The sup{(n-1)/n : n is a natural number} exists?

Question

True or False?

a)The $\sup \{\frac{(n-1)}{n} : n\in\mathbb{B} \}$ exists?

b) If ${x_n}$ is a sequence with $x_n < 3$ for all $n$, then it cannot converge to $3$?

For a) True: By definition, the number $s$ is called the supremum of a set $E$ if 1) $S$ is an upper bound for $E$. And 2) If $t$ is an upper bound for $E$, then $s \leq t$.

Then take $\sup \{1 - \frac{1}{n} : n \in \mathbb{N} \}$. Then we can see that $1$ is an upper bound for the set. And if $t$ is an upper bound for the set, then $1 \leq$ for all $t$.

For b) True : Take $x_n = (-1)^(n)$ , then this is bounded above by $1$ and bounded below by $-1$, then $x_n < 3$. Then it cannot converge to $3$.

Can someone please verify this?

Thank you.

Answer 1

a) is true. 1 is supremum.

b) is not true. For example, let $x_n=3-\frac1n<3$, however, it converges to 3.

Answer 2

a) Observe that $(n-1)/n \leq 1$ for every $n \in \mathbb{N}$. Hence the completeness of $\mathbb{R}$ implies that its supremum exists.

Your argument for (a) is too strong (showing that an upper bound exists would suffice). You also need to justify that if $t$ is an upper bound, then $t \geq 1$.

b) False. Take $x_n = 3 - n^{-1}$.

The problem with your argument is that a theorem cannot be proved by example, but can be proved false with one counterexample.