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I hope the title is clear, because I am Italian and I study calculus exclusively from Italian books.

I had to prove this proposition refusing to look the book (because if I read the proof, I'll forget in two hours and I don't learn). I tried this but I think it's wrong:

For absurd M has not minimum, then $\forall m \in M, \ \exists m'\in M : m'\lt m$

First case, A is open, then $\forall a \in A, \ \exists a'\in A : a'\lt a$

(the two conditions are identical, so M would be an open set)

Can I consider an arbitrary decreasing sequence of $m'$? Let's call it $m_n$.

So, I can also consider an increasing sequence of $a_n$. The two sequences approach to the same value $x$;

$x\notin A$ but is greater of all elements of $A$ (it's an upper bound) and $x\notin M$. Absurd.

Second case, $A$ is closed so it has maximum. Maybe the proof can end here? Boh, I assume again that M has not minimum, so choosing a $m\neq \max A$ I consider $m_n$.

$m_n \to \max A$ that, because of the hypothesis, is not inside $M$. Absurd.

tinlyx
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user3621272
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1 Answers1

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The simple way is to use the least upper bound property. As $A$ is bounded above, there is some $x$ that is an upper bound to $A$. Now consider the set of all upper bounds to $A$. This set is bounded below by any element of $A$, so it has a greatest lower bound $m$. By the way we picked it $m \in M$, so it is the minimum of $M$. We don't care if $A$ is open, closed, or neither.

Ross Millikan
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