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Let $A$ be an infinite set that includes Real numbers and its not bounded. Let $B$ be a set of Real numbers $x$ s.t. the intersection $A\cap[x,\infty)$ is $empty$ or includes $finite$ number of elements.

  • prove or disprove the existence of $\inf B$ if $A$ is not bounded.

Actually this is the third part of the question that I didn't solve. I think there is more than one situation in this case. can someone give me a hint ?

Firas Abd El Gani
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  • What is the difference with http://math.stackexchange.com/questions/1000340/question-about-two-sets? Or http://math.stackexchange.com/questions/1001474/question-about-infimum-conclusion-about-intersection? If you are understanding the given solutions you should be able to get the answer. – mfl Nov 05 '14 at 18:27
  • the answer is wrong in the previous question, and I couldn't delete it. – Firas Abd El Gani Nov 05 '14 at 18:32

2 Answers2

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Hint: Notice that $A$ being bounded means that there exists $M \in \mathbb{R}$ a positive number s.t. $A \subset (-M,M)$, which implies that $A \cap [M,+\infty)= \emptyset$. Try to solve it from there.

Ivo Terek
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Rafael Bordoni
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  • I solved the questions that deal with A being bounded, but now what happens if its not? – Firas Abd El Gani Nov 05 '14 at 18:35
  • being not bounded in this case means that from one side it can be bounded and the other side it can continue to infinity. – Firas Abd El Gani Nov 05 '14 at 18:36
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    Well, you can always think of an example where $A$ is not bounded. Take $A=\mathbb{Z} \cap (-\infty,0]$, for example. You will find that for every real $x$, the set $A \cap [x,+\infty)$ is finite, which means $B=\mathbb{R}$ and then, there is no infimum. That disproves the existence of the infimum. With another counter example, you can disprove the inexistance too, meaning that both cases might happen, it just depends on the $A$ given. – Rafael Bordoni Nov 05 '14 at 18:48
  • is there any example where i can disprove the nonexistence of infimum? ( this is the hard part for me) – Firas Abd El Gani Nov 05 '14 at 19:08
  • Take $A=(-\infty,0]$. $A$ is not bounded and $B=[0,+\infty)$, wich means that $infB = 0$. So in this case, the infimum exists, disproving it's inexistence for all $A$ unbounded. – Rafael Bordoni Nov 05 '14 at 19:11
  • but in this case its not called infB, its minB because the infimum is included in B. Im thinking about the real definition of infimum. – Firas Abd El Gani Nov 05 '14 at 19:29
  • But $\min B$ can be $\inf B$. The definition of $\inf B$ is a real number $x$ s.t. $y \in B \implies x \leq y$ and given a lower quota $m$ for $B$, $m \leq x$. This definition does not require it to be outside $B$. It is also equivalent to $\inf B = x$ is defined as a real number s.t. $y \in B \implies x \leq y$ and $\forall \epsilon >0\ \ \exists y \in B \cap [x, \epsilon)$. This last definition is better for proving other things and solving problems and exercises. Its good to remember that it only always exists in $\mathbb{R}$. In $\mathbb{Q}$, it might not exists for some sets. – Rafael Bordoni Nov 07 '14 at 20:50
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Consider $A=\{-n:n\in \mathbb{N}\}$ and $B=A.$

mfl
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