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Archimedes the genius?
 

Continuity

Definition 1.12.

Our definition of continuity follows the same line of thought.
Definition 1.12. We work in an ordered field F. Suppose that E is a subset of¥ and that f is some function from E to F. We say that f is continuous at x 6 E if given any e > 0 we can find 6o(e, x) [read 'So depending on e and x'] with 5o(e, x) > 0 such that
\f(x) — f(y)\ < e for all у 6 E such that \x — y\ < 6o(e, x).

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Lemma 1.13.

We work in an ordered field F. Suppose that E is a subset of¥, that x 6 E, and that f and g are functions from E to F.
(i) If f(x) = с for all x 6 E, then f is continuous on E.
(ii) If f and g are continuous at x, then so is f + g.

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Lemma 1.15.

We work in an ordered field F. Suppose that E is a subset of¥, that x 6 E, and that f is continuous at x. If xn 6 E for all n and
xn —»• x as n —> oo, then f(xn) —»• f(x) as n —»• oo.
Proof. Left to reader.

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The fundamental axiom

The key property of the reals, the fundamental axiom which makes everything work, can be stated as follows:
The fundamental axiom of analysis. If an £ R for each n > 1, A 6 R and a\ < a2 < аз < ... and an < A for each n, then there exists an a 6K such that an —> a as n —> oo.

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Theorem 1.17.

 (The intermediate value theorem.) /// : [a, b] —>• R
is continuous and /(a) > 0 > f(b) then there exists а с & [a, b] such that /(c) =0.
(The proof will be given as Theorem 1.35.)

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Lemma 1.19.

In Ж every decreasing sequence bounded below tends to a limit.
Proof. Observe that if a\, аг, аз, ... is a decreasing sequence bounded
below then — a\, — аг, —аз, ... is an increasing sequence bounded above.
We leave the details to the reader as an exercise.

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