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)


  • A9: the reals are complete.

Exercise 9, p. 34 hint:

=\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

\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.
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