Proving a certain set is inductive?

Question

Let $m$ be a natural number in a field $F$ and let $$ S_m= \{k:k\in N \mbox{ and } k\leq m \}\cup\{x:x\in F, m<x\} $$

Show that the set $S_m$ is inductive.

Thanks in advance!

Answer 1

Let $J_m= \{k:k\in N ~~~and~~~ k\leq m \} $ and $X_m =\{x:x\in F, m<x\}$ so that $S_m = J_m \cup X_m$

Zero $\le m$ so that $0 \in J_m$ and therefore $ 0 \in S_m$

If $z \in S_m$ then either (1) $z \le m $ or (2) $z \gt m$

If (1) then $z$ must be a natural number in $J_m$ and so $z + 1$ is also a natural number. If $z+1 \le m$, then $z+ 1 \in J_m$ whereas if $z+1 \gt m$ then $z+1 \in X_m$, so $z + 1 \in S_m$.

If (2) then $z + 1 \gt m$ and so $z+1 \in X_m$. (There can be values of $y$ such that $y \notin S_m$ and $y + 1 \in X_m$, but that still allows that if $z\in S_m ($ and $ z \gt m) \implies z+1 \in X_m$)

Putting (1) and (2) together, $z\in S_m \implies z+1 \in S_m$. Taken with $0 \in J_m$ , these are the two conditions for $S_m$ to be an inductive set. then