# Why is (ℕ, $\leq$) considered a well ordered set but (ℕ, $\geq$) is not?

I'm taking a discrete mathematics course right now and I can't quite understand what a well ordered set is. From what I do understand, a set is considered to be well ordered if it's a totally ordered set and all non empty subset has a smallest element. So $$(ℕ, \leq)$$ is considered a well ordered set because it's a totally ordered set and $$0$$ is the smallest element. Isn't this true for the set $$(ℕ, \geq)$$ as well? From what I understand, it's a totally ordered set and $$0$$ would be the smallest element but the notes from the professor says that it's not.

Thanks

(edit : typo)

You're mixing up the terminology. "Smallest" with respect to $$\geq$$ is the "Largest" with respect to $$\leq.$$
In other words, saying $$(\mathbb{N}, \geq)$$ is well-ordered would imply that $$\mathbb{N}$$ has a largest element with respect to the usual order. Does it?
For $$(\mathbb{N},\le )$$ and any set $$A \subseteq \mathbb{N}$$ such that $$A \neq \emptyset$$, the smallest element $$s$$ of $$A$$ satisfies the property that for all $$a \in A$$, $$s \le a$$.
Now, consider $$(\mathbb{N}, \ge )$$ and any set $$B \subseteq \mathbb{N}$$ such that $$B \neq \emptyset$$. The "smallest" element $$s$$ of $$B$$ must satisfy the property that for all $$b \in B$$, $$s \ge b$$. If $$B = \mathbb{N}$$, you already know there is no such element $$s$$.
Note that for a subset $$S \subseteq \mathbb N$$, the "smallest" element with respect to $$\geq$$ would mean an element $$a$$ such that, for all $$s \in S$$ you have $$a \geq s$$.
You can convince yourself that any infinite $$S \subset \mathbb N$$ does not have a smallest element with respect to $$\geq$$.