# Does $\{n,m\}\subset\mathbb{Z}$ imply that $n\neq{m}$?

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)?

• $n\neq m$ 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 $\{n\}$.) – Bungo Nov 8 '17 at 8:31

$\{ 3,3\} = \{3\} \subset \mathbb{Z}$, we can't conclude that $n \neq m$.
• 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.? – theonlygusti Nov 8 '17 at 8:32
• Yes. Notation wise, you might want to write $f_{m,n}$ in stead if you intend to let $f$ to be a function. – Siong Thye Goh Nov 8 '17 at 8:33
• @theonlygusti If you don't like $\{n,m\}$ degenerating to a one-element set when $n = m$, you could instead write $(n,m) \in \mathbb Z^2$. – Bungo Nov 8 '17 at 8:35
• 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. – P. Siehr Nov 8 '17 at 8:38
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 $n\neq m$".