# A sequence in $\mathbb{R}$ that has no Cauchy subsequence

Give an example of a sequence in $$\mathbb{R}$$ which has no subsequence which is a Cauchy sequence.

I can find out a sequence that is not a Cauchy sequence such as $$\{\ln(n)\}$$ once $$|\ln(n)-\ln(n+1)|=0$$ but $$|\ln(n)-\ln(2n)|=|\ln(\frac{1}{2})|>\epsilon$$
$$\forall \epsilon<\ln(\frac{1}{2})$$

I can still find a subsequence of the type $${\ln(2n)}_{2n\in\mathbb{N}}$$ such that $$|\ln(2n)-\ln(2n+1)|=0$$

Question:

What should I do to get a sequence that has no Cauchy subsequence?

• In the first place, you should only be looking at sequences which are not convergent, right?
– MPW
Commented Nov 22, 2018 at 22:57
• I do not understand your question. What do you mean by $|\ln(2n)-\ln(2n+1)|=0$? Commented Nov 23, 2018 at 10:29

Take $$a_n=n$$. Then for any subsequence $$n_k$$, $$|n_k-n_{k-1}|\geq 1$$. So, it has no Cauchy sub-sequence.

Take a sequence $$(a_n)$$ such that $$a_n\to \infty$$. Then each subsequence $$(a_{n_k})$$ is such that $$a_{n_k}\to \infty$$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.

Take the sequence $$1,2,3,4,\ldots$$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $$1$$.

As Foobaz John says, any sequence with $$a_n\to\infty$$ works, and it's easy to see that $$|a_n|\to\infty$$ is also sufficient. In fact this latter condition is necessary and sufficient. If $$|a_n|\not\to\infty$$, then by definition there is some $$c>0$$ for which there are infinitely many values $$n_i$$ such that $$|a_{n_i}| for each $$i$$. Now by Bolzano-Weierstrass the subsequence $$a_{n_i}$$ has an infinite subsequence $$a_{n_{i_j}}$$ which is convergent, and therefore Cauchy.

• Ah, Bolzano-Weierstrass, also known as the theorem about subsubsubscripts. Commented Nov 23, 2018 at 18:32

I didn't understand your analysis but sequence $$a_n = \ln n$$ doesn't have Cauchy subsequence since its limit is $$\infty$$

• I guess this one uses the theorem that every Cauchy sequence converges Commented Nov 23, 2018 at 21:15

Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.

Fact: Every convergent sequence is bounded.

Strategy: Try an unbounded sequence.

Guess: $$a_n=n$$

Conclusion: (I leave it to you)