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I am trying to understand the following theorem:

image of theorem text

I can't understand how the author gets to the conclusion that $\alpha = \sup L$ is $\in L$

I'm ok until the "Our hypothesis about $S$ implies therefore that $L$ has a supremum in $S$, call it $\alpha$" part.

I understand until this part, but how exactly does he conclude that $\alpha$ is in $L$?

AakashM
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Yossi
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2 Answers2

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The author showed that $\alpha$ is a lower bound of $B$, hence by definition in $L$. One way to show that is suppose that $\alpha$ isn't, i.e. take $y\in B$ such that $y<\alpha$, but then since $\alpha=\sup L$ we get $z\in L$ such that $y<z<\alpha$. but $y<z$ is a contradiction.

Jonathan Y.
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  • i am not asking about the last part, my question was about the part that "if y<a then y is not upper bound of L ..... thus a $\in$ L". maybe if you could explain it more intuitively it would be better – Yossi Aug 30 '13 at 10:06
  • And I have referred to that as well: if $y<\alpha$ then--being strictly smaller than the least upper bound of $L$--$y$ isn't an upper bound of $L$; therefore, $y\not\in B$. This shows that for all $y\in B$ we have $\alpha\leq y$ (the negation of $y<\alpha$), hence $\alpha$ is a lower bound for $B$, and by definition $\alpha\in L$. – Jonathan Y. Aug 30 '13 at 10:18
  • y isn't an upper bound of L; therefore, y∉B - could you elaborate how you conclude that. do you refer here that L is the set of all lower bounds of B. and B is the set of some upper bounds of L? – Yossi Aug 30 '13 at 22:26
  • Certainly. $L$ is the set of lower bounds of $B$, hence for all $a\in L$, $b\in B$ we have $a\leq b$. But this implies that while $B$ might not be the set of all upper bounds of $L$, every element of $B$ is an upper bound for $L$. – Jonathan Y. Aug 30 '13 at 22:32
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"If $\gamma<\alpha$ then $\gamma$ is not an upper bound of $L$" - That's because $\alpha$ is the least upper bound of $L$.

"hence $\gamma\notin B$" - as we have seen before that every element of $B$ is an upper bound of $L$

"It follows that $\alpha\le x$ for all $x\in B$" - This is just the contraposition of what we just did: $\gamma<\alpha\rightarrow \gamma\notin B$ becomes $\gamma\in B\to \gamma\ge \alpha$ or after renaming of variables $x\in B\to x\ge \alpha$.

"It follows that $\alpha\in L$" - The previos sentence was just a rephrasing of "$\alpha$ is a lower bound of $B$".

Hagen von Eitzen
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