$$E = \{x \in \mathbb{Q} : x \ge 0, x^2 < 2\}$$
$sup(E)$ = is not contained in $\mathbb{Q}$ but exists in $\mathbb R$
$inf(E) = 0$
So here for me is clear why the inferior exists, but what about the superior, shouldn't it be $2$?
$$E = (0, 1]$$
sup(E) = 1
inf(E) = does not exist
Why the inferior does not exist?
$$E = \{0\}$$
In that case I think that the superior and inferior limit has to be $0$, right?
On
The supremum in your first example is in fact $\sqrt{2}$, as the inequality $x^2<2$ is equivalent to $x \in (-\sqrt{2}, \sqrt{2}).$
The infimum of your second example is incorrect: $\inf(0, 1]=0.$
As for the third example, your assumption is correct. But the infimum and the supremum of $\{0\}$ are $0$.
1. $E = \{x \in \mathbb Q : x \ge 0, x^2 < 2\}$
Notice the statement is $x^2 < 2$, not $x < 2$. For example, $1.9$ is not an element of $E$, because $1.9^2 = 3.61$, which is not less than $2$.
The elements of $E$ are nonnegative rational numbers $r$ between $0$ and $\sqrt{2}$. You can see this because $x \ge 0, x^2 < 2$ is equivalent to $x \ge 0, x < \sqrt{2}$, when we take the square root of both sides.
2. $E = (0, 1]$
What you wrote is not right, the infimum does exist, and is $0$. However, the infimum does not exist as a positive real number, only as a real number. Maybe that's what you meant.
3. $E = \{0\}$
You are right, the infimum and supremum are both $0$.