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If $\{n,m\}\subset\mathbb{Z}$ is $n\neq{m}$ also true as a result of that?

I was thinking that this might be the case because I'm pretty sure sets don't allow duplicate values.

If it is the case how can I show $n\in\mathbb{Z},m\in\mathbb{Z}$ succinctly (both can be the same integer)?

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  • $\begingroup$ $n\neq m$ might be an implicit assumption depending on context, but in general, $\{n,m\}$ is a perfectly valid subset of $\mathbb Z$ even if $n=m$. (In that case, it's equal to $\{n, n\}$, which is simply $\{n\}$.) $\endgroup$ – Bungo Nov 8 '17 at 8:31
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$\{ 3,3\} = \{3\} \subset \mathbb{Z}$, we can't conclude that $n \neq m$.

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  • $\begingroup$ So if I say $f(x)=mx+n,\{n,m\}\subset\mathbb{Z}$ then it could also represent functions like $3x+3$, $5x+5$ etc.? $\endgroup$ – theonlygusti Nov 8 '17 at 8:32
  • $\begingroup$ Yes. Notation wise, you might want to write $f_{m,n}$ in stead if you intend to let $f$ to be a function. $\endgroup$ – Siong Thye Goh Nov 8 '17 at 8:33
  • $\begingroup$ @theonlygusti If you don't like $\{n,m\}$ degenerating to a one-element set when $n = m$, you could instead write $(n,m) \in \mathbb Z^2$. $\endgroup$ – Bungo Nov 8 '17 at 8:35
  • $\begingroup$ It is $(n,n)⊂ℤ^2$. This seems like an xy-question. What is your original interest? If it is a lot different from this question, open a new one. $\endgroup$ – P. Siehr Nov 8 '17 at 8:38
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    $\begingroup$ @Bungo You are correct. Just a typo as I write with neo2. $\endgroup$ – P. Siehr Nov 8 '17 at 8:47
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Sets work different than order sets or sequences. In sets se have {1;1}={1} while in ordered sets they are two completely different things. One usually saves this situations by saying "let $\{n,m\} $ be a set with $n\neq m$".

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