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For example if I plot number line and add $0 + 12$, then subtract $12 -12$ (if I wrote this in one line it would be: $(0+12)-12$), how it would be implemented in set theory? $A = \{x \in R\ |\ x \in [0,12]\}$ $([0] \cup A]) \setminus A = \emptyset$? Or mathematical operations can't touch first element ($(0,12]$)?

Andrés E. Caicedo
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Даниил Шило
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    What do you mean with symbols $[0]$ and $[0,12]$ ? The sets containing $0$ and $0$ and $12$ respectively ? – Mauro ALLEGRANZA Oct 27 '20 at 11:06
  • In set theory numbers are "implemented" as sets, but the number $12$ is not the set ${ 12 }$. There are more than one way to implement numbers as sets; according to another one: $12 = { 11 }$. – Mauro ALLEGRANZA Oct 27 '20 at 11:19
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    Regarding your proposal, consider this simple example. According to Set union we have that: ${ a } \cup { a } = { a }$. Thus: ${ 12 } \cup { 12 } = { 12 }$ – Mauro ALLEGRANZA Oct 27 '20 at 11:29
  • @MauroALLEGRANZA, I have to edit the question. Every number on number line is a range of numbers in set theory (ex.: $1 = {x \in R\ |\ x \in [0;1]}$). My question is: if i sum x and y, should I think about y, something like $[0;y]$ or $(0;y]$? – Даниил Шило Oct 27 '20 at 11:40
  • Maybe you have to avoid speaking of sets and consider segments: the number $n$ will be implemented on the number line with segment $[0,n]$ (that is the same as the interval. – Mauro ALLEGRANZA Oct 27 '20 at 11:42
  • Because, if I add to 0 two ($[0] \cup {x \in R\ |\ x \in [0;12]}$) and then subtract two ($([0] \cup {x \in R\ |\ x \in [0;12]}) \setminus {x \in R\ |\ x \in [0;12]}$) I'll receive $\emptyset$, not a zero – Даниил Шило Oct 27 '20 at 11:45
  • But if $x=[0,x]$ and $y=[0,y]$ you cannot simply take the union, because if $x \le y$ you will have $x+y=[0,x] \cup [0,y]=[0,y]$. You have to "translate" the segment $[0,y]$ in such a way that the "old" $0$ coincide with $x$ to get $[0,x+y]$. – Mauro ALLEGRANZA Oct 27 '20 at 11:46
  • Nothing changes... $(0,y] \setminus (0,y]= \emptyset$ and not $[0]$. – Mauro ALLEGRANZA Oct 27 '20 at 11:49
  • @MauroALLEGRANZA, A line segment is a non-empty set, so when I add something I need to take range(0,10] or range[0,10]? Because If I try $([0] \cup range(0,10]) \setminus range(0,10]$ it will return 0. So when I count, should I include first num or not? If no, then why we are counting a segments from 0, but not from next minimal num? – Даниил Шило Oct 27 '20 at 12:07

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Define $A=\{1,2,3,4,5,6,7,8,9,10,11,12\}$

$0=|\emptyset|,12=|A|$, where the $|X|$ symbol means cardinality of the set $X$

$(0+12)-12$ means, in set operations, $$(A\cup\emptyset) \setminus A$$ which obviously gives $\emptyset$.

Raffaele
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