For homework I was supposed to prove the statement:
"Let $M \subset \mathbb{R}$. If $M$ has a supremum then there exists a sequence $\{ x_n \}$, $n \geq 1$, with $x_n \in M$ such that $\lim_{n \xrightarrow{} +\infty}{x_n} = \sup{M}$."
Here is my incomplete proof:
"Let $a = \sup{M}$. By definition of supremum there exists an element $x_n \in M$ such that $a - \frac{1}{n} < x_n \leq a$, for all $n \geq 1$. the limit of $a - \frac{1}{n}$ as $n$ tends to infinity is $a$, and the limit of $a$ as $n$ tends to infinity is also $a$. Thus we have that $a < \lim_{n \xrightarrow{} +\infty}{x_n} \leq a$."
Now I don't know what to do. My first thought was to apply the squeeze theorem, but I have that $a < \lim_{n \xrightarrow{} +\infty}{x_n} \leq a$ and not the conventional $a \leq \lim_{n \xrightarrow{} +\infty}{x_n} \leq a$. Can I still apply it or do I have to do something else?
Limits do not preserve strict inequalities! Always remember the example $x_n = 1/n$, where all terms are strictly positive, but still $x_n\to 0$ which is not strictly positive. The squeeze theorem will (in your situation) imply that $a\leq \lim_{n\to\infty} x_n \leq a$, as you want.