# Real Analysis

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## 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 $\left.a\ne 0\right.$).
• 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 $\left.\right.(z+\delta)^n \le z^n+\delta\sum_{k=1}^n{n \choose k}\alpha^{n-k} \equiv z^n+\delta\beta$ 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.