- #1

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## Homework Statement

Prove or produce a counterexample: If ##\{a_n\}_{n=1}^\infty\subset \mathbb{R}## is convergent, then

##

\min (\{a_n:n\in\mathbb{N}\} )## and

##\max (\{a_n:n\in\mathbb{N}\} )

##

both exist.

## Homework Equations

## The Attempt at a Solution

I will produce a counterexample.

Let ##a_n = \frac{1}{n}##. First we will show that ##a_n## is converges to 0.

Let ##\epsilon > 0##. By the Archimedian principle there exists an ##N \in \mathbb{N}## such that ##\frac{1}{N} < \epsilon##. If ##n \ge N##, then ##|a_n - 0| = |\frac{1}{n} - 0| = \frac{1}{n} \le \frac{1}{N} < \epsilon##. So ##\lim a_n = 0##.

Let ##S = \{\frac{1}{n}:n\in\mathbb{N}\}##. Then ##\max(S) = \sup(S) = 1##. However, the set does not have a minimum, and this is where I'm stuck. Do I need to show that ##\inf(S) = 0## and that ##0 \not \in S##? Clearly there is no minimum, but I am not sure how to show this rigorously.