I'm now learning about relations of order. This is what I gather:
- First element: Precedes all elements
- Minimal elements: Have no predecessors. The first element is always a minimal.
- Last element: Succeeds all elements
- Maximal elements: Have no successors. The last element is always a maximal.
I have three exercises. I did two of them. I didn't quite get the last one. Can you help check if my answers are right, and advice me regarding the last one?
Over $\mathbb{N}$, $aRb \iff a \le b$
- The first element would be $1$ (or $0$), because it $\le$ all elements in $\mathbb{N}$.
- The minimal element is $1$ (or $0$), because it has no predecessors.
- The last element is… well, I guess there is no last element given that $\mathbb{N}$ is infinite.
- The same goes for maximal element, I think.
Over $\mathbb{Z}\times\mathbb{Z}$, $(m,n)R(r,s) \iff m \le r \land n \le s$
We can visualize this as a graph, right? The relation occurs only if the point $(m,n)$ is neither to the right or above of the point $(r,s)$.
- The first element should be the point to the very bottom-left of the graph. But since ZxZ expands infinitely in that direction, I guess there is no such element?
- The minimal element would have the same problem...
- The last and maximal elements... well, the same thing...
Hm. This is fishy. Did I do something wrong?
Over $E \not = \emptyset$, $ARB \iff A \subseteq B$
And here I am not entirely sure about what to do.