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Let $A,B\subseteq\mathbb{R}$ two non-empty sets such that $a<b$ for all $a\in A$ and $b\in B$.

Which of the following properties hold:

(1) $\sup A<\infty$

(2) $\sup B<\infty$

(3) $\sup A<\inf B$

(4) $\sup A=\inf B$ iff there exist sequences $(a_n)_{\mathbb{N}}$ in $A$ and $(b_n)_{\mathbb{N}}$ in $B$ such that $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n$

(5) $\sup A<\inf B$ iff there exist sequences $(a_n)_{\mathbb{N}}$ in $A$ and $(b_n)_{\mathbb{N}}$ in $B$ such that $\lim\limits_{n\to\infty}x_n<\lim\limits_{n\to\infty}y_n$

My ideas:

(2) is false: Consider $B=\{x+y: x,y\in \mathbb{R}_{>0}\}$ and $A=\{0\}.$ Then $A$ and $B$ satisfy $a<b$ for all $a\in A$ and $b\in B$, but $\sup B$ does not exist in $\mathbb{R}$.

(3) My attempt: Fix $b\in B$, by assumption it is $a<b$ for all $a\in A$. Thus, $b$ is an upper bound for $A$, therefore $\sup A<b$. Or follows $\sup A\le b$ (I'm not sure at this point)? Since $\sup A$ is a lower bound for $B$, $\sup A<\inf B$...or is $\sup A=\inf B$ as well possible?

(1) is correct: $\sup A\le \inf B$ and $\inf B$ exists in $\mathbb{R}$. (Suppose $\inf B$ doesn't exist in $\mathbb{R}$. $\inf B=-\infty$ only can happen if $B$ is not bounded below which is a contradiction to $B$ is bounded below by a non-empty set $A$.)

(4) Suppose $\sup A=\sup B$. Take $(a_n)_{\mathbb{N}}$ in $A$ converging to $\sup A$ and $(b_n)_{\mathbb{N}}$ in $B$ converging to $\sup B$, for the existence see here Let $S$= Sup $M$. Prove there exists a sequence that converges to $S$. . Then: $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n$. However, I don't know about the other direction.

(5) I don't know..

Is everything what I have tried correct? How does it work for (3) and the remaining directions in (4) and (5)? Thank you

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(1) is true. Take a $b\in B$. Then $\sup A\leqslant b<+\infty$.

(2) is true and what you wrote is correct, but you could have defined $B$ as $(0,+\infty)$.

(3) is false. Just take $A=\{0\}$ and $B=(0,+\infty)$ (again).

(4) is true, but your justification doesn't make sense. If $\sup A<\inf B$, take $\varepsilon=\inf B-\sup A$. Then, if $a\in A$ and $b\in B$, $b-a\geqslant\varepsilon$. Take $n\in\mathbb N$ such that $y_n-x_n<\varepsilon$. But $y_n\in B$ and $x_n\in A$. There's a contradiction here.

(5) is false. Again, take $A=\{0\}$ and $B=(0,+\infty)$. Define $x_n=0$ and $y_n=\frac1n$.

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