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

[mfb]

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).

 

[Ssnow]

With the same logic 1+⋯+1=11+ \cdots + 1\,=\, 1 because {1}∪⋯∪{1}={1}\{1\} \cup \cdots \cup \{1\} = \{1\}?

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

 

[mfb]

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.

 

[Ssnow]

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

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