Completeness, epsilon-delta proofs, sequence convergence, Cauchy sequences, compactness and uniform continuity.
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- What is the least upper bound property?
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Every non-empty subset of the reals that is bounded above has a least upper bound (supremum).- Define a complete ordered field.
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An ordered field in which every non-empty subset that is bounded above has a least upper bound.- What does the nested intervals theorem state?
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If {[a_n, b_n]} is a nested sequence of closed intervals with b_n - a_n to 0, then the intersection contains exactly one point.- Which property of the real numbers distinguishes them from the rationals?
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Completeness: the least upper bound property. Suprema of bounded sets always exist in reals, not always in rationals.- State the epsilon-delta definition of the limit of a function f at point c.
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lim(x->c) f(x) = L if for all epsilon > 0, there exists delta > 0 such that 0 < |x - c| < delta implies |f(x) - L| < epsilon.- State the epsilon-delta definition of continuity of f at point c.
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f is continuous at c if for all epsilon > 0, there exists delta > 0 such that |x - c| < delta implies |f(x) - f(c)| < epsilon.- What is the negation of the limit definition (when lim(x->c) f(x) does NOT equal L)?
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There exists epsilon > 0 such that for all delta > 0, there exists x with 0 < |x - c| < delta but |f(x) - L| >= epsilon.- State the epsilon-N definition for sequence limits.
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lim(n->infinity) a_n = L if for all epsilon > 0, there exists N such that n > N implies |a_n - L| < epsilon.- Why is the strict inequality 0 < |x - c| < delta used instead of |x - c| < delta?
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The strict inequality excludes the point x = c itself, allowing the limit to exist even if f is undefined or discontinuous at c.- Define the convergence of a sequence {a_n} to a limit L.
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For all epsilon > 0, there exists N such that for all n > N, |a_n - L| < epsilon.- Prove that limits of sequences are unique.
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Suppose a_n -> L and a_n -> M. For epsilon > 0, choose N so n > N gives |a_n - L| < epsilon/2 and |a_n - M| < epsilon/2. Then |L - M| <= |L - a_n| + |a_n - M| < epsilon for all n > N. Since epsilon is arbitrary, L = M.- State the squeeze theorem for sequences.
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If a_n <= b_n <= c_n for all n and both a_n and c_n converge to L, then b_n also converges to L.- What can be concluded about any convergent sequence?
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Every convergent sequence is bounded: there exists M such that |a_n| <= M for all n.- State the algebra of limits for sequences.
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If a_n -> a and b_n -> b, then a_n + b_n -> a + b, a_n * b_n -> a * b, and (if b != 0) a_n / b_n -> a / b.- What is the monotone convergence theorem?
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A monotone sequence of real numbers converges if and only if it is bounded.
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