755

Proof: First consider claim 23.0.1. Let x ∈U \U. If B(x,r) contains no points of U,then x /∈U . If B(x,r) contains no points of UC, then x ∈U and so x /∈U \U . Therefore,U \U ⊆ ∂U . Now let x ∈ ∂U . If x ∈U, then since U is open there is a ball containingx which is contained in U contrary to x ∈ ∂U . Therefore, x /∈U. If x is not a limit pointof U, then some ball containing x contains no points of U contrary to x ∈ ∂U . Therefore,x ∈U \U which shows the two sets are equal.

Why is K open for K a component of Rp \C? This follows from Theorem 7.13.10and results from open balls being connected. Thus if k ∈ K,letting B(k,r)⊆CC, it followsK ∪B(k,r) is connected and contained in CC and therefore is contained in K because K ismaximal with respect to being connected and contained in CC.

Now for K a component of Rp \C, why is ∂K ⊆C? Let x ∈ ∂K. If x /∈C, then x ∈ K1,some component of Rp \C. If K1 ̸= K then x cannot be a limit point of K and so it cannotbe in ∂K. Therefore, K = K1 but this also is a contradiction because if x ∈ ∂K then x /∈ Kthanks to 23.0.1.

Note that for an open set U ⊆ Rp, and h :U → Rp, dist(h(∂U) ,y) ≥ dist(h(U),y)

because U ⊇ ∂U .The following lemma will be nice to keep in mind.

Lemma 23.0.4 f ∈ C(Ω× [a,b] ;Rp

)if and only if t → f(·, t) is in C

([a,b] ;C

(Ω;Rp

)).

Also∥f∥

∞,Ω×[a,b] = maxt∈[a,b]

(∥f(·,t)∥

∞,Ω

)Proof:⇒By uniform continuity, if ε > 0 there is δ > 0 such that if |t− s|< δ , then for

all x ∈Ω, ∥f(x,t)− f(x,s)∥< ε

2 . It follows that ∥f(·, t)− f(·,s)∥∞≤ ε

2 < ε .⇐Say (xn, tn)→ (x,t) . Does it follow that f(xn, tn)→ f(x,t)?

∥f(xn, tn)− f(x,t)∥ ≤ ∥f(xn, tn)− f(xn, t)∥+∥f(xn, t)− f(x, t)∥≤ ∥f(·, tn)− f(·, t)∥

∞+∥f(xn, t)− f(x, t)∥

both terms converge to 0, the first because f is continuous into C(Ω;Rp

)and the second

because x→ f(x, t) is continuous.The claim about the norms is next. Let (x, t) be such that ∥f∥

∞,Ω×[a,b] < ∥f(x, t)∥+ ε .Then

∥f∥∞,Ω×[a,b] < ∥f(x, t)∥+ ε ≤ max

t∈[a,b]

(∥f(·, t)∥

∞,Ω

)+ ε

and so ∥f∥∞,Ω×[a,b] ≤ maxt∈[a,b] max

(∥f(·,t)∥

∞,Ω

)because ε is arbitrary. However, the

same argument works in the other direction. There exists t such that

∥f(·, t)∥∞,Ω = max

t∈[a,b]

(∥f(·, t)∥

∞,Ω

)by compactness of the interval. Then by compactness of Ω, there is x such that

∥f(·,t)∥∞,Ω = ∥f(x, t)∥ ≤ ∥f∥

∞,Ω×[a,b]

and so the two norms are the same.

755Proof: First consider claim 23.0.1. Let x € U \ U. If B(x,r) contains no points of U,then x ¢ U. If B(x,r) contains no points of US, then x € U and sox ¢ U \ U. Therefore,U\U Cou. Now let x € OU. If x € U, then since U is open there is a ball containingx which is contained in U contrary to x € OU. Therefore, x ¢ U. If x is not a limit pointof U, then some ball containing x contains no points of U contrary to x € QU. Therefore,x € U\U which shows the two sets are equal.Why is K open for K a component of R? \C? This follows from Theorem 7.13.10and results from open balls being connected. Thus if k € K, letting B(k,r) C CS, it followsK UB(k,r) is connected and contained in C© and therefore is contained in K because K ismaximal with respect to being connected and contained in C°.Now for K a component of R? \\C, why is 0K CC? Let x € OK. If x ¢C, thenx € Kj,some component of R? \C. If K, #4 K then x cannot be a limit point of K and so it cannotbe in 0K. Therefore, K = Ky but this also is a contradiction because if x € 0K then x ¢ Kthanks to 23.0.1. ffNote that for an open set U C R?, and h:U + R?, dist(h(0U) ,y) > dist (h (U) ,y)because U D QU.The following lemma will be nice to keep in mind.Lemma 23.0.4 f € C (Q x [a,b];IR”) if and only if t + £(-,t) is in C ([a,b];C (Q;R’)).Also0)Proof: =By uniform continuity, if € > 0 there is 6 > 0 such that if |t — s| < 6, then forall x € Q, ||f(x,t) —f(x,s)|| < §. It follows that ||f (-,¢) —f(-,s) ||, <§ <<.<Say (Xn,tn) > (x,t). Does it follow that f(Xp,t,) > f(x,r)?f\|5o = f(-,tIlion) = max (ICA)IIPXnstn) —£(x0)|| < [I£%nstn) —£(Xn1) | + I] &Xn,t) — £(x,9)< |lfC.m)-£6,0)|].. + Ilf nt) — £(x,1)||both terms converge to 0, the first because f is continuous into C (Q;R? ) and the secondbecause x — f(x,f) is continuous.The claim about the norms is next. Let (x,t) be such that ||fl|,. 9,4.) < llf(«.t)|| +.Thenf\|_ = f < f(-,¢ relIfleaxian} < It) Fe < max ([I€(Dllea) +eand so If]... [a,b] < mMax;efq,p] Max (litlea) because € is arbitrary. However, thesame argument works in the other direction. There exists t such thatIf.) lla = max ((lfCDllea), tela,b] ,by compactness of the interval. Then by compactness of Q, there is x such thatf(t) II. = NEGO S [fll ax{a.5)and so the two norms are the same. §f