推 tianne :感謝! 09/04 12:09
: sup(-Xk)=-inf(Xk)
if sup(-Xk) is +infinity
for every N, we have some -Xk>N
=> Xk<-N
=> inf(Xk) is -infinity
if sup(-Xk) is finite A
for every positive e
there are infinite many -Xk such that -Xk>A-e
=> Xk<-A+e
=> inf(Xk)=B <= -A
if B=-A-C
then for every positive e
there are infinitely many Xk such that Xk<B+e=-A-C+e
=> -Xk>A+C-e
=> sup(-Xk)>=A+C
=> C=0
=> A=-B
: sup(Xk+Yk)<=sup(Xk)+sup(Yk)
if sup(Xk+Yk)= +infinity
for every N, we have some Xk+Yk>N
then (Xk or Yk) >N/2
=> sup(Xk)=+infinity or sup(Yk)=+infinity
if sup(Xk) or sup(Yk) +infinity we are done
if sup(Xk)=B, sup(Yk)=C are both finite
for every k, Xk<B, Yk<C
=>Xk+Yk<B+C
=>sup(Xk+Yk)<=B+C
: inf(Xn) <= inf(Yn) if Xn<=Yn
if inf(Xn)=+infinity, obvious
if inf(Yn)=A is finite
for every positive e
there are infinitely many Yn such that
Yn<A+e => Xn<A+e
=>inf(Xn)<=A
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※ 編輯: JohnMash 來自: 112.104.91.167 (09/04 03:17)