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?

$$m(A)=\{z|z \mbox{ is lower bound}\}$$$$M(A)=\{z|z \mbox{ is upper bound}\}$$

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    $\begingroup$ What is $m(A)$? $\endgroup$ – William Chang Mar 9 '14 at 23:55
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    $\begingroup$ Are you trying to prove that A has a least element, or that inf(A) exists? $\endgroup$ – user84413 Mar 10 '14 at 0:11
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    $\begingroup$ Well first of all the result is not true. The set of real numbers is a complete ordered field. $\{\frac 1 n\}$ for natural numbers $n$ is bounded below by $0$. But it has no least element. $\endgroup$ – Ishfaaq Mar 10 '14 at 1:56
  • $\begingroup$ oh, I'm sorry... I edited my post! It is correct? $\endgroup$ – mle Mar 10 '14 at 8:36
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    $\begingroup$ The result still is not true. You need the additional condition that $A$ is non-empty. Or else $\emptyset$ is a subset of $\Bbb R$ and is bounded below. But its infimum does not exist. $\endgroup$ – Ishfaaq Mar 10 '14 at 8:59

Right. Your result is equivalent to the completeness property. This is how you form your contradiction:- We know that $A$ is bounded below so $m(A) \neq \emptyset$. And since $A \neq \emptyset$ any element $a \in A$ is an upper bound for $m(A)$. Hence $\sup m(A) = c$ exists. By definition $c \in M(m(A))$. Suppose there is $a \in A$ such that $a \lt c$. This is a contradiction since $a$ is an upper bound for $m(A)$ and is less than its supremum. This gives us that $c \le a \;\; \forall a \in A$ and hence $c \in m(A)$.

Q. E. D.

There is an easier way to prove the result though. Consider the set $B = \{- a \ | \ a \in A\}$ and apply the supremum property to B.


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