Real Number System and Elementary Logic
This domain tests whether you can build proofs on the ordered-field and completeness axioms of R rather than on intuition, and whether you can manipulate quantifiers correctly. Expect items on supremum and infimum, the Archimedean property, density of the rationals, and the logical form of the ε-statements used throughout the course.
Axioms and their consequences
R is a complete ordered field: it satisfies the field axioms, the order axioms, and the least-upper-bound property (every nonempty subset bounded above has a supremum). Completeness is exactly what separates R from Q; Q is ordered but not complete, since {q in Q : q^2 < 2} has no rational supremum. Standard consequences include the Archimedean property (for every x there is n in N with n > x), density of Q and of the irrationals, and existence of nth roots of positive numbers.
Working with sup and inf
To prove sup S = M: first show s ≤ M for all s in S; then, for every ε > 0, exhibit s in S with s > M − ε. For infimum, reverse both inequalities. Never assume sup S belongs to S; a maximum exists only when the supremum is attained.
Quantifiers
The negation of 'for all ε > 0 there exists N such that for all n ≥ N, P(n, ε)' is 'there exists ε > 0 such that for all N there exists n ≥ N with not P(n, ε)'. Negation flips each quantifier and negates the inner statement; the order of quantifiers carries meaning.
Worked example
Claim: sup S = 1 for S = {1 − 1/n : n ≥ 1}. Upper bound: 1/n > 0, so 1 − 1/n < 1 for every n. Sharpness: let ε > 0. By the Archimedean property choose n in N with n > 1/ε; then 1/n < ε, so 1 − 1/n > 1 − ε. Thus no number below 1 bounds S, and sup S = 1. Since 1 is not in S, S has no maximum.
Traps
Assuming a supremum is attained; proving only the upper-bound half; invoking density or completeness without stating it; choosing N before ε in an ε-N proof; treating a single example as a proof.