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I will try to prove the theorem in the title:

Suppose S is closed, non-empty then if $b = \sup\{x: x \in S\}$ (least upper bound), $b \in S$.

I also use the following Theorem which we proved in class: S is closed iff every Cauchy sequence in S converges to a point in S.

Either $b \in S$ and we are done, or $b \notin S$ and so $\forall \epsilon_n = \frac{1}{n}$ for $n \in \mathbb{N},$ $\exists x_n \in (b - \epsilon_n, b] \cap S$ because if not then $b - \epsilon_n$ would be the least upper bound. So, clearly $x_n \rightarrow b$ as $n \rightarrow \infty.$ Since $x_n$ converges then $x_n$ is Cauchy and in $S$, so it must converge to some point in $S$, lets say $x_o$, by the above theorem. However, limits are unique, so $b = x_o$, but $b \notin S$ and $x_o \in S$. Thus we have arrived at a contradiction and so $b \in S. $

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  • $\begingroup$ It seems a fine proof. $\endgroup$ – AspiringMathematician Apr 20 '17 at 3:03
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    $\begingroup$ You should probably mention that $S$ is a subset of the real numbers. There are other structures in mathematics which support suprema and/or Cauchy sequences. $\endgroup$ – Bernard W Apr 20 '17 at 3:10
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    $\begingroup$ Presumably it is also assumed that $b$ is finite (otherwise what about $S=\mathbb R$?) $\endgroup$ – stewbasic Apr 20 '17 at 3:23
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Your proof looks fine. Here is an alternative (direct) proof. Recall the definition of the supremum is that for every $\epsilon>0$, there exists $x\in S$ with $b-x<\epsilon$. Now define a sequence $\{x_k\}_{k\in\mathbb{N}}$ as follows. Let $\epsilon_k=2^{-k}$ and let $x_k$ be the point in $S$ obtained by applying the definition of the supremum to $\epsilon_k$. Then for every $k$, we have $b-x_k<2^{-k}$.

You can prove directly that $\{x_k\}$ is Cauchy, since $\sum_{k=1}^\infty 2^{-k}$ converges. The sequence also converges to $b$. Thus, $b\in S$!

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  • $\begingroup$ Cauchy is overkill. You know $S$ is closed, so any sequence from $S$, like the $x_n$, converges to a point in $S$. It converges to $x$, it's constructed that way, so $x \in S$. $\endgroup$ – Henno Brandsma Apr 20 '17 at 3:54
  • $\begingroup$ @HennoBrandsma I know it's overkill. I was trying to use the op's theorem: $S$ is closed iff every Cauchy sequence in $S$ converges to a point in $S$. $\endgroup$ – D Wiggles Apr 20 '17 at 3:56
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You don't need Cauchy. You construct the $x_n$ which exist, as otherwise $x_n - \frac{1}{n}$ would be a smaller upperbound than $b$, which cannot be. Now you already know that $x_n \rightarrow b$ at this point. You know that $S$ is closed (as you're trying to use the theorem about Cauchy-sequences) but closedness already implies what you want; $b$ is a limit point (or adherence point) of $S$ by virtue of the $x_n \in S$ that converge to it. So $b \in S$ already follows.

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  • $\begingroup$ I see. I wanted to arrive at contradiction but I guess that isn't necessary. Thank you this makes things simpler $\endgroup$ – student_t Apr 20 '17 at 11:32

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