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Assuming $\inf \mathbb{R} = - \infty$ and $\inf \emptyset = + \infty$.

Let $A_0 \supseteq A_1 \supseteq \cdots$ and $B_0 \supseteq B_1 \supseteq \cdots$ be decreasing sequences of sets of real numbers, such that $\inf A_n \geq \inf B_n$ for all $n \in \mathbb{N}$. Does the following hold?

$$\inf \bigcap\limits_{n \in \mathbb{N}} A_n \geq \inf \bigcap\limits_{n \in \mathbb{N}} B_n$$

Similarly, let $A_0 \subseteq A_1 \subseteq \cdots$ and $B_0 \subseteq B_1 \subseteq \cdots$ be increasing sequences of sets of real numbers, such that $\inf A_n \geq \inf B_n$ for all $n \in \mathbb{N}$. Does the following hold?

$$\inf \bigcup\limits_{n \in \mathbb{N}} A_n \geq \inf \bigcup\limits_{n \in \mathbb{N}} B_n$$

If not, does it hold if $A_n$ and $B_n$ have certain properties?

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NO $\quad$ for the first statement:

Consider $A_n= [1-\frac 1n,1]$ and $B_n= [-\frac 1n,0) \cup \{2\}$. We have $1-\frac 1n>-\frac 1n$ but $1<2$.

YES $\quad$ for the second statement, even without the increasing chain conditions:

Assume $\inf\bigcup A_n<\inf\bigcup B_n$. Then there exists $c\in\Bbb R$ with $\inf\bigcup A_n<c<\inf\bigcup B_n$, hence for some $n\in\Bbb N$ there exists $a\in A_n$ with $a<c$, whereas for all $b\in B_n$ we have $c<b$, thus $\inf A_n\le a<c\le \inf B_n$.

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  • $\begingroup$ For the first statement, a more extreme example could be $A_n = (-1/n, 0]$, $B_n = (-1/n, 0)$, so the individual infima are equal, but $\bigcap B_n = \emptyset$, hence has infimum $+\infty$. $\endgroup$ – B. Mehta Apr 17 '18 at 14:21
  • $\begingroup$ @B.Mehta Yes, but I picked my example to specifically avoid infinities and also to have $\lim_{n\to\infty} \inf A_n>\lim_{n\to\infty}\inf B_n$ $\endgroup$ – Hagen von Eitzen Apr 17 '18 at 14:38
  • $\begingroup$ Sure, there's absolutely nothing wrong with your example - I just personally thought this one was easier to visualise, and so might help other users. $\endgroup$ – B. Mehta Apr 17 '18 at 14:39

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