These  Real Analysis appendices continue the decimal viewpoint of limits, continuity and convergence in chapter 14. and this further lesson

A. What's Next
B. Pigeon Hole Principle
B. Bolzano-Weierstrass
C1. Triangle Inequality
C2. Triangle Inequality
C. More T.Inequality
D. Sets & Sequences
D. Monotone Sequences
E. Limits,  Properties
E Limits & Error Control
F. Continuous Functions
F. Closed Range Thm
F. Intermediate Val. Thm
F. Compactness Thm
F. Equicontinuity Thm
F Extreme Value Thm
G. Rolle's Theorem etc
G. Mean Val. Thm.
G. Constant Difference Thm
G. Lipschitz Continuity I
PS: One Sided Range Theorems
G. Velocity Revisited
G. Sufficient Conditions
H. Riemann Sums Conv
H. Lipschitz Continuity II

Proofs of  one-sided theorems could be of interest in the study of 2D topology.

Closed Range Theorem

A set W is set to be closed if and only if it contains all its limit points. Recall, a real number A to be a limit point of a infinite set of numbers S if and only if every interval I centered at A, no matter how small, contains infinitely many elements of the set S. Equivalently, a real number A to be a limit point of a set S if and only if every interval I centered at A contains at least one element s ¹ A of the set S.

Theorem F.2 [Closed Range Theorem] Suppose f(x) is a real-valued function which is continuous at each point x in the interval [a,b]. Suppose if L is a limit point of the function's range

range(f) = {y | there exists a point x in [a,b] such that y = f(x)}.

Then there exists a point w in [a,b] such that L = f(w). The proof relies on the second triangle inequality

|a-b| ³

ê ê ê

|a|-|b|

ê ê ê

Proof of Closed Range Theorem.

Observe L is a limit point of the range of the function f, that is, the set

W = {y |  there exists a point x in [a,b] such that y = f(x)}

Therefore, each interval of length [1/2]10^-k centered at L contains at least one point of W. Choose one and label it w_k. Then w_k = f(x_k) for some x_k in the interval [a,b]. Further, the infinite set of points x_k has a greatest lower bound A = inf{x|  k ³ 1} in the interval [a,b].

For the sake of a contradiction, suppose that f(A) ¹ L. Then there exists an integer K > 0 such that |f(A)-L| > [1/2] [1/(10^K)] for some whole number K. Now in every interval centered at A, there exists a w_k with k > K. This w_k has the property that

 |f(w_k)-L| £ [1/2]10^-k < [1/2]10^-K.

Therefore

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