We leave it as an exercise to prove that these definitions are equivalent. Note that c must belong to the domain A of f in order to define the continuity.
A function f is continuous at x0 in its domain if for every sequence (xn) with xn in To show a function is continuous we can do one of three things:.
Theorem 21. A continuous function on a compact metric space is bounded and uniformly continuous. Proof. If X is a compact metric space and f :
(a) Show that f is continuous at. 0. x = (b) For. 0 x ? express ( ). f x. ? as a piecewise-defined function. Find the value of x for which ( ).
j=1Fj. The function f is continuous on F which is a finite union of Proof. Let us first show that f2 is integrable on [a
zero set of f. Proof Since Z(f) = f?1({0}) and {0} is a closed set in R
x0 /? E. Show that there is an unbounded continuous function f : E ? R. uniform continuity without actually using showing the dependence of ? on ?
The Intermediate Value Theorem will be used in the following problems. (a) Show that the equation 2x = 3x has a solution c ? (14). Proof . Let g(
Show that the compactness of Y cannot be omitted from the hypotheses even when. X and Z are compact. Proof (a) Suppose f is not uniformly continuous. Then
x where nis a positive integer then f(x) is continuous on the interval [0;1) We can use symmetry of graphs to extend this to show that f(x) is continuous on the interval (1 ;1) when nis odd Hence all n th root functions are continuous on their domains Trigonometric Functions In the appendix we provide a proof of the following Theorem :
First let us show that a continuous extension of thefunctionf to the set is unique (assuming it exists) Suppose ghSince the set ? isdense inE for any c??E0 converging toc Sincec we get g(xn)?g(c) and However g(xn) =h(xn) =f(xn) for all : E are two continuous extensions of f E0 sequence {xn} continuous at ? ? Hence g(c) =h(c) there is a
Clearly F is continuous on (a;b):To prove continuity at a let fx ngbe a sequence in (a;b) converging to a We need to show that F(x n) = f(x n) !F(a) = A Let ">0 There exists N 1such that for all n>N 1 jA f(a n)j< " 2 : 4 The proof will be complete if we can show that for nlarge enough jf(x n) f(a n)jcan be made smaller than "=2
1 Consider the sequence of functions f n(x) = xnon [0;1] (a)Show that each function f nis uniformly continuous on [0;1] Solution: Any continuous function on a compact set is uniformly continuous (b)For a sequence of functions f n on [0;1] write what it means for them to NOT be an equicontinuous sequence
to show that feis continuous at 0 We need to show that if (x n) is a sequence in (0;1] such that x n!0 then fe(x n) !fe(0) which is equivalent to that f(x n) !0 From x n!0 we get x2 n!0 Since jsin(1=x n)j 1 we have jf(x n)j= jx2n sin(1 xn)j jx nj2 By squeeze lemma we have f(x n) !0 as desired Thus feis continuous on [0;1] Thus
That means to show that f is not continuous at 0, it is sufficient to exhibit one sequence (x n) which converges to 0 for which the sequence (f (x n) does not converge to f (0)=1. So the suggestion was take x n =1/n.
Consider f:[?1,3]?R, f(x)= ! 2x, ?1? x ?1 3?x,1
If f (x)=x and the domain is R, then f is uniformly continuous. (We can take delta=epsilon.) The range of this f is all of R, an unbounded set. If f (x)=sin (1/x) and the domain in (0,1), then f is not uniformly continuous.
If f is continuous on [ 1, 2] (i.e., its graph can be sketched as a continuous curve from ( 1, ? 10) to ( 2, 5)) then we know intuitively that somewhere on [ 1, 2] f must be equal to ? 9, and ? 8, and ? 7, ? 6, …, 0, 1 / 2, etc. In short, f takes on all intermediate values between ? 10 and 5.