Convergence Theorems in the extended real numbers

Question

I'm studying measure theory and usually functions take values in the extended real line. For most of the theorems, pointwise limits of such functions are considered.

Now I was wondering whether some limit theorems that hold for $\mathbb{R}$ also hold in the extended real numbers since the proofs of these I know most of the time rely on the absolute value metric in $\mathbb{R}$. Are there alternative proofs that can be generalized more easily?

In particular, I would like to know if the following theorems hold and what the best way to prove them is, ideally without studying a lot of topology since I am familiar with metric spaces but not with topology.

For sequences in $\mathbb{R}$:

(1) Let $\lim \limits_{n \to \infty} a_{n}=a$ and $\lim \limits_{n \to \infty} b_{n}=b$. Then $a_{n} \leq b_{n}$ $\forall n \implies a \leq b$

(2) Let $\lim \limits_{n \to \infty} a_{n}=x$ and $\lim \limits_{n \to \infty} b_{n}=x$. Then $a_{n} \leq x_{n} \leq b_{n}$ $\forall n \implies \lim \limits_{n \to \infty} x_{n}=x$

(3) Let $x_{n}$ be monotonically increasing and bounded above, then $\lim \limits_{n \to \infty}= \sup x_{n}$.

Any hints or references are much appreciated.

Thanks very much!

Edit: I've recently found a note on the extended real number system. It states that there exists a bijective, order preserving function $f:[-\infty,+\infty] \to [-1,+1]$ defined by $\varphi(c)=\frac{c}{1+\lvert c \rvert}$ for $c \in \mathbb{R}$, $\varphi(-\infty)=-1$ and $\varphi(\infty)=1$.

I'm pretty sure this is even a homeomorphism, so it preserves convergence. Am I correct that this proves that all the theorems above for $\mathbb{R}$ carry over to the extended real numbers?

Answer 1

Yes, they all hold on the extended real line $\overline{\mathbb R}$ and you do not need topology to prove them. Actually, you prove them basically as in $\mathbb R$. Consider the third statement, for instance. If $\sup_nx_n\in\mathbb R$, you prove that $\lim_nx_n=\sup_nx_n$ as in $\mathbb R$. If each $x_n$ is equal to $-\infty$, then it is clear that both $\lim_nx_n$ and $\sup_nx_n$ are equal to $-\infty$. Otherwise, both $\lim_nx_n$ and $\sup_nx_n$ are equal to $\infty$

Answer 2

Give $\overline{R}$ the order topology, as I describe here.

Then in any ordered topological space $(X,\le)$, the set $U=\{(x,y) \in X^2: x \le y \}$ is closed in $X^2$ (in the product topology on $X^2$ of course), the proof for $X=\overline{\Bbb R}$ is not hard: if $(x,y) \notin U$, this means that $x > y$. Find $z \in \overline{\Bbb R}$ such that $x > z > y$ and note that $O_x=\{u \in \Bbb R: u > z\}$ is an open neighbourhood of $x$ in $\overline{\Bbb R}$, $O_y=\{u \in \Bbb R: u < z\}$ is an open neighbourhood of $y$ in $\overline{\Bbb R}$ and $(O_x \times O_y) \cap U = \emptyset$, showing that $U$ is closed in the product topology.

A closed set is sequentially closed, so $\forall n: a_n \le b_n$ means $(a_n,b_n) \in U$ for all $n$ and so if $a_n \to a$ and $b_n \to b$, $(a_n,b_n) \to (a,b)$ in $X^2$ and so $(a,b) \in U$ as well, so $a \le b$.

It follows completely from standard order topology facts.