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Does $1 \cup 2$ has any meaning?

In this answer I used something similar, but it was pointed out $1 \cup 2$ is not same as $\{1,2\}$

Is this only due to convention or is there a deeper reason for it?

Added what about $a \cup b \equiv\{a,b\}$?

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    $\begingroup$ Did you read the tag description before using it? How is that set-theory? There is a tag called elementary-set-theory, FYI. $\endgroup$ Commented Mar 24, 2017 at 13:32

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Where did you get the notion that $a\cup b$ means $\{a,b\}$? What we do have is $$\{a\}\cup\{b\}=\{a,b\}$$ But that is not the same thing at all.

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Just to be the devil's advocate, you can give a meaning to $1\cup 2$ by considering the construction of $\mathbb{N}$ in ZFC. That is, $$0 = \emptyset$$ $$1 = \{\emptyset\}$$ $$2 = \{\{\emptyset\},\emptyset\}$$ Then $$1\cup 2 = 2\ .$$ But unless you specify what you are doing very precisely, I would advise against writing things like $1\cup 2$.

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The union of two numbers is not defined. If you want to express $\{1,2\}$ you can write down $\{1\} \cup \{2\}$ where those are singlets. In fact the union is between sets, not between numbers.

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  • $\begingroup$ added not numbers to question as well $\endgroup$
    – jimjim
    Commented Mar 24, 2017 at 13:00
  • $\begingroup$ It's the same. As I said, union is between sets. If you write down $\{a\}\cup\{b\}$ it has sense, otherwise it hasn't. $\endgroup$ Commented Mar 24, 2017 at 13:03
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Set theoretically it has meaning and it is defined, remember that the reals (in one of its representations) are Dedekind cuts, so joining two numbers has meaning. Also you can think 1 and 2 as ordinals, so by computing the union you have $1\cup2=2$.

And working in ZFC, $a\cup b=\{a,b\}$ violates foundation.

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  • $\begingroup$ You also violate Foundation if $a=\{b\}$ and $b=\{a\}$. $\endgroup$ Commented Mar 24, 2017 at 13:33
  • $\begingroup$ @martin.koeberl You're right, i didn't saw it when i wrote it. $\endgroup$ Commented Mar 25, 2017 at 13:19
  • $\begingroup$ No worries. Correcting it would be good though. $\endgroup$ Commented Mar 25, 2017 at 20:17
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Since $1=\{\emptyset\}$ and $2=\{\emptyset,\{\emptyset\}\}$ by definition, yes.

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  • $\begingroup$ This is only in ZFC, in normal use of mathematics I would say this is false. $\endgroup$
    – Paul
    Commented Mar 24, 2017 at 19:23

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