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Since f iscontinuous at every point x ∈ E, it follows (see 60 in Sect. 4.1.1) that, knowing ε > 0we can find a δ-neighborhood U δ (x) of x such that the oscillation ω(f ; UEδ (x))of f on the set UEδ (x) = E ∩ U δ (x), consisting of the points in the domain ofdefinition E lying in U δ (x), is less than ε. For each point x ∈ E we construct aneighborhood U δ (x) having this property. The quantity δ may vary from one pointto another, so that it would be more accurate, if more cumbersome, to denote theneighborhood by the symbol U δ(x) (x), but since the whole symbol is determined bythe point x, we can agree on the following abbreviated notation: U (x) = U δ(x) (x)and V (x) = U δ(x)/2 (x).The open intervals V (x), x ∈ E, taken together, cover the closed interval[a, b], and so by the finite covering lemma one can select a finite coveringV (x1 ), .
. . , V (xn ). Let δ = min{ 12 δ(x1 ), . . . , 12 δ(xn )}. We shall show that |f (x ) −f (x )| < ε for any points x , x ∈ E such that |x − x | < δ. Indeed, since the system of open intervals V (x1 ), . . . , V (xn ) covers E, there exists an interval V (xi ) ofthis system that contains x , that is |x − xi | < 12 δ(xi ). But in that case x − xi ≤ x − x + x − xi < δ + 1 δ(xi ) ≤ 1 δ(xi ) + 1 δ(xi ) = δ(xi ).2224.2 Properties of Continuous FunctionsConsequently x , x ∈ UEδ(x )ω(f ; UE i (xi )) < ε.δ(xi )163(xi ) = E ∩ U δ(xi ) (xi ) and so |f (x ) − f (x )| ≤The examples given above show that Cantor’s theorem makes essential use of acertain property of the domain of definition of the function.
It is clear from the proofthat, as in Theorem 3, this property is that from every covering of E by neighborhoods of its points one can extract a finite covering.Now that Theorem 4 has been proved, it is useful to return once again to theexamples studied earlier of functions that are continuous but not uniformly continuous, in order to clarify how it happens that sin(x 2 ) for example, which is uniformlycontinuous on each closed interval of the real line by Cantor’s theorem, is nevertheless not uniformly continuous on R. The reason is completely analogous to thereason why a continuous function in general fails to be uniformly continuous.
Thistime we invite our readers to investigate this question on their own.We now pass to the last theorem of this section, the inverse function theorem.We need to determine the conditions under which a real-valued function on a closedinterval has an inverse and the conditions under which the inverse is continuous.Proposition 1 A continuous mapping f : E → R of a closed interval E = [a, b]into R is injective if and only if the function f is strictly monotonic on [a, b].Proof If f is increasing or decreasing on any set E ⊂ R whatsoever, the mappingf : E → R is obviously injective: at different points of E the function assumesdifferent values.Thus the more substantive part of Proposition 1 consists of the assertion that every continuous injective mapping f : [a, b] → R is realized by a strictly monotonicfunction.Assuming that such is not the case, we find three points x1 < x2 < x3 in [a, b]such that f (x2 ) does not lie between f (x1 ) and f (x3 ).
In that case, either f (x3 ) liesbetween f (x1 ) and f (x2 ) or f (x1 ) lies between f (x2 ) and f (x3 ). For definitenessassume that the latter is the case. By hypothesis f is continuous on [x2 , x3 ]. Therefore, by Theorem 2, there is a point x1 in this interval such that f (x1 ) = f (x1 ).We then have x1 < x1 , but f (x1 ) = f (x1 ), which is inconsistent with the injectivity of the mapping. The case when f (x3 ) lies between f (x1 ) and f (x2 ) is handledsimilarly.Proposition 2 Each strictly monotonic function f : X → R defined on a numericalset X ⊂ R has an inverse f −1 : Y → R defined on the set Y = f (X) of values of f ,and has the same kind of monotonicity on Y that f hαs on X.Proof The mapping f : X → Y = f (X) is surjective, that is, it is a mapping of Xonto Y .
For definiteness assume that f : X → Y is increasing on X. In that case∀x1 ∈ X ∀x2 ∈ X x1 < x2 ⇔ f (x1 ) < f (x2 ) .(4.1)1644Continuous FunctionsThus the mapping f : X → Y assumes different values at different points, andso is injective. Consequently f : X → Y is bijective, that is, it is a one-to-one correspondence between X and Y .
Therefore the inverse mapping f −1 : Y → X isdefined by the formula x = f −1 (y) when y = f (x).Comparing the definition of the mapping f −1 : Y → X with relation (4.1), wearrive at the relation∀y1 ∈ Y ∀y2 ∈ Y f −1 (y1 ) < f −1 (y2 ) ⇔ y1 < y2 ,(4.2)which means that the function f −1 is also increasing on its domain of definition.The case when f : X → Y is decreasing on X is obviously handled similarly.
In accordance with Proposition 2 just proved, if we are interested in the continuity of the function inverse to a real-valued function, it is useful to investigate thecontinuity of monotonic functions.Proposition 3 The discontinuities of a function f : E → R that is monotonic on theset E ⊂ R can be only discontinuities of first kind.Proof For definiteness let f be nondecreasing. Assume that a ∈ E is a point ofdiscontinuity of f .
Since a cannot be an isolated point of E, a must be a limit pointof at least one of the two sets Ea− = {x ∈ E | x < a} and Ea+ = {x ∈ E | x > a}.Since f is nondecreasing, for any point x ∈ Ea− we have f (x) ≤ f (a), and therestriction f |Ea− of f to Ea− is a nondecreasing function that is bounded from above.It then follows that the limitlim (f |Ea− )(x) =Ea− x→alimEx→a−0f (x) = f (a − 0)exists.The proof that the limit limEx→a+0 f (x) = f (a + 0) exists when a is a limitpoint of Ea+ is analogous.The case when f is a nonincreasing function can be handled either by repeatingthe reasoning just given or passing to the function −f , so as to reduce the questionto the case already considered.Corollary 1 If a is a point of discontinuity of a monotonic function f : E → R,then at least one of the limitslimEx→a−0f (x) = f (a − 0),limEx→a+0f (x) = f (a + 0)exists, and strict inequality holds in at least one of the inequalities f (a − 0) ≤f (a) ≤ f (a + 0) when f is nondecreasing and f (a − 0) ≥ f (a) ≥ f (a + 0) whenf is nonincreasing.
The function assumes no values in the open interval defined bythe strict inequality. Open intervals of this kind determined by different points ofdiscontinuity have no points in common.4.2 Properties of Continuous Functions165Proof Indeed, if a is a point of discontinuity, it must be a limit point of the set E,and by Proposition 3 is a discontinuity of first kind. Thus at least one of the basesE x → a − 0 and E x → a + 0 is defined, and the limit of the function overthat base exists. (When both bases are defined, the limits over both bases exist.)For definiteness assume that f is nondecreasing. Since a is a point of discontinuity,strict inequality must actually hold in at least one of the inequalities f (a − 0) ≤f (a) ≤ f (a + 0).
Since f (x) ≤ limEx→a−0 f (x) = f (a − 0), if x ∈ E and x < a,the open interval (f (a − 0), f (a)) defined by the strict inequality f (a − 0) < f (a)is indeed devoid of values of the function. Analogously, since f (a + 0) ≤ f (x) ifx ∈ E and a < x, the open interval (f (a), f (a + 0)) defined by the strict inequalityf (a) < f (a + 0) contains no values of f .Let a1 and a2 be two different points of discontinuity of f , and assume a1 < a2 .Then, since the function is nondecreasing,f (a1 − 0) ≤ f (a1 ) ≤ f (a1 + 0) ≤ f (a2 − 0) ≤ f (a2 ) ≤ f (a2 + 0).It follows from this that the intervals containing no values of f and correspondingto different points of discontinuity are disjoint.Corollary 2 The set of points of discontinuity of a monotonic function is at mostcountable.Proof With each point of discontinuity of a monotonic function we associate thecorresponding open interval in Corollary 1 containing no values of f .
These intervals are pairwise disjoint. But on the line there cannot be more than a countablenumber of pairwise disjoint open intervals. In fact, one can choose a rational number in each of these intervals, so that the collection of intervals is equipollent witha subset of the set Q of rational numbers. Hence it is at most countable. Therefore,the set of points of discontinuity, which is in one-to-one correspondence with a setof such intervals, is also at most countable.Proposition 4 (A criterion for continuity of a monotonic function) A monotonicfunction f : E → R defined on a closed interval E = [a, b] is continuous if andonly if its set of values f (E) is the closed interval with endpoints f (a) and f (b).6Proof If f is a continuous monotonic function, the monotonicity implies that allthe values that f assumes on the closed interval [a, b] lie between the values f (a)and f (b) that it assumes at the endpoints.
By continuity, the function must assumeall the values intermediate between f (a) and f (b). Hence the set of values of afunction that is monotonic and continuous on a closed interval [a, b] is indeed theclosed interval with endpoints f (a) and f (b).Let us now prove the converse. Let f be monotonic on the closed interval [a, b].If f has a discontinuity at some point c ∈ [a, b], by Corollary 1 one of the open intervals ]f (c − 0), f (c)[ and ]f (c), f (c + 0)[ is defined and nonempty and contains6 Heref (a) ≤ f (b) if f is nondecreasing, and f (b) ≤ f (a) if f is nonincreasing.1664Continuous Functionsno values of f .
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