How to prove $a \lt b$ for all $a \in A, b \in B$ implies $\sup A \leq \inf B$?

If $A, B$ are two non empty sets of real numbers and for every $a$ from $A$ and $b$ from $B$

$a < b$,

how can I prove that

$\sup A \leq \inf B$?

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This is not true; you can only say that $supA \le infB$. – Gadi A Sep 13 '10 at 18:21
This question is part of a series that sound like homework problems. – whuber Sep 13 '10 at 18:22
Looking at your history this is a homework problem! – alext87 Sep 13 '10 at 18:24
-1. Do your own homework, or ask conceptual questions. – Qiaochu Yuan Sep 13 '10 at 20:57

I don't think this is true. Let $A=(0,1)$ and $B=(1,2)$ then $supA=1$ and $infB=1$ but $A$ and $B$ satisfy your conditions.

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The conditions are violated because 1 is in A and is not strictly less than all elements of B. – whuber Sep 13 '10 at 18:48
@whuber 1 is in A=(0,1)? Shurely shome mishtake? – Bob Durrant Sep 13 '10 at 19:27
@Bob: Thank you: I was misreading these as doubleton sets, not intervals. Alext87's answer is a good one. – whuber Sep 13 '10 at 20:37
@alext87, this question is absolutely correct. In A=(0,1) and B=(1,2), supA=infB. The question wants us to prove that supA<=infB. (It includes '=' case). – Nipun Malhotra Sep 11 '11 at 13:47
@Nipun: The question was edited after alext87's answer was given (and presumably accepted), as shows the edit history of the question. – Asaf Karagila Sep 11 '11 at 14:54

With the edited problem: if $a\lt b$ for all $a\in A$ and $b\in B$, why is it that $\sup(A)\leq\inf(B)$? Is every element of $A$ a lower bound for $B$? If so, what does that tell you about the elements of $A$ relative to $\inf(B)$? And given that, what does that tell you about $\sup(A)$?

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Here's the solution:

Let us suppose that $\sup(A) > \inf(B)$. So, there exists some $x∈A$, such that $\inf(B) < x$. Again, as $\inf(B)< x$, there exists $y∈B$, such that $y<x$. which is a contradiction as we have been given that $A<B$ (i.e, $x<y$ for all $x∈A$, $y∈B$). So, our assuption that $\sup(A) > \inf(B)$ is WRONG. So, $\sup(A) \leq \inf(B)$. Hence Proved...... :)

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I have converted the first part of your answer into a comment to alext87 (look above). Because you do not have 50 reputation points yet, you can only comment on your own questions and answers. So, you didn't do anything wrong; the "add comment" button will only appear for you once you gain 50 points. Here is an explanation of reputation points. – Zev Chonoles Sep 11 '11 at 15:15

Consider $b\in B$, by hypothesis is a upper bound of $A$, so by definition of sup (the least upper bound), $\sup A\leq b,$ $\forall b\in B$. Now the $\sup A$ is a lower bound of $B$, so by defnition of inf(the grater lower bound), $\sup A\leq \inf B$ .

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