Infimum of Subsets in R: True or False?

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The statement that if a is the infimum of subsets A and B in R, then a is also the infimum of A∩B is false. A counterexample is provided with A as the set of positive rational numbers and B as the set of positive irrational numbers, both having an infimum of 0, while their intersection is empty. The discussion highlights that the infimum does not need to belong to the subsets, and thus the infimum of A∩B can be ∞. Participants also discuss the implications of empty intersections and the nature of infimums. Overall, the conversation emphasizes the importance of understanding infimums in the context of set intersections.
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Homework Statement



If a is both the infimum of A\subseteq \mathbb{R} and of B\subseteq \mathbb{R} then a is also the infimum of A\capB

Is this statement true or false? If true, prove it. If false, give a counterexample.

Homework Equations





The Attempt at a Solution



I think it's true because let's say A={1,2,3,4} and B={1,2,3} then A\capB = {1,2,3}.

Then inf {A}= 1 and inf {B} = 1.
And inf {A\capB} = 1.

However, I think it's false because, and correct me if I'm wrong, the infimum doesn't necessarily have to belong to the subsets A nor B to be an infimum. The infimum can also be a value outside of those sets. Which would imply that the infimum of A and B doesn't have to be equal to the infimum of A\capB.
 
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Bolz said:

Homework Statement



If a is both the infimum of A\subseteq \mathbb{R} and of B\subseteq \mathbb{R} then a is also the infimum of A\capB

Is this statement true or false? If true, prove it. If false, give a counterexample.

Homework Equations





The Attempt at a Solution



I think it's true because let's say A={1,2,3,4} and B={1,2,3} then A\capB = {1,2,3}.

Then inf {A}= 1 and inf {B} = 1.
And inf {A\capB} = 1.

However, I think it's false because, and correct me if I'm wrong, the infimum doesn't necessarily have to belong to the subsets A nor B to be an infimum. The infimum can also be a value outside of those sets. Which would imply that the infimum of A and B doesn't have to be equal to the infimum of A\capB.

What happens if A \cap B is empty? Nothing in the problem statement says that they have to intersect, so long as they have the same infimum which, as you point out, does not have to be a member of either A or B.

Is it possible to have two subsets A and B with \inf A = \inf B and A \cap B = \varnothing?
 
Hm, I don't think that last part is possible. Both sets have something in common, i.e. the infimum, which would imply A \cap B is not empty. Is my reasoning correct?
 
Have you heard of Zeno's paradox (the well-known one I mean)?
 
verty said:
Have you heard of Zeno's paradox (the well-known one I mean)?

Yes. Why?
 
Bolz said:
Hm, I don't think that last part is possible. Both sets have something in common, i.e. the infimum, which would imply A \cap B is not empty. Is my reasoning correct?
No it is not. Let A be the set of all positive rational numbers. It's infimum is 0. Let B be the set of all positive irrational numbers. Its infimum is also 0. But their intersection is empty.
 
HallsofIvy said:
No it is not. Let A be the set of all positive rational numbers. It's infimum is 0. Let B be the set of all positive irrational numbers. Its infimum is also 0. But their intersection is empty.

So this would fit as a counterexample because you've found the exact same infimum for set A and set B, i.e. 0, and this infimum does not equate to the infimum of their empty intersection?
 
Bolz said:
So this would fit as a counterexample because you've found the exact same infimum for set A and set B, i.e. 0, and this infimum does not equate to the infimum of their empty intersection?

Indeed, ##inf(ø) = ∞## and ##sup(ø) = -∞##.
 
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Zondrina said:
Indeed, ##inf(ø) = ∞## and ##sup(ø) = -∞##.

Thanks! Unrelated question : Any advice to someone learning this on his own? I love physics and I know I have to grind through the mathematical details because they matter too but sometimes I get a bit frustrated if I don't immediately get the answer correct.
 

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