Q:

Suppose. a∈R. Show that sup{r∈Q:r<a}=a.

Suppose $a\in\mathbb R$. Show that $\sup\{r\in\mathbb Q\ :\ r<a\}=a$.
My attempt: $S:=\{r\in\mathbb Q, r<a\}$. From this, we can see that $S$ is certainly bounded
from above. And I tried to use a new point $b$ to show that $a$ is the Supremum but got stuck...

A:

Hint: Use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, meaning that for any two real numbers $a$ and $b$ where $a<b$, there is a rational number $r$ such that $a<r<b$.

A:

Let $E = \{ r\in\Bbb Q: r < a\}$.

$\forall r\in E \ \ \ r \le a \implies \sup E \le a $
Consider $\epsilon >0$. As $\Bbb Q$ is a dense set in $\Bbb R$, there is an
$r\in \Bbb Q, \ \ \  a-\epsilon < r < a$. In particular, $r\in E$ and so $\sup E > r-\epsilon$.
This is true for every $\epsilon$, hence $\sup E \ge a$.

A:

Hint: 

Show that $a$ is an upper bound for $S$
Show that if $b$ is an upper bound for $S$, then $a\le b$. 

Extra hint for $2$:

If $b$ is an upper bound for $S$ and $b<a$, then $b \in S$. Can you use the density of $\mathbb Q$ in $\mathbb R$ to find an element of $S$ that is larger than $b$? Can this happen if $b$ is an upper bound?

