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Let $A$ be an infinite set that includes Real numbers and is 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.

  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.

I started searching for examples to see the whole picture, like sequences of the form "$\frac{1}{n}"$ and others, but I didn't know to to do the intersection and why I must be sure that B has an infimum.

its easier if someone can give an example of two sets $A$ and $B$, and make it clear why the intersection must be a finite set. (after choosing "x" from set B)

Firas Abd El Gani
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1 Answers1

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Hint: if you do not impose finiteness of the intersection, then you essentially lose control on $B$, which can be anything you want since the intersection condition is vacuous now.

  1. if $\inf B=-\infty$, then $[x,+\infty)\cap A = A$ for some $x$ since $A$ is bounded...

  2. $A=(0,1)$, $B=(5,6)$...

  3. $A:=\{-1,-2,-3,-4,\cdots\}$, $B=\mathbb{R}$

Milly
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  • In $3$, you defined $A$ as a group of negative integers? and by saying that $B=\mathbb{R}$ it means that B doesn't have infimum at this case? – Firas Abd El Gani Nov 01 '14 at 19:23
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    Yes, $A$ is the negative integers, and $\inf B =-\infty$, which sometime is said "$B$ does not have infimum" (but others say infimum always exists, be it finite or infinite - I think this is not your case since the problems asks to prove existence of inf in step 1). – Milly Nov 01 '14 at 19:25
  • sorry, the down vote was by mistake. I thought ur solution wasn't right! – Firas Abd El Gani Nov 05 '14 at 19:13