Is continuous at 0 c what can we say about the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f x  x. Some additional exercises on continuity can be found by clicking on the icon . Further Exercises on Continuity 1. Suppose that f : R R. Prove that the following conditions are equivalent: a. The function f is continuous on R. b. For every open set V the set f 1 c. For every closed set H the set f V is also open. 1 H is also closed. To prove that condition a implies condition b we assume that condition a holds. Suppose that V is an open set. Suppose that x f 1 V . We need to show that f 1 V is a neighborhood of x. Using the fact that f is continuous at the number x and the fact that V is a neighborhood of f x we choose a number   0 such that the condition f t V holds whenever |t x |  . Since x , x   f1V we have shown, as promised, that x is an interior point of the set f 1 V . To prove that condition b implies condition c we assume that condition b holds. Suppose that H is a closed set. Since R H is open, the set f 1 R H is open. Therefore, since R f1H f1R H, the set R f 1 H is open and we conclude that f 1 H is closed. The fact that condition c implies condition b follows in exactly the same way. Finally we shall show that condition b implies condition a. We assume that condition b holds. Suppose that x is a real number. We want to show that f is continuous at x. Suppose that  0. Since the set f x , f x  is open, so is the set f 1 f x ,f x  . Therefore the latter set is a neighborhood of thenumber x. Choose   0 such that x , x   f1 fx ,f x  and observe that if t is any number satisfying the inequality |t x |   we have |f t f x |  . 2. Give an example of a continuous function f on the set R of all real numbers and a closed set H such that the set f H fails to be closed. 196 We define fx  1 x for every number x 1. This function f is continuous from the closed set 1, Ý onto the set 0, 1 which fails to be closed. 3. Given that f is a continuous function from R to R and that S is a set of real numbers, prove that fS. fS Solution: Suppose that f is continuous from R to R and that S we know that the set f 1 fS R. Sinse the set f S is a closed set is closed. Therefore since S f 1 fS S f 1 fS we have and we conclude that fS fS Now we prove the “if” part of the exercise. Suppose that the inequality fS fS holds for every subset S of R. To prove that f is continuous, suppose that H is closed. We shall show that the set f 1 H is closed. Now ff 1 H ff 1 HH H and therefore f and we have shown that f 1 1 H f 1 H H is closed. 4. Given that f : R R and that for all numbers x and t we have |f x f t | |x t | 2 , prove that the function f must be constant. Note that although this exercise is quite difficult right now, it will become considerably easier after we have studied the concept of a derivative. I am resisting the urge to write a direct solution of this exercise. Those students who wish to attempt it now will probably want to be left alone. All others can wait until after the mean value theorem when the fact that f is constant...
View Full Document

This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

Ask a homework question - tutors are online