If $A$ is a non-empty bounded subset of $\mathbb{R}$, and $B$ is the set of all upper bounds of $A$, prove that $$glb(B)= lub(A)$$

My reasoning and thinking:

The set of upper bounds may be infinite countable set, right? Also all the upper bounds (elements of set $B$) are greater than or equal to $x$ for all $x$ in $A$, using definition of upper bound. Then it is $glb(B)$, I am not getting what does it mean? As $y$ is an element of $B$, then is $glb$ of $y$ is $y$ itself?

Any help/hint please.

  • $\begingroup$ Since $A$ has an upper bound $b$, any member of $S= \{c : c\geq b \}$ is an upper bound for $ A ,$ and $S$ is not countable.And it's irrelevant to your question $\endgroup$ – DanielWainfleet Oct 17 '15 at 7:36
  • $\begingroup$ The set of upper bounds of A is infinite and in fact uncountable, and as @user254665 said, that has nothing to do with it. $\endgroup$ – BrianO Oct 17 '15 at 7:45
  • $\begingroup$ Then, what fact will help me out? It was my duty to explain what I know about the terms related to question. I am still waiting for the proper reasoning and answer. $\endgroup$ – Kavita Oct 17 '15 at 7:53
  • $\begingroup$ @user254665. Let me clear one more thing, whether or not we are able to write uncountable set in { } ? $\endgroup$ – Kavita Oct 17 '15 at 8:04
  • $\begingroup$ @ Gudson Chou. Well, the number l is said to be sup of A if it is upper bound of A and no number less than l can be upper bound of A. $\endgroup$ – Kavita Oct 17 '15 at 8:09

First, observe that for any $b \in B$, $lub(A) \le b$, because $b \in B$ is an upper bound of $A$ and $lub(A)$ is the least of them. So $lub(A)$ is a lower bound of B, hence: $$lub(A) \le glb(B) \text{.} $$

Because $lub(A)$ is an upper bound of $A$, $lub(A) \in B$ by definition of $B$. Thus, $$glb(B) \le lub(A)\text{.} $$ Hence $lub(A) = glb(B)$.

  • $\begingroup$ Logical and perfect answer. $\endgroup$ – Kavita Oct 17 '15 at 8:14
  • $\begingroup$ This sort of comment is discouraged [this one, not yours], but: thank you :) $\endgroup$ – BrianO Oct 17 '15 at 8:18
  • 1
    $\begingroup$ @ user5273591 I simplified it - no more proof by contradiction (hope you don't miss it ;/ ) $\endgroup$ – BrianO Oct 17 '15 at 9:01
  • $\begingroup$ Yes, definitely not. Thank you. $\endgroup$ – Kavita Oct 17 '15 at 9:06

(1)For any $S\subset R$, if glb$(S) \in S$ then glb$(S)=\min (S)$.....(2) Let $z=$glb$(B)$. Suppose $z$ is not an upper bound for $A $ . Then $z<x$ for some $x\in A$. But then (by the definition of $B$), no member of $B$ is less than $x $, so $x$ is a lower bound for $B$. So $B$ has a lower bound (namely, $x$) that is greater than the greatest lower bound of $B$ (namely, $z$),which is absurd..... (3.) Therefore the supposition in (2) is false : So it must be true that $z$ is an upper bound for $A$,that is, glb$(B)=z\in B$. So, by (1), $z$ is the least member of $B $ That is, $z$ is the least of the upper bounds of $A$. $$\text { So we have lub}(A)=z=\text {glb}(B).$$

  • $\begingroup$ Not so concise as BrianO....Conciseness is not my best talent. $\endgroup$ – DanielWainfleet Oct 17 '15 at 8:32

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