June 24, 2017 22:28:28
Math
The sooner all the animals are extinct, the sooner we'll find their money. -- Ed Bluestone
 .:  Math

Note: If you cannot view some of the math on this page, you may need to add MathML support to your browser. If you have Mozilla/Firefox, go here and install the fonts. If you have Internet Explorer, go here and install the MathPlayer plugin.

Math > Number theory > Real Numbers

Real Numbers

Properties of the real numbers.

Sibling topics:

Contents:

Definition of upper and lower bounds [tags: setBound]
For `x in RR` and `S sube RR`, "`x` is an upper bound for `S`" means:
`(AA y in S)(x >= y)`
Similarly, "`x` is a lower bound for `S`" means:
`(AA y in S)(x <= y)`
An upper of a set does not need to be a member of that set, but if the set does contain its own upper bound, it is called an extremum of the set.

If a set has an upper bound, it is said to be bounded above, and if it has a lower bound, it is said to be bounded below. A set with both an upper and lower bound is bounded. If it does not have both, it is unbounded.
Definition of extrema [tags: extrema maximum minimum setBound]
Given `S sube RR`, `x` is said to be the maximum (or largest element) of `S`, and we write `x="max "S`, if `x in S` and `x` is an upper bound for `S`.

Similarly, `y` is said to be the minimum (or smallest element) of `S`, and we write `x="min "S`, if `y in S` and `y` is a lower bound for `S`.

An extremum (plural "extrema") is either a minimum or a maximum. Not all sets have a minimum or a maximum. Note the difference between an extremum and an upper or lower bound. An extremum of `S` is in `S`, but an upper or lower bound need not be.
Definition of lub and glb [tags: setBound lub glb]
For `x in RR` and `S sube RR`, `x` is the least upper bound (lub) (or supremum) of `S` (denoted "lub of `S`" or "`"sup " S`") if:
`(AA y in S)(x >= y) and (AA z in RR)(z" is an upper bound for "S => z >= x)`
Similarly, "`x` is the greatest lower bound (glb) (or infimum) of `S`" (denoted "glb of `S`" or "`"inf " S`") means:
`(AA y in S)(x <= y) and (AA z in RR)(z" is a lower bound for "S => z <= x)`
If a set contains its own least upper bound, then the set has a minimum equal to the lub. Similarly, if a set contains its own greatest lower bound, the set has a extrema equal to the glb.
Definition of the completeness property [tags: completeness]
For `S sube RR`, `S` has the completeness property if:
  1. Every nonempty subset `T` of `S` which is bounded above by an element of `RR` has a lub in `S`, and
  2. Every nonempty subset `T` of `S` which is bounded below by an element of `RR` has a glb in `S`.
Note that the lub and glb must be in `S`, though they need not be in `T`.
Theorem: Some non-empty, bounded sets do not have extrema [tags: setBound]
Some non-empty subsets of the real numbers do not have extrema, even if they are bounded. That is, even though the set may be bounded above or below, they lack a maximum or minimum element (or both). This of course can only be the case for infinite sets. Consider the set `S=[0,1]`. Clearly, the infimum of this set is 0, and the supremum is 1. Both 0 and 1 are also in the set, so the set has both extrema. Now consider the set `T=[0,1)`. 1 is clearly an upper bound for this set, but is it the least upper bound? In fact it is, and `T` has no maximum element.
Proof: By definition, the set `T` contains every number `n` where `0 <= n < 1`. Assume by way of contradiction that there does exist an upper bound for `T` (call it `u`) that is less than 1. Since `u` would have to be very close to 1, we won't consider `u <= 0`. Then we have `0 < u < 1`. But given any such `u`, we can always find another number `v` such that `u < v < 1`, which would contradict `u` being an upper bound. For instance, let `v=(u+1)/2`. If we can prove that `u < (u+1)/2 < 1`, then we will have shown that any number less than 1 cannot be an upper bound for `T`. Similarly, we will have shown that `T` has no maximum element, since a greater element that is also within `T` can always be found. We will now prove that `u < (u+1)/2 < 1`. Because `u < 1`, `u+u=2u < u+1 < 2`. Then, dividing by 2, we can see that `u < (u+1)/2 < 1`.
Copyright 2003-2016 Adam Milazzo. Verbatim copying and redistribution of this entire page are permitted without royalty in any medium provided this notice is preserved.