Weierstrass... thing There is in my maths text-book this property/theorem given under the name of Weierstrass property/theorem:
Let $ (a_n) $ be a sequence of real numbers.
a)If $ (a_n) $ is monotonically increasing and has an upper bound, then $ (a_n) $ is convergent.
b)If $ (a_n) $ is monotonically decreasing and has a lower bound, then $ (a_n) $ is convergent.
When searching on Google, I can't find anything related to this, so my first question is:
What is the name of this thing?
What I want to do is to attempt to prove this property/theorem, I would like some hints on what to start with and what should I end with.
Thanks!
 A: HINT: Use the fact that $\Bbb R$ has the least upper bound property: if $\varnothing\ne A\subseteq\Bbb R$, and $A$ is bounded above, then $A$ has a least upper bound. If your increasing sequence is $\langle x_n:n\in\Bbb N\rangle$, let $x$ be the least upper bound of $\{x_n:n\in\Bbb N\}$, and show using the definitions of least upper bound and limit of a sequence that $x=\lim\limits_{n\to\infty}x_n$.
The proof for the other result is similar. Alternatively, if $\langle x_n:n\in\Bbb N\rangle$ is decreasing and bounded below, then $\langle-x_n:n\in\Bbb N\rangle$ is increasing and bounded above, so by the first result it converges to some $x$. Now show that $\langle x_n:n\in\Bbb N\rangle$ converges to $-x$.
Forgot to say: in my experience it’s usually called the monotone convergence theorem.
A: See monotone convergence theorem
A: This is often referred to as the Monotone Convergence Theorem.
The proof of each part will be similar. Let me give you an idea of how to prove the first part.
Suppose that $(a_n)_{n=1}^\infty$ is an increasing sequence of real numbers with an upper bound. In particular, that means that the set $\{a_n:n\ge 1\}$ is a non-empty set of real numbers that is bounded above, and so has a least upper bound, say $a$. Use the definitions of least upper bound and sequence limit, together with the fact that $(a_n)_{n=1}^\infty$ is increasing, to show that $a_n\to a$.
