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I Proof that 0 + 0 +...+ 0 +... = 0

  1. Nov 26, 2017 #26

    WWGD

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    But you have a result for Real numbers that a Cauchy sequence converges. A Cauchy sequence is one where the difference between terns beyond a fixed one becomes "As small as you want", and it converges to the greatest lower bound.
     
  2. Nov 27, 2017 #27

    PeroK

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    Let ##s_k = \sum_{n=1}^{k} 0##.

    By definition: ##\sum_{n=1}^{\infty} 0 = \lim_{k \rightarrow \infty} s_k##

    (Note that ##\forall k: s_k = 0##)

    Let ##\epsilon > 0##. Take ##N =1##:

    ##k > N \Rightarrow |s_k - 0| = |0 - 0| = 0 < \epsilon##

    Hence ##\lim_{k \rightarrow \infty} s_k = 0##

    QED
     
  3. Nov 27, 2017 #28

    lavinia

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    Any constant sequence in a metric space converges. There is nothing special about a sequence of finite sums of zeros. Also, one does not need the usual metric on the real numbers. Any metric will work.

    The metric is used to interpret an infinite summation as a limit of a sequence. Without the metric one might still ask whether one can make sense of it.
     
    Last edited: Nov 28, 2017
  4. Nov 28, 2017 #29

    Svein

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    Well, the standard way is to go by limits.

    Let [itex] S_{N}=\sum_{n=0}^{N}[/itex]. Then [itex](\forall N>0)(S_{N}=0) [/itex]. Thus [itex]\sum_{n=0}^{\infty}=\lim_{N\rightarrow \infty}S_{N}=\lim_{N\rightarrow \infty}0=0 [/itex].
     
  5. Dec 3, 2017 #30

    Ssnow

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    If we correspond ##0 \mapsto \emptyset## and the binary operation ##+\mapsto \cup## then by the algebra of sets:

    ## \emptyset \cup \cdots \cup \emptyset \,=\, \emptyset ##

    that is

    ## 0+ \cdots + 0\,=\, 0##.
    Ssnow
     
  6. Dec 4, 2017 #31

    mfb

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    With the same logic ##1+ \cdots + 1\,=\, 1## because ##\{1\} \cup \cdots \cup \{1\} = \{1\}##?
    That approach brings more problems than it is supposed to solve, especially as it is trivial to show the limit directly with the definitions (as done in posts 27 and 29).
     
  7. Dec 5, 2017 #32

    Ssnow

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    No beacuse defining the successor function for sets ##S(A)=A\cup\{A\}## you have that ##1=S(\emptyset)=\emptyset \cup \{\emptyset\}=\{\emptyset,\{\emptyset\}\}## so ##1+1 \not = 1## but ##\{\{\emptyset,\{\emptyset\}\},\{\{\emptyset,\{\emptyset\}\}\}\}=2##.
    Ssnow
     
  8. Dec 5, 2017 #33

    mfb

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    That is different from what you suggested before, and it is still wrong. If you identify addition with the union of sets then x+x=x for all x because ##s \cup s=s\quad \forall s##.

    If you identify the addition with ##f(s,t)=\{s,\{t\}\}## then 1+1 works but nothing else works any more.
     
  9. Dec 5, 2017 #34

    Ssnow

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    Yes, I was thinking of "classes" rather than "sets" ...
     
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