Extended real value

From Maths
(Redirected from Extended real valued)
Jump to: navigation, search

If a function is described as "extended real valued" it means: [math]f:X\rightarrow \mathbb{R}\cup\{-\infty,+\infty\}[/math]

Definition

The set [math]\mathbb{R}\cup\{-\infty,+\infty\}[/math] refers to "extended real values".

The following algebraic relations are defined on [ilmath]-\infty[/ilmath], [ilmath]+\infty[/ilmath] and [ilmath]x\in\mathbb{R} [/ilmath][1]

Pitfalls

Note that the property [math]-\infty+\ +\infty[/math] cannot be sensibly defined, only the properties listed below are allowed.

Proof that [math]-\infty[/math] is not the additive inverse of [math]+\infty[/math]


Notice that [math]\forall x\in\mathbb{R}[x+(\pm\infty)=\pm\infty][/math] - this would suggest [ilmath]x[/ilmath] is the additive inverse but yet it works for all [ilmath]x[/ilmath] and is clearly not unique.

This would also define [math]x=\infty-\infty[/math] which would be anything (including 0) and this makes no sense. Thus showing we cannot have this property.


One must be careful when doing proofs with extended real valued entities in play to make sure only to use the properties below.

Properties

Note that [math]\pm x+\pm y[/math] refers to [math]+x + +y[/math] or [math]-x + -y[/math] NOT [math]-x + y[/math] or something, the order of [math]\pm[/math] compared to [math]\mp[/math] matters!

Relation Note
[math](\pm\infty)+(\pm\infty)=\pm\infty[/math] Note: This says nothing about [math](-\infty)+(+\infty)[/math]
[math]x+(\pm\infty)=(\pm\infty)+x=\pm\infty[/math] Commutative - as you'd expect for addition
[math]x(\pm\infty)=(\pm\infty)x=\left\{\begin{array}{lr} \pm\infty & \text{if }x>0 \\ 0 & \text{if }x=0\\ \mp\infty & \text{if }x<0 \end{array}\right.[/math] Commutative - as you'd expect, also defined as you'd expect
[math](\pm\infty)(\pm\infty)=+\infty[/math] positive * positive = positive, negative*negative=positive, as expected
[math](\pm\infty)(\mp\infty)=-\infty[/math] positive*negative = negative, negative*positive = negative
[math]\frac{x}{\pm\infty}=0[/math] a number divided by an absolutely huge number is an absolutely tiny number, regardless of sign
[math]-\infty< x< +\infty[/math] As you'd expect.

References

  1. Halmos - Measure Theory - Page 1 - Spring - Graduate Texts in Mathematics (18)