I have got the set:
$M_2 = \{l \in \mathbb{N} \mid \forall y(y \in \mathbb{N} \rightarrow y \leq l)\}.$
Does this mean that M2 includes all elements $l \in {N}$ that are bigger than every number in $\mathbb{N}$ because there is for every number a bigger one? So M2 would have an infinite amount of elements whose exact value is undefined.
Or is M2 an empty set because there is no element $l \in \mathbb{N}$ that is bigger than all numbers in $\mathbb{N}$, because $l$ is also only an element of $\mathbb{N}$?
Thank you that you took the time to read my question.
I am really locking forward to your answers.
Kindest regards =)
Your second interpretation is correct, because, as you said, there is no element $l\in \mathbb N$ such that $l$ is larger than all natural numbers. So $M_2= \varnothing$
If the set were defined as follows: $$M_3 = \{l\in \mathbb N\mid \exists y(y\in \mathbb N \land y \leq l\}$$ then, indeed, this set would be all of $\mathbb N$, because for every $l\in \mathbb N$, there is some $y \in \mathbb N$ such that $y\leq l$.