Proofs Exam 3 Definitions and Theorems

Injective (1-1)

∀x1,x2∈A, f(x1)=f(x2) ⇒ x1=x2

Surjective (onto)

function f: A→B
∀y∈B, ∃x∈A such that f(x)=y

image

Let f:A→B and let S⊆A
f(S) = {f(x) ∣ x∈S}

preimage

Let f:A→B and let T⊆B.
f−1(T)={x∈A ∣ f(x)∈T}
*The preimage does not require an inverse function*

convergence

A sequence (aₙ) converges to L ∈ R if ∀ε > 0, ∃N ∈ N s.t. n ≥ N,
|aₙ - L| < ε

monotone

(aₙ) is monotone if it is either increasing or decreasing.
Increasing is defined as: (aₙ) ≤ (aₙ+1) ∀n ∈ N
Decreasing is defined as: (aₙ) ≥ (aₙ+1) ∀n ∈ N

bounded

∃M ∈ R, where M > 0, such that |(aₙ)​| ≤ M, ∀n∈N

MCT (monotone convergence theory)

If a sequence is monotone & bounded then it converges

ALT (Algebraic Limit Theorem)

Suppose lim⁡(aₙ) = a and lim⁡(bₙ) = b.
Then:

(i) ∀ c∈R lim⁡(caₙ) = ca
(ii) lim⁡(aₙ+ bₙ) = a+b
(iii) lim⁡(aₙ⋅bₙ) = a⋅b
(iv) lim⁡(aₙ/bₙ) = a/b provided b≠0

Injective examples

f(x) = 2x satisfies injectivity as if f(x₁) = f(x₂) then 2x₁ = 2x₂ so x₁ = x₂
f(x) = x² fails injectivity as f(2) = f(−2) = 4 but 2 ≠ −2

Surjective examples

f(x) = 2x satisfies surjectivity as given any y ∈ ℝ, take x = y/2, then f(y/2) = y
f(x) = x² fails surjectivity as no x satisfies x² = −1 so −1 ∈ ℝ has no preimage

Image examples

f(x) = x² with S = [−2, 3] gives f(S) = [0, 9] as x² achieves minimum 0 at x = 0 and maximum 9 at x = 3

Preimage examples

f(x) = x² with T = [1, 4] gives f⁻¹(T) = [−2, −1] ∪ [1, 2] as these are all x with 1 ≤ x² ≤ 4,
but T = [-2,-1] gives an f⁻¹(T) that fails as no number from -2 to -1 is mapped to by this function as it can’t go negative.

Convergence examples

aₙ = 1/n satisfies convergence as given ε > 0, choose N > 1/ε, then n ≥ N ⟹ |1/n − 0| < ε so aₙ → 0
aₙ = (−1)ⁿ fails convergence as it oscillates between −1 and 1 and never settles near any single L

Monotone examples

aₙ = 1/n satisfies monotonicity as 1/n ≥ 1/(n+1) for all n so the sequence is decreasing
aₙ = (−1)ⁿ/n fails monotonicity as it alternates sign so is neither increasing nor decreasing

Bounded examples

aₙ = sin(n) satisfies boundedness as |sin(n)| ≤ 1 for all n so take M = 1
aₙ = n fails boundedness as for any M > 0 we can choose n > M so |aₙ| = n > M