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I am trying to understand the proof of "Supremum of lower bounds is infimum":

Let A be bounded below, and define B = {b in R : b is a lower bound for A}. Show that sup B = inf A.

In particular, I have shown that $\alpha = \sup B$ exists so far, and want to show $\alpha = \inf A$. However, that means $(\forall a \in A, a \geq \alpha)$ However, I do not understand why that's necessarily true. I know that $\forall a \in A, b \in B, b \leq a$. However, $\alpha$ doesn't necessarily be in $B$, and thus might not satisfy that? Am I mistaken somewhere, or is this by definition of something?

Thank you.

Gareth Ma
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1 Answers1

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HINT: Suppose that there is an $a\in A$ such that $a<\alpha$. Now remember that $\alpha=\sup B$, so there must be a $b\in B$ such that ...?

Brian M. Scott
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  • such that $b<\alpha$? Am I missing something here, we have $a<\alpha,b<\alpha,$ want to find a contradiction... I think I need a break after this haha – Gareth Ma Apr 05 '20 at 14:19
  • WAit no em is this correct - by definition of supremum, if $a<\sup B$ then $a$ is not an upper bound of $B$, which means $\exists b \in B$ such that $b \geq a$, contradicts that $b$ is a lower bound of $A$? – Gareth Ma Apr 05 '20 at 14:21
  • @GarethMa: Now you’ve got it, except that it means that there is a $b\in B$ that is strictly greater than $a$, and you need that strict inequality in order to get the contradiction. – Brian M. Scott Apr 05 '20 at 14:22
  • I see now. Thank you so much, when it "clicked" everything made sense. I should check the definition more often and closely. Thank you! – Gareth Ma Apr 05 '20 at 14:23
  • @GarethMa: You’re welcome! – Brian M. Scott Apr 05 '20 at 14:23
  • Sorry just to confirm, the negation of the statement $\forall a \in A (x \geq a)$ is $\exists a \in A (x < a)$, correct? (Definition of upper bound) – Gareth Ma Apr 05 '20 at 14:28
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    @GarethMa: Yes, that’s right. – Brian M. Scott Apr 05 '20 at 14:36