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Suppose that $S$ is a nonempty subset of $\mathbb{R}$ and $S$ is bounded above. Also suppose that $T=\{x\in\mathbb{R}:x$ is an upper bound of $S\}$ Show that $\sup(S)=\inf(T)$.

My proof: Since $S$ is nonempty bounded above subset of $\mathbb{R}$, $\sup(S)$ exists. Let $\sup(S)=\alpha$. It is clear that $\alpha\in T$ because $\alpha$ is an upper bound of $S$. Also, $\alpha\leq x$ for all $x\in T$ because $\alpha$ is a supremum. Since $\alpha\leq x$ and $\alpha\in T$, we can conclude that $\inf(T)=\alpha=\sup(S)$ as required.

Is this proof ok?

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    $\begingroup$ Your proof is fine. $\endgroup$ – Kabo Murphy Mar 9 '18 at 9:17
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    $\begingroup$ The ideas are all there, it seems to me. Two points you might want to go into more depth (since this is apparently an introductory course). Clarify how $\alpha \leq x$ because "$\alpha$ is a supremum". Also clarify how $\alpha \leq x$ and $\alpha \in T$ implies that $\alpha = \inf T$. $\endgroup$ – user357980 Mar 9 '18 at 9:18
  • $\begingroup$ You should use \mathbb{R} to get the correct symbol for the real numbers $\endgroup$ – Tony Mar 9 '18 at 9:28
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Your work is fine, rephrasing a bit:

$S \subset \mathbb{R}$, $S$ nonempty, is bounded above : $\sup(S)$ exists.

$T \subset \mathbb{R}$, $T$ nonempty (why?), is bounded below by $s \in S$: $\inf(T)$ exists.

$\sup(S)$ is the least upper bound of $S$, i.e. an upper bound, hence $\sup(S) \in T.$

Assume there is an $x \in T$ with $x \lt \sup(S)$.

Since $x$ is an upper bound of $S$ : contradiction, $\sup(S)$ is the least upper bound of $S$.

Hence for all $x \in T$: $x \ge \sup(S)$,

and since $\sup(S) \in T$, we have

$\inf(T)=\sup(S).$

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