# Definition 513 let a r and f a r we say that f is

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Definition 5.13.LetAR, andf:A!R. We say thatfis continuousata2Aif8">09δ>0so that|f(x)-f(a)|<",8x2Awith|x-a|<δ.We note that in the definition,a2Ais in the domain offand sof(a) has been defined.And, as long asais a limit point of the setA, the condition given in the definition matcheswith the existence of the limit, limx!af(x) =f(a).We thus have the equivalent definition of continuity at a limit pointa2A:Theorem 5.14.Letf:AR!Randa2Aa limit point. Thenfis continuous ataifand only iflimx!af(x) =f(a).Ifa2Aisnota limit point, in other words, ifais an isolated point ofA, then anyfis continuous ata.So generally we only really care about continuity at limit points ofthe domainA.As with limits, it will often be convenient to think of continuity in termsof sequences.Applying Theorem 5.7 to Theorem 5.14, we have yet one more equivalentcondition for continuity at a limit point:Theorem 5.15.Letf:AR!Randa2Aa limit point. Thenfis continuous ataifand only iflimn!1f(xn) =f(a)holds for everysequence(xn)n2Nwithxn---!n!1a.The best functions are continuous everywhere on their domainA:Definition 5.16.Letf:AR!R. We sayfis continuous onAiffis continuous ateach pointx2A.Continuity on a setAcan be written in a very satisfying way: a continuous function isone for whichlimn!1f(xn) =flimn!1xnholds for every convergentsequence (xn)n2NA.As in the previous discussion of limits, the use of sequences makes for a simple conditionfor a function to be discontinuous:Corollary 5.17.Letf:AR!Randa2A. If there exists a sequence(xn)n2NAwithxn---!n!1asuch thatlimn!1f(xn)6=f(a), thenfis discontinuous ata.47
Note that if there is a sequence (xn)n2Nwithxn---!n!1afor which (f(xn))n2Ndoesn’tconverge at all, thenfis discontinuous ata. One way to exploit this is to find a sequencexn---!n!1afor whichf(xn) has two dierent subsequential limits. For example, iff(x) = sgn (x) =8><>:-1,ifx <0;0,ifx= 0;1,ifx >0,we takexn= (-1)n1n---!n!10, thenf(xn) = (-1)n, which is divergent since it has twosubsequential limit points. Thus, the signum function is discontinuous atx= 0.Using the properties of the limit from Theorem 5.10 we immediately derive the usualproperties of continuous functions from calculus:Theorem 5.18.Assumef, g:AR!Rare each continuous ata2A. Then(f+g),fg, andf/g(providedg(a)6= 0) are all continuous ata2A.The other natural combination between continuous functions iscomposition. To definethe composition of two functionsh(x) =gf(x) =g(f(x)) we have to be sure that theirdomain and range are compatible.Theorem 5.19.Letf:AR!Randg:BR!Rwithf(A)B.Iffiscontinuous ata2Aandgis continuous atb=f(a)2B, thenh=gfis continuous ata.Proof.Using the definition of continuity forgaty=b, for any">0 there existsγ>0 forwhich|g(y)-g(b)|<"for ally2Bwith|y-b|<γ.