Übungsaufgaben zur Vorlesung Analsis1

1
Aufgabe

(i) 22 .
f(AVB7 f(A)rf(B)
=
FA,BCX


Sei
yef(AvB)c=> 7 xeArB:f(x)=Y


1 =

A:f(x) y)v(7x
c(7x = =
B:
-

f(x) y) =




1 3
=

yef(A) vyef(B)

2
xyef(A)rf(B)
=




Somit gilt f(AvB) f(A)uf(B) =


VA,BCK 4.e.d.

(ii) 2.2. f(AnB) 2 f(A)nf(B) FA,BCX


yeflAnB) c 5xeAnB:f(x)=Y
=




Sei



c
B1y f(x))
Ix:(xe An x =
=


=




1
(7x(xe Any f(x))n(7x(xeBry f(x)))
=


=
=




1
ye f(A) 1y=f(B)
=




c
3ye f(A)n f(B)
=




=(f(AnB)<f(A)nf(B) p.e.d.

(ii) Nein! Gegenbeispiel:

f:R-R, x x
-




A (13,B { 13
=
= -




f(AnB) f(0) 0 f(A)nf(B) [13e[13 [13
=
=




p.e.d.
= = =




(iv) 22. f(MUN) f (M) wfIN)
=



F
MINCY


Seiy ef"(MUN) ==> 5 xeMUN.f(x) Y =




(5 f(x) x)v(5 eN:f (x) y)
-




1 > M: x
=




x =
=
=




1 3
=




y e f (M) vy = f (N)

c 2
=


x f(M) Uf (N)
=

=cf (MUN) f-(M) rf-IN)
=




p.e. C.

, (V) 2.2. f(MN):f(M) ef(N) FMIN CY


Sei yefTMuN) cc
=



5xeMmN:fixl=y

1 Ix:(XeM1xeN ny fix))
=


=




1
(7 x(xeM1y f(x))n (7 x(xeNny f(x)))
=


=
=




1 2
=




ye fTM) 1 x fiN)
=




bijektiv, da
umkehrfunktion
=3
xe fi) n fiN
existiert ==
-




=L f "(MmN) f-r (M) f- (N) p.e.d.
-




=

e