CONTINUOUS FUNCTIONS
Section 5.1 Continuous Functions
In this section, which is very similar to Section 4.1, we will define what it means to say that a function is continuous at a point, or on a set. This notion of continuity is one of the central concepts of mathematical analysis, and it will be used in almost all of the following material in this book. Consequently, it is essential that the reader master it.
5.1.1 Definition LetAR, letf :A!R, and letc2A. We say thatfiscontinuous at cif, given any numbere>0, there existsd>0 such that ifxis any point ofAsatisfying jxcj<d, thenjfðxÞ fðcÞj<e.
Ifffails to be continuous atc, then we say thatf isdiscontinuous atc.
As with the definition of limit, the definition of continuity at a point can be formulated very nicely in terms of neighborhoods. This is done in the next result. We leave the verification as an important exercise for the reader. See Figure 5.1.1.
5.1.2 Theorem A function f:A!Ris continuous at a point c2A if and only if given anye-neighborhood VeðfðcÞÞof f(c)there exists ad-neighborhood VdðcÞof c such that if x is any point of A\VdðcÞ,then f(x)belongs to VeðfðcÞÞ,that is,
fðA\VdðcÞÞ VeðfðcÞÞ:
Remarks (1) Ifc2Ais a cluster point ofA, then a comparison of Definitions 4.1.4 and 5.1.1 show thatf is continuous atcif and only if
ð1Þ fðcÞ ¼lim
x!cfðxÞ:
Figure 5.1.1 GivenVeðfðcÞÞ, a neighborhoodVdðcÞis to be determined
5.1 CONTINUOUS FUNCTIONS 125
Thus, ifcis a cluster point ofA, then three conditions must hold forfto be continuous atc: (i) f must be defined atc(so thatf(c) makes sense),
(ii) the limit offatcmust exist inR (so that lim
x!c fðxÞmakes sense), and (iii) these two values must be equal.
(2) Ifc2Ais not a cluster point ofA, then there exists a neighborhoodVdðcÞofcsuch thatA\VdðcÞ ¼ fcg. Thus we conclude that a functionfis automatically continuous at a pointc2Athat is not a cluster point ofA. Such points are often called ‘‘isolated points’’ of A. They are of little practical interest to us, since they have no relation to a limiting process.
Since continuity is automatic for such points, we generally test for continuity only at cluster points. Thus we regard condition (1) as being characteristic for continuity atc.
A slight modification of the proof of Theorem 4.1.8 for limits yields the following sequential version of continuity at a point.
5.1.3 Sequential Criterion for Continuity A function f :A!R is continuous at the point c2A if and only if for every sequence(xn)in A that converges to c, the sequence ðfðxnÞÞconverges to f(c).
The following Discontinuity Criterion is a consequence of the last theorem. It should be compared with the Divergence Criterion 4.1.9(a) with L¼f(c). Its proof should be written out in detail by the reader.
5.1.4 Discontinuity Criterion Let AR, let f :A!R, and let c 2 A. Then f is discontinuous at c if and only if there exists a sequence(xn)in A such that(xn)converges to c, but the sequenceðfðxnÞÞdoes not converge to f(c).
So far we have discussed continuity at a point. To talk about the continuity of a function on aset, we will simply require that the function be continuous at each point of the set. We state this formally in the next definition.
5.1.5 Definition Let AR and let f :A!R. If Bis a subset of A, we say thatf is continuous on the setBiffis continuous at every point ofB.
5.1.6 Examples (a) The constant function fðxÞ:¼bis continuous onR. It was seen in Example 4.1.7(a) that ifc2R, then lim
x!c fðxÞ ¼b. SincefðcÞ ¼b, we have lim
x!c fðxÞ ¼fðcÞ, and thus f is continuous at every point c2R. Therefore f is continuous onR.
(b) gðxÞ:¼xis continuous onR.
It was seen in Example 4.1.7(b) that ifc2R, then we have lim
x!cg¼c. SincegðcÞ ¼c, thengis continuous at every point c2R. Thusg is continuous onR.
(c) hðxÞ:¼x2 is continuous onR.
It was seen in Example 4.1.7(c) that if c2R, then we have lim
x!ch¼c2. Since hðcÞ ¼c2, thenhis continuous at every point c2R. Thushis continuous onR. (d) wðxÞ:¼1=xis continuous onA:¼ fx2R :x>0g.
It was seen in Example 4.1.7(d) that if c2A, then we have lim
x!cw¼1=c. Since wðcÞ ¼1=c, this shows thatwis continuous at every pointc2A. Thuswis continuous onA.
(e) wðxÞ:¼1=xis not continuous atx¼0.
Indeed, if wðxÞ ¼1=xfor x>0, then w is not defined for x¼0, so it cannot be continuous there. Alternatively, it was seen in Example 4.1.10(a) that lim
x!0wdoes not exist inR, sowcannot be continuous atx¼0.
(f) The signum function sgn is not continuous at 0.
The signum function was defined in Example 4.1.10(b), where it was also shown that
x!lim0sgnðxÞdoes not exist inR. Therefore sgn is not continuous atx¼0 (even though sgn 0 is defined). It is an exercise to show that sgn is continuous at every pointc6¼0.
Note In the next two examples, we introduce functions that played a significant role in the development of real analysis. Discontinuities are emphasized and it is not possible to graph either of them satisfactorily. The intuitive idea of drawing a curve in the plane to represent a function simply does not apply, and plotting a handful of points gives only a hint of their character. In the nineteenth century, these functions clearly demonstrated the need for a precise and rigorous treatment of the basic concepts of analysis. They will reappear in later sections.
(g) LetA:¼R and letfbe Dirichlet’s ‘‘discontinuous function’’ defined by fðxÞ:¼ 1 if xis rational;
0 if xis irrational:
We claim thatfisnot continuous at any point ofR. (This function was introduced in 1829 by P. G. L. Dirichlet.)
Indeed, ifcis a rational number, let (xn) be a sequence of irrational numbers that converges to c. (Corollary 2.4.9 to the Density Theorem 2.4.8 assures us that such a sequence does exist.) Since fðxnÞ ¼0 for all n2N, we have limðfðxnÞÞ ¼0, while fðcÞ ¼1. Therefore fis not continuous at the rational numberc.
On the other hand, ifb is an irrational number, let (yn) be a sequence of rational numbers that converge tob. (The Density Theorem 2.4.8 assures us that such a sequence does exist.) Since fðynÞ ¼1 for all n2N, we have limðfðynÞÞ ¼1, while fðbÞ ¼0.
Thereforef is not continuous at the irrational numberb.
Since every real number is either rational or irrational, we deduce that f is not continuous at any point inR.
Figure 5.1.2 Thomae’s function
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(h) Let A:¼ fx2R:x>0g. For any irrational number x > 0 we define hðxÞ:¼0.
For a rational number in A of the form m=n, with natural numbers m, n having no common factors except 1, we definehðm=nÞ:¼1=n. (We also definehð0Þ:¼1.)
We claim thath is continuous at every irrational number in A, and is discontinuous at every rational number in A. (This function was introduced in 1875 by K. J. Thomae.)
Indeed, if a > 0 is rational, let (xn) be a sequence of irrational numbers in Athat converges toa. Then limðhðxnÞÞ ¼0, whileh(a)>0. Henceh is discontinuous at a.
On the other hand, ifbis an irrational number ande>0, then (by the Archimedean Property) there is a natural numbern0such that 1=n0 <e. There are only a finite number of rationals with denominator less thann0 in the intervalðb1;bþ1Þ. (Why?) Hence d>0 can be chosen so small that the neighborhoodðbd;bþdÞcontains no rational numbers with denominator less than n0. It then follows that for jxbj<d;x2A, we have jhðxÞ hðbÞj ¼ jhðxÞj 1=n0<e. Thus h is continuous at the irrational numberb.
Consequently, we deduce that Thomae’s function h is continuous precisely at the
irrational points in A. (See Figure 5.1.2.) &
5.1.7 Remarks (a) Sometimes a function f:A!R is not continuous at a point c because it is not defined at this point. However, if the functionfhas a limitLat the pointc and if we defineFonA[ fcg !R by
FðxÞ:¼ L for x¼c;
fðxÞ for x2A;
thenFis continuous atc. To see this, one needs to check that lim
x!cF¼L, but this follows (why?), since lim
x!c f ¼L.
(b) If a functiong:A!R does not have a limit atc, then there is no way that we can obtain a functionG:A[ fcg !R that is continuous atcby defining
GðxÞ:¼ C for x¼c;
gðxÞ for x2A:
To see this, observe that if lim
x!cGexists and equals C, then lim
x!cg must also exist and equal C.
5.1.8 Examples (a) The functiongðxÞ:¼sinð1=xÞforx6¼0 (see Figure 4.1.3) does not have a limit atx¼0 (see Example 4.1.10(c)). Thus there is no value that we can assign atx¼0 to obtain a continuous extension ofgatx¼0.
(b) LetfðxÞ:¼xsinð1=xÞforx6¼0. (See Figure 5.1.3.) It was seen in Example 4.2.8(f) that lim
x!0ðxsinð1=xÞÞ ¼0. Therefore it follows from Remark 5.1.7(a) that if we define
F:R !R by
FðxÞ:¼ 0 for x¼0; xsinð1=xÞ for x6¼0;
thenFis continuous atx¼0.
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Exercises for Section 5.1
1. Prove the Sequential Criterion 5.1.3.
2. Establish the Discontinuity Criterion 5.1.4.
3. Leta<b<c. Suppose thatfis continuous on [a,b], thatgis continuous on [b,c], and that fðbÞ ¼gðbÞ. Definehon [a,c] byhðxÞ:¼fðxÞforx2 ½a;bandhðxÞ:¼gðxÞforx2 ½b;c.
Prove thathis continuous on [a,c].
4. Ifx2R, we definevxbto be the greatest integern2Zsuch thatnx. (Thus, for example, v8:3b¼8;vpb¼3;vpb¼ 4.) The functionx7!vxbis called thegreatest integer function.
Determine the points of continuity of the following functions:
(a) fðxÞ:¼vxb; (b) gðxÞ:¼xvxb;
(c) hðxÞ:¼vsinxb; (d) kðxÞ:¼v1=xb ðx6¼0Þ:
5. Let fbe defined for allx2R;x6¼2, byfðxÞ ¼ ðx2þx6Þ=ðx2Þ. Canfbe defined at x¼2 in such a way thatfis continuous at this point?
6. LetARand letf:A!Rbe continuous at a pointc2A. Show that for anye>0, there exists a neighborhoodVdðcÞofcsuch that ifx;y2A\VdðcÞ, thenjfðxÞ fðyÞj<e.
7. Letf :R!Rbe continuous atcand letfðcÞ>0. Show that there exists a neighborhoodVdðcÞ ofcsuch that ifx2VdðcÞ, thenfðxÞ>0.
8. Letf :R!Rbe continuous onRand letS:¼ fx2R:fðxÞ ¼0gbe the ‘‘zero set’’ off. If ðxnÞis inSandx¼limðxnÞ, show thatx2S.
9. LetABR, letf:B!Rand letgbe the restriction offtoA(that is,gðxÞ ¼fðxÞfor x2A).
(a) Iffis continuous atc2A, show thatgis continuous atc.
(b) Show by example that ifgis continuous atc, it need not follow thatfis continuous atc.
10. Show that the absolute value functionfðxÞ:¼ jxjis continuous at every pointc2R.
11. LetK>0 and letf :R!Rsatisfy the conditionjfðxÞ fðyÞj Kjxyjfor allx;y2R.
Show thatfis continuous at every pointc2R.
12. Suppose thatf:R!Ris continuous onRand thatfðrÞ ¼0 for every rational numberr. Prove thatfðxÞ ¼0 for allx2R.
13. Defineg:R!Rby gðxÞ:¼2xforx rational, andgðxÞ:¼xþ3 forxirrational. Find all points at whichgis continuous.
Figure 5.1.3 Graph of fðxÞ ¼xsinð1=xÞ ðx6¼0Þ
5.1 CONTINUOUS FUNCTIONS 129
14. LetA:¼ ð0;1Þand letk:A!Rbe defined as follows. Forx2A;xirrational, we define kðxÞ ¼0; forx2Arational and of the formx¼m=nwith natural numbersm,nhaving no common factors except 1, we definekðxÞ:¼n. Prove thatkis unbounded on every open interval inA. Conclude thatkis not continuous at any point ofA. (See Example 5.1.6(h).)
15. Let f:ð0;1Þ !R be bounded but such that lim
x!0f does not exist. Show that there are two sequencesðxnÞandðynÞin (0, 1) with limðxnÞ ¼0¼limðynÞ, but such thatðfðxnÞÞandðfðynÞÞ exist but are not equal.