$$E = \{x \in \mathbb{Q} : x \ge 0, x^2 < 2\}$$

$sup(E)$ = 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?

  • $\begingroup$ The first one is the set of non-negative rationals whose square is less than $2$. $2$ is an upper bound, but it's not the least upper bound. For instance, $1.5$ is an upper bound as well, because $1.5^2=2.25$. $\endgroup$ – Arthur Oct 9 '16 at 21:06

1. $E = \{x \in \mathbb Q : x \ge 0, x^2 < 2\}$

what about the superior, shouldn't it be 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 $0$.


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 $0$.


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