a subset of the natural numbers. Right? $\endgroup$ – codetalker Apr 4 '17 at 5:56
  • $\begingroup$ @Siddhant Yes the sequence is in the metric space $(\Bbb N,d)$. $\endgroup$ – PropositionX Apr 4 '17 at 5:58
1
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – PropositionX Apr 5 '17 at 6:15
  • $\begingroup$ 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 $\epsilon$. $\endgroup$ – PropositionX Apr 5 '17 at 6:34
  • $\begingroup$ @PropositionX Edited my answer, hope is clearer $\endgroup$ – Luis Vera Apr 5 '17 at 9:26
  • $\begingroup$ 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. $\endgroup$ – PropositionX Apr 5 '17 at 9:45
  • $\begingroup$ @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 $\varepsilon$ $\endgroup$ – Luis Vera Apr 5 '17 at 9:52
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks! An explicit $\epsilon$ is exactly what I expect. But could you please tell me how to show that the sequence is bounded? $\endgroup$ – PropositionX Apr 5 '17 at 0:22
  • $\begingroup$ 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 $\endgroup$ – Rafa Budría Apr 5 '17 at 5:02
  • $\begingroup$ In fact diverging sequences exist that aren't Cauchy, but, as you stated correctly, they don't contradict the affirmation we've proved. $\endgroup$ – Rafa Budría Apr 5 '17 at 5:13
  • $\begingroup$ 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 $(\exists \epsilon\in \Bbb N)(\forall N\in \Bbb N)(\exists j\in\Bbb N)(n_j\le \epsilon)$ $\endgroup$ – PropositionX Apr 5 '17 at 5:14
  • $\begingroup$ Oh, no, no, diverging is directly affirm that $\forall M\in\mathbb N,\;\exists n, x_n>M$ $\endgroup$ – Rafa Budría Apr 5 '17 at 5:21
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.