# Prove that a convergent real sequence always has a smallest or a largest term

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.

• 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? Sep 14, 2019 at 3:43
• 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. Sep 14, 2019 at 3:49
• @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. Sep 14, 2019 at 3:53
• 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. Sep 14, 2019 at 4:26
• 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 Sep 14, 2019 at 17:13

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.

• 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." Sep 14, 2019 at 22:01

$$A:=$${$$a_n| n \in \mathbb{Z^+}$$}

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.

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

we have $$I , i.e .

$$I , a contradiction.

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

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.

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.

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.