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Übungsaufgaben zur Vorlesung Analsis1
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Übungsaufgaben zur Vorlesung Analsis1
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Aufgabe 1(i) sei MEO undX P(M).
=
Für A,BEX
gelte AmB, genau dann,
wenn AnB 0.
=
Reflexivitat:FAc gilt AvA, da AnA A=
=
O
AnB= 0
(kommutativität)
symmetrie: FA,B CX
giltA-B B-A =
, da BrA =
Transitivität:FA,B, CCX:ArBrBrC = Arc, da (AnBInlBnC
An(BuB)nC
=
=
ArBrC
=On
-
⑧
(Assoziativität (
=>
A-C
~
Aquivalenzklassen:Kl.):alle einelementigen Teilmengen
K(2) alle
=
zweielementigen Teilmengen
:
(n)
* alle
=
n-elementigen Teilmengen
·
Reflexivitat:x -x 0 2
= =
xx
=
r
·
symmetrie: 2.3. x
wy y2x
=
x -
ycz 3y
=
-
x -
= (x -
x) 2
+
3xx
= v
·
Transitivität: 2.2.x
-
y1y z 3x
=
-
2
x -
yezny
-
ze 2 ((x
=
-
y) (y z)
+
-
x
=
-
ze xzv
=
Aquivalenzklassen:K( -n) {x
=
-
y -
=
n3,
:
k( 1 -
{x y
=
- -
=
13,
k(0) {x-y 03,
= =
k(1) 4x
= -
y 13,
=
:
k(n) =
{x-y m3 =
,
+
Xx,yeR:x -
y-2,n 2
=
, (ii) MCI, für A.BCM sei ArB,
genau dann, wenn
und B
A
gleichviele Elemente besitzen
gilt:(A1
Reflexivitat:VAcM = (A) AnA
=
w
gilt:(Al=1B)
Symmetrie:FA,BCM 1B1 (A) AUBE BA
c => =
=
Transitivität.FA, B, CCM (A) (B( n1B1 (C) (A) 1B1 C (A) (C) Arc v
gilt:
)
=
=
= =
= =
= =
Aquivalenzklassen:
Jede Äquivalenzklasse enthält
die
Teilmengen von M, die die
gleiche Anzahl an Elementen
haben. Es sind endlich viele, da M endlich ist.
k() alle
einelementigen Teilmengen von M
=
K(2) alle M
zweielementigen Teilmengen von
=
"
kin) = ...
für neN
·Y: {f: =
N +
M3 " für endlich viele k, also o
↑
Reflexivität:Vf: N-M
gilt f(k) f(k)
=
FreN=> fik)-f(k) w
symmetric:fug cf(h) g(k) = = nur für endl. viele keN
f(k)=g(k)< g(k) =f(h) =
grfv
=
Transitivität:frg -grh =(f(x) g(4)
= nur für endlich viele k
q(h) h(h) = nur für endlich viele k
=(f(h) g(h)
= nur für endl. viele h
fun,
=>
(bitte das Thema nochmal besprechen!!!)
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