If max｢a⸤n⸥｣ does not exist, then for all a⸤n⸥ there exits a natural number N such that a⸤n⸥ < a⸤N⸥.
This means that there exists an increasing subsequence.
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A bounded monotone sequence is convergent.
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Since ｢a⸤n⸥｣ is bounded, the increasing subsequence is convergent.
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From this, we can say
"If a bounded sequence does not have a maximum value, it contains an increasing convergent subsequence."
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If max｢a⸤n⸥｣ exists, then set
s⸤1⸥ = ｢a⸤n⸥:n≥N where a⸤N⸥=max(｢a⸤n⸥｣)｣
s⸤i+1⸥
= ｢a⸤n⸥:n≥N where a⸤N⸥=max｢s⸤i⸥⸤j⸥:j>1｣｣.
If max｢s⸤i⸥⸤j⸥:j>1｣ does not exist, then the subsequence ｢s⸤i⸥⸤j⸥:j>1｣ contains a convergent subsequence.
┆Refer to the above.┆
If ｢s⸤i⸥｣ is an infinite sequence,
then ｢s⸤i⸥⸤1⸥｣ is a decreasing sequence.
Since ｢a⸤n⸥｣ is bounded and ｢s⸤i⸥⸤1⸥｣ is a subsequence of ｢a⸤n⸥｣, the sequence ｢s⸤i⸥⸤1⸥｣ is bounded.
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A bounded monotone sequence is convergent.
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Thus ｢s⸤i⸥⸤1⸥｣ is convergent.
┆The proof for min｢a⸤n⸥｣ is similar.┆
