I am reading classic set theory by D.C. Goldrei. I am stuck on understanding an argument for proving $\mathbb{N}$ is well ordered (every non-empty subset is of a least element). The author assumed that $B$ is a nonempty subset with no least element. Then the following subset of $\mathbb{N}$ is constructed:
\begin{equation}
A = \left\{\boldsymbol{n} \in \mathbb{N}: \boldsymbol{m} \not\in B\ \textrm{for all}\ \boldsymbol{m} \leq \boldsymbol{n}\right\},
\end{equation}
and the author proved that $A = \mathbb{N}$, so that $B$ is empty, which leads to a contradiction. I am not quite sure about how this argument leads to the fact that $B$ is empty, and what is the relation between the assumption that $B$ is nonempty with no least element and the construction of set $A$. I have no idea why $A$ has to be constructed as such. Could anyone provide hints?
Prove that $\mathbb{N}$, the set of natural numbers, is well ordered
elementary-set-theory
Best Answer
Assume $k \in B$. Then $k+1 \notin A$ because there is a natural number ($k$) that is less than $k+1$ that is also in $B$. If $A= \Bbb N$, then $k+1 \notin A$ can't happen, so your initial assumption that $k \in B$ must be false. But $k$ was an arbitrary natural number, so that means no natural numbers can be elements of $B$; i.e., $B$ is empty.