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

Question

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)

Answer 1

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?

Answer 2

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 $s$.

Answer 3

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 $\geq$.