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<x<\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<b\le \infty,

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

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)<f(y) whenever x<y; 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<a}f(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


[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.