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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.

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  • $\begingroup$ 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? $\endgroup$ Sep 14, 2019 at 3:43
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    $\begingroup$ 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. $\endgroup$
    – saulspatz
    Sep 14, 2019 at 3:49
  • $\begingroup$ @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. $\endgroup$
    – spkakkar
    Sep 14, 2019 at 3:53
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    $\begingroup$ 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. $\endgroup$
    – saulspatz
    Sep 14, 2019 at 4:26
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    $\begingroup$ 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 $\endgroup$
    – Henry
    Sep 14, 2019 at 17:13

4 Answers 4

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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.

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  • $\begingroup$ 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." $\endgroup$ Sep 14, 2019 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.

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  • $\begingroup$ Thanks, for case-wise analysis. $\endgroup$
    – spkakkar
    Sep 14, 2019 at 6:52
  • $\begingroup$ spkakkar. Welcome.:) $\endgroup$ Sep 14, 2019 at 6:53
  • $\begingroup$ The first contradiction is unnecessary. You started with $S \neq I$, then showed that $S = I$ without using your assumption. $\endgroup$
    – Wood
    Sep 14, 2019 at 14:49
  • $\begingroup$ Wood. Don't get your point. $\endgroup$ Sep 14, 2019 at 17:08
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    $\begingroup$ @Wood Is your idea to delete the lines "1) $\sup A \neq \inf A$" and "Hence a contradiction"? $\endgroup$
    – David K
    Sep 14, 2019 at 17:55
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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.

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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 $l<q$.

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