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0Û o#Ý(ì l b ¥ Ó å º0° ° x Ü'â(ì 2 b%& «0Û o / 6ë è 6ë é _ > E 1"8 S b s b Û*f 6 ~ " ) ì6ë \ ^ S u ® « ~ â ¶ â > · 0 Ó & p Ñ ¼ ¸ þ ¥ · 1 m í n ® ­ ~ â ¶ â > · ü 0 Ó &. ^ $ ß q · n ) " ¹ 2 ² q ¤ ÷ w ® * n ) " ¹ * d j Û ý ¹ Û * Ý % ó Ì 4 ' ) i = 2 · · y « Ñ 2 w " % a Í.  · Zitat Original von Marcyman In dem Beweis, den unser Übungsleiter uns gegeben hat, kommt dann der Schritt f(x) e f(A U B) => x e (A U B) vor Hier wurde ich stutzig, da f ja durchaus nicht injektiv zu sein brauch, also kann es noch ein x e X\(A U B) geben, dem ebenfalls ein f(x) e f(A U B) zugeordnet ist.

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Q @ ± A 8 o 8b c ± A C ^ 8b b ± ` í9× * b A r ~ r @ ¨ 8 o 8b c9× C ^ ¢ Û Ç _/² K Z s $>8 ö b8b$ $ $ \ L9× I ± A b \ L ± A I 9× b ~ ` I C Z * b m' p º #&É >3>>0>3>>0>7 s >/ º#Õ b *f >0 º ) ¶ ¡ ® 7 ¦ 7 I ´ Ñ Ô µ K >&>/>' b c S } A ½ /$× ^ í"@ / í"@ / í"@ / b >&>0>' b U0(ò %® 3 b G /² M ?. 1n*o » c 6 ~ r o g } b*Ë c 8 n v m*ñ*Ë ( 0Á 2 v w ^ m ( 0Á"g b d w c ²0 \ i z > ~ r o f 1n*f b w1n _ x 8 z c Å » \ h%&1/ c t i 8 ¦ ¢& 1 7h c e4 &É Û%,>f>n & 1 a r m kwws zzz phw jr ms dbphqx vkrwrx nrxvklq lqgh kwp ¦ m ( 0Á f @ ²0 ?. Alle neuen Fragen Zeigen sie f(A∩B) ⊆ f(A) ∧ f(B) Nächste » 0 Daumen 819 Aufrufe Wie ist der Beweis zu führen?.

VF· ¥ H v>ÌH H >Ô u>Õ >Ì ² \ ú ã& &t& GwG GF%44E* >Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ì>Ô& ­ w0Y /*ñ9 >Õ>Ì>Ý>å>Ô ¥>Õ. Ich habe Probleme zu erkennen, wie viel man denn beweisen muss, sprich wann der Beweis endet Mein Ansatz wäre der folgende gewesen f(A∩B) ⇔ f(A∧B) ⇔ f(A) ∧ f(B) (Distributivgesetz) ⇔ f(A) ∧ f(B. Analysis I (WS 08/09) Denk/Rheinl¨ander Gruppe 1 (Sylvia Lange) Universit¨at Konstanz FB Mathematik & Statistik Alternative L¨osung zur Aufgabe 2.

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