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Theorem Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$.

I have proved the existence of $\sup L = \alpha$.

Question I am not able to prove that $\alpha = \inf B$?

Rudin has provided the following explanation - "If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $L$, hence $\gamma \notin B$. It follows that $\alpha \leq x$ for every $x \in B$. Thus $\alpha \in L$."

Question I am not able to follow this argument? Specifically, shouldn't $\alpha < x$ for every $x \in B$?

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  • $\begingroup$ Not necessarily true that $\alpha <x$, because the set $B$ could possibly contain the infimum of $S$, which is it's own infimum. It is always safer (and correct) to say $\alpha \leq x$. $\endgroup$ – астон вілла олоф мэллбэрг Sep 14 '16 at 8:00
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If $\gamma \lt \alpha$, then $\gamma$ is less than the least upper bound of $L$, hence $\gamma$ is not an upper bound of $L$. But since $L$ is the set of all lower bounds of $B$, we see that every $ x \in B$ is an upper bound of $L$, because every $x \in B$ is greater than any of the lower bounds of B. Since $\gamma$ is not an upper bound of $L$, $\gamma \not\in B$.

We also have stated that every $x \in B$ is an upper bound of $L$. Since $\alpha$ is the least upper bound of $L$, $\alpha \le x$ for all $ x \in B$.

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