Extended Real Number System: $\bar{\mathbb R}=\mathbb R\cup\{-\infty, \infty\}$, and to extend the usual ordering on $\mathbb R$ by declaring that $-\infty for all $x\in\mathbb R$. The completeness of $\mathbb R$ can be stated as: Every subset of $A$ of $\bar{\mathbb R}$ has a least upper bound, or supremum, and a greatest lower bound, or infimum, which are denoted by $sup A$ and $inf A$. If $A=\{a_1, a_2, ..., a_n\}$, we also write:

$\max (a_1, ..., a_n)=\sup A$, $\min(a_1, ..., a_2)=\inf A$.

From completeness it follows that every sequence $\{x_n\}$ in $\bar{\mathbb R}$ has a limit superior and a limit inferior:

$\lim\sup x_n=\inf_{k\ge 1}(\sup_{n\ge k}x_n)$, $\lim\inf x_n=\sup_{k\ge 1}(\inf_{n\ge k} x_n)$

The sequence $\{x_n\}$ converges (in $\mathbb R$) iff these two numbers are equal (and finite), in which case its limit is their common value. One can also define $\lim\sup$ and $\lim\inf$ for functions $f: \mathbb R\to\bar{\mathbb R}$, e.g. $\lim\sup_{x\to a} f(x)=\inf_{\delta>0}(\sup_{0<|x-a|<\delta}f(x))$.

The arithmetical operations on $\mathbb R$ can be partially extended to $\bar{\mathbb R}$:

$x\pm\infty=\pm\infty (x\in\mathbb R)$, $\infty+\infty=\infty$, $-\infty-\infty=-\infty$,

$x\dot(\pm\infty)=\pm\infty(x>0)$, $x\dot(\pm\infty)=\mp\infty(x<0)$, $0\dot(\pm\infty)=0$.

We employ the following notation for intervals in $\bar{\mathbb R}$: If $-\infty\le a,

$(a, b)=\{x: a, $[a, b]=\{x: a\le x\le b\}$, $(a, b]=\{x: a, $[a, b)=\{x: a\le x.

If $X$ is an arbitrary set and $f: X\to[0, \infty]$, we define $\sum_{x\in X}f(x)$ to be the supremum of its finite partial sums:

$\sum_{x\in X}f(x)=\sup\{\sum_{x\in F}f(x): F\subset X, F$   is finite $latex \}$.

Proposition. Given $f: X\to[0, \infty]$, let $A=\{x: f(x)>0\}$. If $A$ is uncountable, then $\sum_{x\in X}f(x)=\infty$. If $A$ is countably infinite, then $\sum_{x\in X}f(x)=\sum_1^\infty f(g(n))$ where $g:\mathbb N\to A$ is any bijection and the sum on the right is an ordinary infinite series.

Proof. We have $A=\cup_1^\infty A_n$ where $A_n=\{x: f(x)>\frac 1 n\}$. If $A$ is uncountable, then some $A_n$ must be uncountable, and $\sum_{x\in F}f(x)>card(F)/n$ for $F$ a finite subset of $A_n$. If follows that $\sum_{x\in X}f(x)=\infty$. If $A$ is countably infinite, $g: \mathbb N\to A$ is a bijection, and $B_N=g(\{1, ..., N\})$, then every finite subset $F$ of $A$ is contained in some $B_N$. Therefore,

$\sum_{x\in F}f(x)\le \sum_1^Nf(g(n))\le\sum_{x\in X}f(x)$.

Taking the supremum over $N$, then

$\sum_{x\in F}f(x)\le\sum_1^\infty f(g(n))\le\sum_{x\in X}f(x)$,

and then taking the supremum over $F$, we obtain the desired result. $\Box$

Some terminology concerning (extended) real-valued functions: A relation between numbers that is applied to functions is understood to hold pointwise. Thus $f\le g$ means that $f(x)\le g(x)$ for every $x$, and $\max(f, g)$ is the function whose value at $x$ is $\max(f(x), g(x))$. If $X\subset\bar{\mathbb R}$ and $f: X\to\bar{\mathbb R}$, $f$ is called increasing if $f(x)\le f(y)$ whenever $x\le y$ and strictly increasing if $f(x) whenever $x; similarly for decreasing. A function is either increasing or decreasing is called monotone.

If $f:\mathbb R\to\mathbb R$ is an increasing function, then $f$ has right- and left-hand limits at each point:

$f(a+)=\lim_{x\searrow a}f(x)=\inf_{x>a}f(x)$, $f(a-)=\lim_{x\nearrow a}f(x)=\inf_{x.

$f$ is called right continuous if $f(a)=f(a+)$ for all $a\in\mathbb R$ and left continuous if $f(a)=f(a-)$ for all $a\in\mathbb R$.

For points $x\in\mathbb R$ or $\mathbb C$, $|x|$ denotes the ordinary absolute value or modulus of $x$, $|a+ib|=\sqrt{a^2+b^2}$. For points $x\in\mathbb R^n$ or $\mathbb C^n$, $|x|$ denotes the Euclidean norm:

$|x|=[\sum_1^n|x_j|^2]^{\frac 1 2}$.

We recall that a set $U\subset\mathbb R$ is open if for every $x\in U$, $U$ includes an interval centered at $x$.

Proposition. Every open set in $\mathbb R$ is countable disjoint union of open intervals.

Proof. If $U$ is open, for each $x\in U$ consider the collection $\mathcal J_x$ of all open intervals $I$ such that $x\in I\subset U$. It is easy to check that the union of any family of open intervals containing a point in common is again an open interval, and hence $\mathcal J_x=\cup_{I\in\mathcal J_x}$ is an open interval. It is the largest element of $\mathcal J_x$. If $x, y\in U$ then either $J_x=J_y$ or $J_x\cap J_y=\emptyset$, for otherwise $J_x\cup J_y$ would be a larger open interval that $J_x$ in $\mathcal J_x$. Thus if $K=\{J_x: x\in U\}$, the (distinct) members of $K$ are disjoint, and $U=\cup_{J\in K}\mathcal J$. For each $J\in K$, pick a rational number $f(J)\in J$. The map $f: K\to\mathbb Q$ thus defined is injective, for if $J\not=J'$ then $J\cap J'=\emptyset$; therefore $K$ is countable. $\Box$

“Nobody needs a smile so much as the one that has none to give. So get used to smiling heart-warming smiles, and you will spread sunshine in a sometimes dreary world.” ~ Lawrence G. Lovasik

References:

[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 10-12.

[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.