# How to prove that ($x_n$) converges and lim $x_n$ is also an upper bound of A

Let $A \subset \Bbb R$ be nonempty and bounded above, and let $(x_n)$ be a decreasing sequence of upper bounds of $A$. How do we prove that $(x_n)$ converges and lim $(x_n)$ is also an upper bound of A?

## 1 Answer

Let ${x_n}$ be a decreasing sequence of upper bounds of $A$. That means, that $\{ x_n\}$ is lower bounded, by $\sup A$. Hence $x_1 \geq \{ x_n\}\geq \sup A$ for all $n$.

Using the Bolzano Weierstrass theorem, $\{ x_n\}$ has a convergent subsequence, say $\{ x_{n_k}\}$.

However, let $x_{n_k} \to L$. I claim two things:

$\lim x_n = L$.

Proof: Let $\epsilon > 0$. Note that there exists $K$ such that $k \geq K \implies |x_{n_k} - L| < \epsilon$. Now, note that $x_n < x_{n_K}$ for all $n > n_K$, so that $|x_{n} - L| < \epsilon$ whenever $n > n_K$. Hence $x_n \to L$.

$L \geq \sup A$.

Note that $x_n > \sup A$ for all $n$. Hence, the subsequence $\{ x_n - \sup A\}$ is a non-negative sequence, hence must have non-negative limit. That is , $\lim x_n = L \geq \sup A$. As $L \geq \sup A$, $L$ is greater than every element of $A$, so is an upper bound of $A$.