Let A be a nonempty subset of R and let B = {b : b is an upper bound of A}. Assuming that B is nonempty, show that B has a minimum element.

I am trying to solve like this: " A $\subset$ R , A $\subset$ [$-\infty$, b] --> bounded above b

Is B has a minimum element means least upper bound or supremum of B? If it is,

A: from x $\le$ M --> $M-x$ is non-negative(*not positive) But , if it’s 0, then clearly M itself is a least upper bound for A, other point $M-x >0$ . Use that ;

$B=\left\{i\in\Bbb N:x_0+\frac{i}n\text{ is an upper bound for }A\right\}$

and set $i=\min B : K\in B so B\ne\varnothing$ and the well-ordering principle applies. "

But I'm not sure going right and that is enough for showing. Thanks :)

  • $\begingroup$ By R do you mean $\Bbb R$, the set of real numbers? $\endgroup$ – Brian M. Scott Nov 14 '16 at 18:43
  • $\begingroup$ Yes, sorry about that I mean set of real numbers @BrianM.Scott $\endgroup$ – cematrix Nov 14 '16 at 19:33


Since $B$ is non-empty there exists $b\in\mathbb{R}$ such that $a\le b,\forall a\in A.$ That is, $A$ is bounded from above. Thus we have that $a\le \sup A,\forall a\in A$ (and $\sup A<\infty$). Show that $\sup A\in B$ and $x<\sup A\implies x\not\in B.$

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