Real Analysis

Set Theory

If you're not in a union, you're in none; if not in an intersection, you're not in all -- you might not be in one!

Axioms of the Reals

• A1: + and · are closed binary operations on the reals.
• A2: + and · are associative.
• A3: + and · are commutative.
• A4: Distributivity holds: $\left.a\cdot (b+c)=(a\cdot b)+(a\cdot c)\right.$.
• A5: ∃ identities: $\left.0+a=a\right.$ and $\left.1\cdot a=a\right.$.
• A6: ∃ additive inverses.
• A7: ∃ multiplicative inverses (for $\left.a\neq 0\right.$).
• A8: ∃ non-empty subset P ∈ IR such that the following hold:
  1. a, b ∈ P → a + b ∈ P
  2. a, b ∈ P → a · b ∈ P
  3. a ∈ IR → (a ∈ P) ∨ (−a ∈ P) ∨ (a = 0)

Completeness

• A9: the reals are complete.

Exercise 9, p. 34 hint:

$\left.\right.(z+\delta )^{n}=\sum _{k=0}^{n}{n \choose k}z^{n-k}\delta ^{k}=z^{n}+\sum _{k=1}^{n}{n \choose k}z^{n-k}\delta ^{k}=z^{n}+\delta \sum _{k=1}^{n}{n \choose k}z^{n-k}\delta ^{k-1}$

Now everything in the sum multiplied by $\left.\delta \right.$ in the final term can be bounded above. For example, we can always demand that $\left.\delta <1\right.$, and we have an upper bound (call it $\left.\alpha \right.$) on $\left.z\right.$. So we can assert that

$\left.\right.(z+\delta )^{n}\leq z^{n}+\delta \sum _{k=1}^{n}{n \choose k}\alpha ^{n-k}\equiv z^{n}+\delta \beta$ where $\left.\beta \right.$ is just some number, and $\left.\delta \right.$ must be chosen to be less than 1.

Now choose $\left.\delta \right.$ appropriately.

Some comments on proofs

Watch for some of these problems:

• Assuming the theorem that you're in the process of proving.
• Assuming that the "arbitrary" sets you're dealing with are denumerable, or even finite.
• Forgetting to prove an "iff" proof in both directions.