Find the supremum and infimum of {x $\in$ [0,1]: x $\notin$ $\mathbb Q$}. Prove why your assertions are correct

Question

Ok I am lost from this question. Does that mean $x$ can only be $0$ or $1$? And it can't be any rational?

Answer 1

Does that mean x can only be 0 or 1? And it can't be any rational?

No, it means $x$ is any irrational between $0$ and $1$.

$\{x\in [0;1]: x\notin \Bbb Q\}$ is: the set of numbers from the real interval $0$ to $1$ inclusive, such that these numbers are also not in the rationals.

Now, find the supremum of this set.   First, what is your definition of a supremum?

A set $E ⊂ R$ is bounded above iff $∃ b ∈ R$ s.t. $a \leq b, ∀ a ∈ E$ ($b$ is an upper bound for $E$)

No, that is not the definition of supremum. That is the definition of upper bound. A supremum is a special kind of upper bound. It's the least such.

A value $s\in R$ is the supremum of $E$ if $\;\nexists b\in R: \forall a\in E: ((a\leq b) \wedge (b< s)\wedge (a\leq s))$

Now, can you find such an upper bound for your set? Can you show that this is the supremum.

Answer 2

I show that $1$ is the supremum of the set $A := \mathbb{R}\setminus \mathbb{Q} \cap [0,1]$, and you may try the infimum case.

The proof relies on the fact that between every two real numbers there is some irrational. (Because $\mathbb{R}$ is the closure of $\mathbb{R}\setminus \mathbb{Q}$.)

By definition, it is equivalent to prove that for every $\varepsilon > 0$ there is some $x \in A$ such that $1 - \varepsilon < x \leq 1$. But this is true, since between any two real numbers there is some irrational.

Answer 3

Remember that $[0,1]$ is an interval on the real line which contains all of the numbers between $0$ and $1$ including $0$ and $1$.

Now, $\{x \in [0,1] \mid x \not \in \Bbb Q \}$ is a subset of the interval $[0,1]$, but it is the subset we get by removing all of the rational points from $[0,1]$.

So, you have to think about all of the irrational numbers in $[0,1]$, and find the supremum and infimum of that set.

We know since $\Bbb R$ is complete that since the set in question is bounded above by $5$, for example, it has a least upper bound which is the supremum. Do you know what the smallest upper bound is? It is $1$, since $1$ is larger than every irrational in $[0,1]$ but it is smaller than every upper bound of the irrationals in $[0,1]$ (why?).

Now you have to find the infimum of the set of irrationals in $[0,1]$ on your own. If you get stuck, let me know.