# 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: ${\displaystyle \left.a\cdot (b+c)=(a\cdot b)+(a\cdot c)\right.}$.
• A5: ∃ identities: ${\displaystyle \left.0+a=a\right.}$ and ${\displaystyle \left.1\cdot a=a\right.}$.
• A6: ∃ additive inverses.
• A7: ∃ multiplicative inverses (for ${\displaystyle \left.a\neq 0\right.}$).
• 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:

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

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

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

Now choose ${\displaystyle \left.\delta \right.}$ 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.