$Proof$. Let $A=\mathbb{Z}\cup\{a\}$, where $\mathbb{Z}$ is ordered in the usual strict linear order, and $a\notin\mathbb{Z}$. Then we have a partial order on $A$. Now since there is no $x\in A$ such that $x<a$, $a$ is minimal, and since for any $x\in\mathbb{Z}$, $x-1<x$, $a$ is the only minimal. Finally, since $a<x$ is undefined for any $x\in\mathbb{Z}$, $a$ is not the smallest element.
What does $x<a$ even mean? Since we are talking about the partial order on $A$, i think it means that $x$ and $y$ are elements of $A$. Im confused because someone told me that by definition $x<y$ tells that $x,y\in\mathbb{Z}$, but can't the $<$ symbol be used for other relations aswell?
Or is it because for $x\in\mathbb{Z}$, $x<y$ is only valid if $y\in\mathbb{Z}$, e.g you cannot compare an integer with some other type of object.
In general, is $x<y$ only valid if $x$ and $y$ are elements of the same set?
There is no unwritten rule or convention you are missing. The partial order in the proof is just not explained explicitly and clearly.
What’s intended is that, for $x,y\in A$, $x<_A y$ if and only if $x,y\in\mathbb Z$ and $x<_{\mathbb Z} y$, where $<_{\mathbb Z}$ is the standard order on $\mathbb Z$. (I call the partial order we’re defining on $A$ “$<_A$“ just to avoid ambiguity.)
With that definition, you can show $<_A$ is a partial order on $A$, and that $a$ is uniquely minimal, but not least in $(A,<_A)$.