# Supremum of two unequal sets are equal

I am reading the first chapter of Principles of Mathematical Analysis by Walter Rudin. On page 4, Example 1.9 says as follows.

Let $E_1$ be the set of all $r \in \mathbb{Q}$ with $r < 0$. Let $E_2$ be the set of all $r \in \mathbb{Q}$ with $r \le 0$. Then $\sup E_1 = \sup E_2 = 0$.

$E_1$ and $E_2$ are not the same sets but share the same $\sup$. Is it because there is no element in $E_1$ which can be identified as the largest in the set?

If $N_1$ is the set of all $n \in \mathbb{N}$ with $n < 5$ and $N_2$ is the set of all $n \in \mathbb{N}$ with $n \le 5$, then is the following is true?

$$\sup E_1 \ne \sup E_2\\ \sup E_1 = 4 \\ \sup E_2 = 5$$

$E_1$ does not have a maximum, only a supremum. $E_2$ has a maximum, which is thus equal to its supremum.
You are right about the suprema of $N_1$ and $N_2$. In this case, each set has a maximum, which is thus equal to its supremum.
Note that for the sets $E_1$ and $E_2$, every upper bound for one is an upper bound for the other, (i.e. $\{s \in \mathbb{R} \mid s\ \text{is an upper bound for}\ E_1\} = \{s \in \mathbb{R} \mid s\ \text{is an upper bound for}\ E_2\}$). In particular, the least upper bound for $E_1$ is the same as the least upper bound for $E_2$ (i.e. $\sup E_1 = \sup E_2$).
You are correct when it comes to the suprema of $N_1$ and $N_2$.