question about two sets.

Question

Let $A$ be an infinite set that includes rational numbers and is bounded. Let $B$ be a set of rational numbers $x$ s.t. the intersection $A\cap[x,\infty)$ is empty or includes finite number of subsets.

1. prove that $\inf B$ exists.

2. prove or disprove $\inf B=\min B$

3. prove or disprove that $\inf B$ exists if we don't ask for $A$ to be bounded.

Answer 1

1. Since $A$ is bounded we may assume $A\subset(-a,a)$ for some $a>0$. If $\inf B$ did not exist (by which I assume you mean $\inf B=-\infty$) then take a rational $x$ in $B$ with $x<-a$. Then $A\cap[x,\infty) = A$ which is infinite, a contradiction.

2. It could happen that $\inf B = 0$ (for example) but $\min B$ does not exist. Take $A = B = \{ \frac1n: n\ge1\}$ (where $n$ runs through the positive integers).

3. Somebody just posted their correct (well, almost:) answer and beat me: I will post my answer anyway, since it seems to me their answer to (3) does not satisfy that $A\cap[x,\infty)$ is finite for each $x\in B$. Take $A = B = \{ -n: n\ge1\}$ (where $n$ runs through the positive integers). That is $A=B=$ the negative integers.

Answer 2

Hints

To $1):$

For any $x\in B$ the set $A\cap [x,\infty)$ is finite. Since $A$ is bounded there exists $\inf A>-\infty.$ If $\inf B$ doesn't exists, consider $x\in B, x\le \inf A.$ What can you get from $A\cap [x,\infty)\subset A\cap [\inf A,\infty)$ being finite?

To $2):$

Consider $A=(\mathbb{Q}\cap (0,1))\cup \{2\}$ and $B=\{q\in\mathbb{Q}: x>0,x^2>2\}.$

To $3):$

Consider $A=B=\mathbb{Z}\cap (-\infty,0].$