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Recall that, back in section 8.3, I proved that f : R Ris contin-uous ifff1 ()for every open set . Its time to generalize. First we

need a lemma, though. This says that there is a statement about rela-tively open sets which is analogous to the definition of a (plain vanilla)open set. This is part of Remark 8.27, section 8.3, which we didntneed until now.

Lemma: UEis relatively open inEiffaUthere exists >0so that B(a) EU.

Proof: "=" First suppose that U is relatively open in E. Thenthere exists an open set A so that U=E A. Since A is open and Uis a subset ofA, for any aU there is an > 0 so that B(a) A.But thenB(a) EA E=U.

"=" Conversely, suppose that U has the property that aUthere exists > 0 so that B(a)E U. For each aU, findthe appropriate , and call it (a). Set A =

aA

B(a)(a). This iscertainly an open set, since its the union of open sets. I claim thatU =A E. Certainly ifa U, then a B(a)(a), hence in A. Also,since UE, a E. Thus a A E, and UA E. For the othercontainment, take x A E. Since x A, x must be in B(a)(a) forsome a A. Since x Eas well, x B(a)(a) E. But by choice of, we must have x U.

Theorem: f :E Rm is continuous on E iff f1 () is relativelyopen in Efor every open set in Rm.

Proof: First suppose that f is continuous on E, and that isopen. If f1 () is empty, its open, so assume that f1 () is non-empty, and take a f1 (). Then f(a) . Since is open, there is >0 so that B(f(a)) . Since f is continuous at a, there is >0so thatx a< , x Eimpliesf(x) f(a)< . In other words,everything in B(a) E gets taken by f to within of f(a), henceinto . This says that B(a) Ef

1 (). This is true for arbitrarya f1 (), so by the lemma, f1 ()is relatively open in E.

Now suppose that inverse images of open sets are relatively open.Take an arbitrary a E and an > 0. We have that B(f(a))is an open set, hence f1 (B(f(a))) is relatively open in E. Sincea f1 (B(f(a))), by the lemma, there is a >0 so thatB(a)Ef1 (B(f(a))). Now untangle this statement. What its saying is thatif you take f of any point of E thats within of a, you must endup in the ball around f(a). This is just the statement that xE,x a< implies f(x) f(x)< , i.e., that f is continuous at a,and hence on Esince a was arbitrary.

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Proposition: f1 () is relatively open for each open set Rm

iff f1 (C) is relatively closed for each closed set C Rm. This gives

a third way of characterizing continuity, but you dont see it nearly asmuch as the statement about relatively open sets.

Proof: "=" Take a closed set C. Then Cc is open, thus f1 (Cc)is relatively open in E. But I claim that f1 (Cc) =E f1 (C). Its

just chasing the definition: a f1 (Cc) iff f(a) Cc, i.e., f(a) / C,which is the same asa /f1 (C), i.e., a E f1 (C). So,E f1 (C)is relatively open in E, which weve seen is the same as f1 (C) isrelatively closed in E.

"=" Take an open set and look at complements, just as in theprevious part.

Note: Images of open sets under continuous maps need not beopen. Heres an example.

Example: f : R R given by f(x) = sin x. Thenf((0, 2)) =[1, 1].

We can say something about images of compact and connected sets,however.

Theorem: IfH is compact, f :H Rm continuous, then f(H) iscompact.

Proof: Suppose that {V : A} is an open cover of f(H). Wewant to show that there is a finite subcover. I claim that the collection{f1 (V) : A}coversH. The reason: ifx H, thenf(x) f(H),so f(x) V0 for some 0, thus x f

1 (V0). Now, the collection{f1 (V) : A} is not an open cover ofH, since all we know abouteach f1 (V) is that they are relatively open in H, but this will beenough. Since each f1 (V) is relatively open in H, there there is anopen set U so that Ua H = f

1 (V). The collection {U: A}is an open cover ofH. Therefore, there is a finite subcover U1, ,Uk . Since these coverH, clearly their intersections with Hcover H,sof1 (V1), , f

1 (Vk)coverH. But thenV1, ,Vk coverf(H):y f(H)iffxHso that f(x) =y (maybe more than 1 x), so thereis Vj with x f

1

Vj

, so yVj .Note:

Continuous images of closed sets need not be closed: let f(x) =1x

. IfC= [1,) (closed, not compact), then f(C) = (0, 1] isnot closed.

Continuous images of bounded sets need not be bounded: takethe same fas before, and look at f((0, 1]). This is [1,).

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