Let $S$ be an ordered set with the $L.U.B$ Property, $S \supset B \neq \varnothing$, $B$ is bounded below. Write $L = \{ l : l \; \text{is a lower bound of } \; B \} $. Then, it follows that $\alpha = \sup L $ exists in $S$ and $\alpha = \inf B $.

My Try:

(1) $L \neq \varnothing$: We are given that $B$ is bounded below which implies $L \neq \varnothing$.

(2) $L$ is bounded above: Let $l \in L$ be arbitrary, hence $ l \leq b $ for all $b$. In particular, every $b$ is an upper bound for $l$. It follows that $L$ is bounded above

(3) $L$ has a L.U.B: follows from $S$ possess the Least Upper Bound property, $\alpha = \sup L $ exists.

(4) $\alpha$ is a lower bound for $B$: If there is $y \in B$ with $y < \alpha$, then $y$ is not an upper bound of $L$, a clear contradiction since we have established above that every element of $B$ is an upper bound of $L$. Hence, $\alpha \leq x$ for all $x \in B$.

(5) This implies $\alpha \in L$

(6) $\alpha$ is the G.L.B (infimum) of $B$. Suppose $\beta$ is another arbitrary lower bound of $B$ with $\alpha < \beta$. I need to show that $\beta$ is not a lower bound of $B$: $\beta \notin L$. This is true since any number greater than the supremum of $L$ is not in $L$ and we have $\alpha < \beta$.

  • $\begingroup$ Looks OK to me. I took the liberty of setting out the steps a little more clearly. $\endgroup$ – Tom Collinge Nov 27 '14 at 10:45
  • $\begingroup$ Thank you very much. Just one question to see if I have understood the concept. This problem in fact shows that $S$ also has the greatest lower bound property. Correct? $\endgroup$ – user195835 Nov 27 '14 at 13:42
  • $\begingroup$ Yes, the two are equivalent. See math.stackexchange.com/questions/470895/… $\endgroup$ – Tom Collinge Nov 27 '14 at 13:57

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