Infimum and supremum of a set between 0 and 1

Question

I am a little confused on what the infimum and supremum would be for the set S of all rational number between (0,1) not including 0 and 1.

If 0 and 1 were included the answer is quite obvious.

Answer 1

I am guessing that your confusion is about the "prerequisites" for the definition of infimum and supremum, by which I mean the "given objects" in that definition. The concepts of infimum and supremum require to be given an ordered set, a subset, and an element:

Given an ordered set $X$, a subset $A \subset X$, and an element $x \in X$, we say that $x$ is the supremum of $A$ in $X$ if (insert exactly what you expect here).

If you had asked more specifically

What are the infimum and supremum of $\mathbb{Q} \cap (0,1)$ in $(0,1)$?

then the answer would be

They do not exist.

The other responses give answers to the question

What are the infimum and supremum of $\mathbb{Q} \cap (0,1)$ in $\mathbb{R}$?

Answer 2

The supremum are still $0$ and $1$, consider the sequences $$a_n:=\frac{1}{n+1}\quad \text{and} \quad b_n := 1-\frac{1}{n+1},$$ then $a_n,b_n \in S$ for every $n \in \mathbb{N}$ and $$\lim_{n \to \infty} a_n= 0\quad \text{and} \quad \lim_{n \to \infty} b_n= 1.$$

Answer 3

Infimum and supremum don't have to belong to the set. $$\text{inf}\{t:\ 0<t<1 \ , \ t\in \mathbb{Q}\}=0$$ $$\text{sup}\{t:\ 0<t<1 \ , \ t\in \mathbb{Q}\}=1$$ Because $\mathbb{Q}$ is dense you can't pinpoint the first or the last element in the set $\{t:\ 0<t<1 \ , \ t\in \mathbb{Q}\}$.