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My attempt:

Suppose not, i.e., suppose there exists a convergent sequence $(a_n)$ that does not have a smallest or largest term.
$\implies (a_n)$ is bounded sequence.
$\implies A=\{a_n:n\in\mathbb{Z}^+\}$ is a bounded subset of $\mathbb{R}$.
$\implies \sup{A},\inf{A}$ exists in $\mathbb{R}$.
Now, we need to show that at least one of $\sup{A},\inf{A}\in A$.
Equivalently, we need to show that the following case is impposible:
"Both $\sup{A}\notin A$ and $\inf{A}\notin A$".

I don't know how to proceed or if I am working it out correctly.

spkakkar
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  • Take some local max/min you find, and consider the limit. If the limit is different from this local max/min, consider $\delta$ less than the difference, and find the point in the sequence where all elements are within $\delta$ of the limit. What does this tell you? – Rushabh Mehta Sep 14 '19 at 03:43
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    If it doesn't have a largest term, it must have an infinite strictly increasing subsequence. If it doesn't have a smallest term, it must have an infinite strictly decreasing subsequence. Such a sequence can't converge. – saulspatz Sep 14 '19 at 03:49
  • @saulspatz, this argument requires $A$ to be infinite set and further, the argument says there exists an increasing sequence in $A$ converging to $\sup{A}$. Such sequence need not be subsequence of $(a_n)$, as far as I know. – spkakkar Sep 14 '19 at 03:53
  • @DonThousand, Your comment is hard for me to follow. Could you be more precise please? – spkakkar Sep 14 '19 at 03:54
  • If we can always inductively create subsequence of $(a_n)$ that converges to $\sup{A}$, then I am done. I just want to confirm this fact. – spkakkar Sep 14 '19 at 04:00
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    If there is no largest term, then there must be a term larger than $a_1.$ Call it $a_{n_1}$. Then there must be an $n_2>n_1$ such that $a_{n_2}>a_{n_1}$, for otherwise, the largest term of the sequence occurs among $a_1,a_2,\dots,a_{n_1}$. And so on. – saulspatz Sep 14 '19 at 04:26
  • @spkakkar: any finite sequence has a largest and smallest term, so your question requires an infinite sequence. – Ross Millikan Sep 14 '19 at 05:22
  • Thanks @saulspatz for the detailed argument. You might right it as answer, so that I could accept it. Thanks all for their valuable time. – spkakkar Sep 14 '19 at 06:22
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    I think the negative of "has a smallest or a largest term" is "does not have a smallest term and does not have a largest term". For example $f(n)=\frac1{1+n^2}$ converges to $0$ as $n \to \infty$ but does not have a smallest term – Henry Sep 14 '19 at 17:13

4 Answers4

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Let $c=\sup A$ and $d=\inf A$. If these values are not attained by the sequence then there is a subsequence $a_{n_k}$ strictly increasing to $c$ and a subsequence $a_{m_k}$ strictly decreasing to $d$. But the sequence is convergent so we must have $c=d$. But then $a_n$ is independent of $n$ contradicting the assumption that sup and inf are not reached.

Kavi Rama Murthy
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  • I love this answer, and voted for it! But I think it's not written in an entirely correct way. Since you start with "If these values are not attained by the sequence", you must reach a contradiction, instead of concluding that "$a_n$ is independent of $n$". --- I would add "This implies $d<c$." after your second sentence, and change your third sentence to "But the sequence is convergent so we must have $c=d$, a contradiction." – Pierre-Yves Gaillard Sep 14 '19 at 22:01
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$A:=${$a_n| n \in \mathbb{Z^+}$}

Since $a_n$ is convergent, it is bounded,

$S:= \sup A$, $I:=\inf A$ exist.

Assume $S, I \not \in A$.

1)$ \sup A \not = \inf A$ ;

There exists a subsequence $a_{n_k}$ of $a_n$ converging to $S$.

There exist a subsequence $a_{n_l}$ of $a_n$ converging to $I$.

Since $a_n$ is convergent every subsequence converges to the same limit.

A contradiction.

2) $S=I$ , and by assumption $S,I \not \in A$,

we have $I <a_n<S$ , i.e .

$I <S$, a contradiction.

Peter Szilas
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1

Case $\bf{1}$:

If there is a $k$ so that $a_k\lt\lim\limits_{n\to\infty}a_n$, then $\inf\limits_{n\ge0}a_n\lt\lim\limits_{n\to\infty}a_n$. Since the limit exists, there is an $n_0$ so that $$ n\ge n_0\implies a_n\ge L=\frac12\left(\inf_{n\ge0}a_n+\lim_{n\to\infty}a_n\right)\tag1 $$ Thus, there are only a finite number of terms where $a_n\lt L$. Since an infimum over a compact set is attained, there must be an $n_1$ so that $a_{n_1}=\inf\limits_{n\ge0}a_n$.

That is, the infimum is attained.

Case $\bf{2}$:

If there is a $k$ so that $a_k\gt\lim\limits_{n\to\infty}a_n$, then $\sup\limits_{n\ge0}a_n\gt\lim\limits_{n\to\infty}a_n$. Since the limit exists, there is an $n_0$ so that $$ n\ge n_0\implies a_n\le L=\frac12\left(\sup_{n\ge0}a_n+\lim_{n\to\infty}a_n\right)\tag2 $$ Thus, there are only a finite number of terms where $a_n\gt L$. Since a supremum over a compact set is attained, there must be an $n_1$ so that $a_{n_1}=\sup\limits_{n\ge0}a_n$.

That is, the supremum is attained.

Case $\bf{3}$:

If neither Case $1$ nor Case $2$ hold, then for all $k$, $a_k=\lim\limits_{n\to\infty}a_n$.

That is, both supremum and infimum are attained.

robjohn
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The sequence is convergent, hence bounded; if $q=\sup\{a_n:n\ge0\}$ and $p=\inf\{a_n:n\ge0\}$, then we know that $l=\lim_{n\to\infty}a_n\in[p,q]$. The case $p=q$ is obvious, so we can assume $p<q$.

Suppose $l>p$. Then, for $n>N$, $a_n>(l+p)/2>p$ and so $p=\inf\{a_n:0\le n\le N\}$ is actually a minimum.

Similarly, if $l<q$.

egreg
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