The Statement of the Problem:

Suppose $E \subset \mathbb R$ is a nonempty set and bounded above, and let $\alpha = \text{sup}(E)$. Prove that there exists a sequence of points $x_n \in E$ such that $x_n \to \alpha$ as $n \to \infty $.

My Approach:

Choose $x_1 \in E $ such that $d(x_1, \alpha) = r >0$. Now, choose $x_2 \in E $ such that $d(x_2, \alpha) = \frac{r}{2}$. Now, choose $x_3 \in E $ such that $d(x_3, \alpha) = \frac{r}{3}$. Continue this process $n $ times by choosing $x_n \in E $ such that $d(x_n, \alpha) = \frac{r}{n}$. We see that $\{x_n \} \to \alpha$ as $n \to \infty$, as desired.

Is this okay?

EDIT: Attempt #2:

Choose $x_1 \in E $ such that $\alpha - 1 \le x_1$. Now, choose $x_2 \in E $ such that $\alpha - \frac{1}{2} \le x_2$. Now, choose $x_3 \in E $ such that $\alpha - \frac{1}{3} \le x_3$. Continue this process $n $ times by choosing $x_n \in E $ such that $\alpha - \frac{1}{n} \le x_n$. We see that $\{x_n \} \to \alpha$ as $n \to \infty$, as desired.

  • 2
    $\begingroup$ This is great! We may not be able to chose $d(x_n,\alpha) = r/n$, however. We can only choose an $x_n$ so that $\alpha - r/n <x_n$. Also rather than just picking an $x_1$ in $E$ and letting $r = d(x_1, \alpha)$, we can pick an $x_1$ so that $\alpha - 1 < x_1$. This gets rid of the $r$ running around. Depending on what your aesthetics are, some people might like this better. $\endgroup$ – BigMathTimes Oct 16 '15 at 20:13
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    $\begingroup$ You don't have the hypothesis that E is connected, so you can't guarantee that E contains points of these particular distances r, r/2, etc. from alpha (see comment from BigMathTimes above). Also, you need to use the fact that alpha is the supremum of E... namely, use connected segments/intervals joining alpha to your sequence points, and argue why these segments must have intermediate points in their intersection with E by using the fact that no upper bound on E is lower than alpha. $\endgroup$ – Sinister Cutlass Oct 16 '15 at 20:15
  • $\begingroup$ Ok. Thanks for the feedback, both of you! So, I'm gathering for your comments that my x's should also be upper bounds of E, that get "lower" with each iteration, making the "lowest one" (as n approaches infinity) alpha? (For some reason, I was picturing the x's as being below alpha.) I'll add an edit to my post, but I'm still trying to incorporate the second half of your comment, Sinister Cutlass; I'm still not too sure I understand it. $\endgroup$ – thisisourconcerndude Oct 16 '15 at 20:39
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    $\begingroup$ Also, if $E=\{\alpha\}$ then you don't have $r>0$ so your method doesn't apply. Of course, you should just eliminate this scenario by noting that you can take all $x_n\equiv \alpha$ (constant sequence) in this case. $\endgroup$ – MPW Oct 16 '15 at 20:43

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