I must proof the following:
- Prop.: let be $\Bbb{R}$ a complete ordered field $$\emptyset \neq A \subset \Bbb{R} \wedge A \mbox{ is bounded below } \to \exists x (x \doteq \inf(A))$$
- Proof: by contradiction I have $ A \mbox{ has not least element }$ therefore $$\nexists x (x \doteq \inf(A)) \equiv$$$$\equiv \nexists x(x=\max(m(A)))\equiv$$$$\equiv \nexists x(x \in m(A) \wedge x \in M(m(A))\equiv$$$$\equiv \forall x(x \notin m(A) \vee x \notin M(m(A)))$$ but if $x \notin m(A)$ I have an absurd because by hypothesis $A$ is bounded below ($m(A) \neq \emptyset$), if $x \notin M(m(A))$ I have an absurd because $A \neq \emptyset$ and $\forall r \in A (r \in M(m(A)))$ therefore $M(m(A))\neq \emptyset$.
It is correct?
$$m(A)=\{z|z \mbox{ is lower bound}\}$$$$M(A)=\{z|z \mbox{ is upper bound}\}$$