MATH1026 LEEDS

Injective: F(x1) = F(x2) => x1 = x2
Surjective: Range is the codomain, find inverse to prove.
Composition: If f: A -> B and g: B -> C then g(f(x)): A -> C.
If f and g both injective/surjective so is the composition.

Let A be a subset of R.
An upper bound on A is any K in R such that, for all x in A,
x ≤ K.
A lower bound on A is any L in R such that, for all x in A,
x ≥ L.
If A has an upper bound, we say that A is bounded above.
If A has a lower bound, we say that A is bounded below.
A is bounded if it is bounded both above and below.

Given any real number K, there is some positive integer n such that n > K (this also applies to Q).

The supremum of a subset A of R is its least upper bound (if this exists), denoted supA. So supA is a real number with two properties:
(i) supA is an upper bound on A.
(ii) No number less than supA is an upper bound on A.
The infimum of A is its greatest lower bound (if this exists), denoted inf A.

Every nonempty subset of R which is bounded above has a supremum in R. The same is true of bouned below and an infimum.

Q is dense in R: between any pair of distinct real numbers, there is a rational number.
R\Q is dense in R: between any pair of distinct real numbers, there is an irrational number.
R\Q is uncountable.

Let In = [an, bn] be a nested sequence of closed intervals. Then there exists x in R such that x in In for all n in Z+.

For all x, y in R, |x + y| ≤ |x| + |y|

A real sequence (an) converges to a real number L if, for each " > 0 there exists some positive integer N such that for all n ≥ N, |an − L| < ". In this case, we will write an → L.

If a sequence converges, its limit is unique.

We say that a real sequence (an) is bounded above if its range, the set A = {an : n in Z+} is bounded above (as in Definition 6), that is, if there exists K in R such that an ≤ K for all n in Z+.
(an) is bounded below if A is bounded below.
(an) is bounded if it is bounded both above and below.
If a sequence converges then it is bounded.
If K ≤ an ≤ M for all n in Z+ and (an) converges to L, then
K ≤ L ≤ M.

If an → A and bn → B then
(i) an + bn → A + B,
(ii) an.bn → AB.
If, in addition, an doesn't equal 0 for all n and A doesn't equal 0, then
(iii) 1/an →1/A

Let (an), (bn) and (cn) be sequences such that (an) and (cn) converge to the same limit L. If there exists k in Z+ such that an ≤ bn ≤ cn for all n ≥ k, then (bn) converges to L also.

A sequence (an) is increasing if an+1 ≥ an for all n in Z+. It is decreasing if an+1 ≤ an for all n ∈ Z+. It is monotone if it is increasing or decreasing.

Every bounded monotone sequence converges.