Suppose that $B$ is a nonempty set of $ \mathbb{R} $ that is bounded above. Let b = sup($B$). Prove there is a sequence $b_n \in B$ that converges to b.
So I know we can use the Squeeze Theorem, and all I could think of using was the convergence theorem where we say for all $\epsilon > 0$, there exists an $N \in \mathbb{N}$ where if $N > n$, then $|b_n - b| < \epsilon$, but I didn't know how to continue from there, or whether that was how to start the proof.
Any help appreciated!
On
Use definition of supremum. If we take $\epsilon=\frac{1}{n}>0$ (you can choose any sequence that is positive and converges to zero) then there exists $b_{n}\in B$ s.t. $b_{n}>b-\frac{1}{n}$. If not, then $b_{m}\leq b-\frac{1}{n}$ for all $m$ and this implies $b-\frac{1}{n}<b$ bounds $B$, which contradicts to the fact that $b$ is supremum of $B$ (remember that supremum means least upper bound). So we have a sequence $b_{n}\in B$ with $b-\frac{1}{n}<b_{n}\leq b$ for each $n\geq $, and here we can use squeeze theorem to get $\lim_{n\to \infty}b_{n}=b$ (or just because of definition of convergence).
Without losing generality, let us assume that $B = [b-1,b]$. Consider the sequence defined as $s_n = b - \frac{1}{n}$. I claim that $s_n \in B$ for all $n\in \mathbb{N}$, and $s_n$ converges to $b$.
To show that $s_n\in B$, notice that since $n\in \mathbb{N}$, $\frac{1}{n} > 0$ for all $n$. As a result, $b - \frac{1}{n} < b$. This implies that $s_n \in B$ for all $n$.
To show that $s_n \to b$, for any $\epsilon>0$, we need to find $N\in\mathbb{N}$ such that whenever $n\ge N$, we have $|b-s_n| < \epsilon$. Let us take out the cheat sheet:
Going back, we choose $N$ such that when $n\ge N$, we have $\frac{1}{n} \le \epsilon$. The result follows.