$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

I am doing exercise 2.5.2 of Understanding Analysis by Stephen Abbott. Is my justification for the below propositions sound?

Decide whether the following propositions are true or false, providing a short justification for each conclusion.

(a) If every proper subsequence of $(x_n)$ converges, then $(x_n)$ converges as well.

(b) If $(x_n)$ contains a divergent subsequence, then $(x_n)$ diverges.

(c) If $(x_n)$ is bounded and diverges, then there exist two subsequences of $(x_n)$ that converge to different limits.

(d) If $(x_n)$ is monotone and contains a convergent subsequence, then $(x_n)$ converges.


(a) This proposition is false. As a counterexample, consider the sequence \begin{align*} \left(1,1,\frac{1}{2},1,\frac{1}{2},\frac{1}{3},1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right) \end{align*}

The subsequences $(1,1,1,1,\ldots)$, $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\ldots\right)$, $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3},\ldots\right)$ are all convergent subsequences with different limits. Hence, $(x_n)$ is not a sequence.

The Monotone Convergence Theorem only guarantees: a bounded sequence of real numbers has atleast one convergent subsequence. These subsequences may or may not converge to the same limiting value.

If every proper subsequence of $(x_n)$ converges to the same limit $a$, then $(x_n)$ converges as well.

(b) This proposition is true.

Justification. If a subsequence of $(x_n)$ diverges, it implies that, given any $\epsilon >0$, there are an infinite number of real numbers in the tail that do not lie in an $\epsilon$-neighbourhood of a limit point. So, the original sequence is divergent.

(c) This proposition is true.

Justification. Assume that the sequence $(x_n)$ is bounded, but divergent. By definition, there exists large number $M$, such that $\absval{x_n} \le M$ for all $n \in \mathbf{N}$. Consider the subsequences \begin{align*} (x_1,x_2,x_3,x_4,\ldots) \\(x_2,x_3,x_4,\ldots) \\(x_3,x_4,\ldots) \\\vdots \end{align*} Each of these subsequences are bounded subsets of $\mathbf{R}$ and therefore by AoC, it is reasonable to talk about their supremum and infimum. Define the subsequences,

\begin{align*} u_n := \sup \{x_k:k \ge n\}\\ d_n := \inf \{x_k:k \ge n\} \end{align*}

$u_n$ is bounded and monotonically decreasing. $d_n$ is bounded and monotonically increasing. By the Montone Convergence theorem, $u_n$ and $d_n$ are convergent.

(d) This proposition is true.

Justification. Assume that the subsequence $(x_{n_k})$ converges to $x$. Then, given an $\epsilon > 0$, there exists $N$ such that $x_{n_k}$ lies in the interval $(x-\epsilon,x+\epsilon)$ for all $n \ge N$. Because $(x_n)$ is monotone, the values of the sequence $(x_n)$ between $x_{n_k}$ and $x_{n_{k+1}}$ also lie in $(x-\epsilon,x+\epsilon)$. Consequently, the same $N$ can be used for the original sequence.

  • 1
    $\begingroup$ (a) is trivially true because every sequence is a subsequence of itself (similar to this i guess: math.stackexchange.com/questions/1990948/…) but i think what it should have said is like every subsequence EXCEPT POSSIBLY the original sequence itself. $\endgroup$
    – BCLC
    Dec 23, 2020 at 22:54
  • $\begingroup$ wait this is weird. the original revision of the post indeed says 'proper subsequence'. must've missed that. lol. ok so (a) is not trivially true. never mind. lol. $\endgroup$
    – BCLC
    Dec 25, 2020 at 16:46

1 Answer 1


For a) I think it's true. First you counter-example does not work : it has many non-convergent subsequences, for example $(1, 1/2, 1, 1/2, ...)$. Then it's actually true even if you only suppose that $(x_{2n})$, $(x_{2n+1})$ and $(x_{3n})$ converge. In fact, then if we note $a, b, c$ the limit of each, we have $(x_{6n})$ that can be extracted from both $(x_{3n})$ and $(x_{2n})$, so $a=c$, and considering $(x_{6n+3})$, we also have $b=c$. So $a=b$. Thus $(x_n)$ converges.

For b) you can also proceed by contraposition : if $(x_n)$ converges, then all subsequences converge.

For c), as $(x_n)$ is bounded, we can extract $(x_{\varphi(n)})$ that converges. Let's call $\ell$ its limit. As $(x_n)$ does not converge to $\ell$, there exists an $\varepsilon$ such that for all $N\in \mathbb{N}$, there exists a $n \geqslant N$ such that $|x_n - \ell| > \varepsilon$. So we can extract another subsequence that does not converge to $\ell$. This subsequence being bounded too, we can extract from it a another subsequence that converges to $\ell ' \neq \ell$. So we found two subsequences that converge to different limits.

For d), you proof is fine.

  • $\begingroup$ I was thinking about the sequence $x_n := \sin \left(\frac{n\pi}{4}\right)$. If one samples the function values at small enough intervals, the subsequences are divergent. $(x_{4n})$ is convergent, but $(x_{2n})$ is not. If and only if, all proper subsequences converge, the sequence $(x_n)$ converges. $\endgroup$
    – Quasar
    Dec 23, 2020 at 18:39
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    $\begingroup$ For your first remak @BCLC I fell quite stupid not having seen it. But I agree, the right question should have suppose all the proper subsequences converge. $\endgroup$
    – math
    Dec 24, 2020 at 8:39
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    $\begingroup$ Maybe I should have used the word 'strict'. I'm sure this one only excludes the original sequence. As for 'proper', I think it's the same, because when you want to exclude both original and empty set, we say 'non trivial'. (But actually I'm French so maybe we don't have the same words...) $\endgroup$
    – math
    Dec 24, 2020 at 13:21
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    $\begingroup$ It's only true if your 3rd subsequence has an infinite number of terms that are odd, and an infinite number of terms that are even. $\endgroup$
    – math
    Jan 1, 2021 at 13:58
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    $\begingroup$ And I agree for your previous comment @BCLC $\endgroup$
    – math
    Jan 1, 2021 at 14:01

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