Proof Verification: Let $A = [0 , a), \ a \in \Re$ with $a > 0$, then Sup A = a

Question

Let $A = [0 , a),\ a \in \Re$ with $a > 0$. Suppose c = sup A. Clearly a is an upper bound of A, since $a \geq x, \ \forall x \in A$.

First, let us suppose that $c > a$. Then $c > a \geq x, \ \forall x \in A$ which is a contradiction, since there exists an upper bound of A less then the supremum of A. Thus $c \leq a$.

Now suppose that $c < a$. It remains to show that if $c < a$, then there exists an $b \in A$ such that $c < b$. There are two cases to consider.

First, consider the case when $0 < c < a$ . Let $b = \frac{c+a}{2}$ such that that $c < b$ (using that fact that $c < a$). Now consider the case when $c \leq 0$. Let $b = \frac{a}{2}$ such that $c < \frac{a}{2} = b$ once again. Hence $c \geq a$.

Thus $ c \leq a \leq c$ implying that c = a, as required. $ \ \Box$

Answer 1

An idea: since clearly $\;\forall\,x\in[0,a)\;,\;\;x\le a\;,\;\;a\;$ is an upper bound of $\;A\;$ .

Take now any $\;\epsilon>0\;$ ( and in order to avoid trivial cases assume $\;\epsilon<a\;$), and take $\;x_\epsilon=a-\frac\epsilon2\;$ . Then

$$\;a-\epsilon<x_\epsilon=a-\frac\epsilon2\le a$$

and thus $\;a\;$ is the least upper bound of $\;A\;$ ...