# Show that if A is bounded above, then it contains its supremum…

..........Question.........
Suppose $A \subseteq \mathbb{R}$ is closed and nonempty. Show that if $A$ is bounded above, then it contains its supremum, and if it is bounded below, then it contains its infimum.

If $sup(A) \notin A$, then it must be in the compliment of $A$, $sup(A) \in \mathbb{R}-A$. A ball around $sup(A)$ would then not be contained in the compliment of $A$. This implies that the compliment of A is not open, and thus A is not closed. RAA.

The same argument holds for $inf(A)$

• I think it is right. Maybe you can explain more why a ball around $\sup(A)$ can not be contained in the compliment of $A$. I think it is not that obvious and you need to apply the definition of supremum. – Brian Ding Aug 19 '15 at 22:48
• As Brian indicates, you need to prove that a ball about $\sup(A)$ is not contained in the complement of $A$. – Paul Sinclair Aug 19 '15 at 22:59

Suppose $A \subset \Bbb{R}$, non-empty, bounded above, and closed. Then, $\sup(A)$ exists.

Suppose in order to derive a contradiction that $\sup(A) \notin A$.

Then, $\sup(A) \in \Bbb{R} \setminus A$.

Since $A$ is closed, $\Bbb{R} \setminus A$ is open.

Thus, $\Bbb{R} \setminus A$ entirely contains $B(\sup(A), \varepsilon)$, the ball of center $\sup(A)$ and of some radius $\varepsilon > 0$.

Then, $B(\sup(A), \varepsilon)$ contains the number $\sup(A) -\frac{\varepsilon}{2}$.

Since $\sup(A) - \frac{\varepsilon}{2} \in B(\sup(A), \varepsilon)$, $\sup(A) - \frac{\varepsilon}{2} \geq x, \forall x \in A$.

But $\sup(A) - \frac{\varepsilon}{2} < \sup(A)$.

This contradicts our assumption that $\sup(A)$ was the least upper bound.

So, we must have $\sup(A) \in A$.

By definition of $\sup A$, for all $\epsilon>0$ you can find $x_{\epsilon}$ in $A$ such that $\sup A-\epsilon<x$. Now let $\epsilon=\frac{1}{n}$ and pick $x_{n} \in A$ such that $\sup A-\frac{1}{n}<x_{n}$. Then we have $\sup A -\frac{1}{n} <x_{n} \leq \sup A$. Take $n \to \infty$ then$x_{n} \to \sup A$ (squeeze) so $\sup A$ is in A.