Infimum of Subsets in R: True or False?

[Bolz]

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$\cap$B

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$\cap$B = {1,2,3}.

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

 

[pasmith]

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$\cap$B

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$\cap$B = {1,2,3}.

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

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 $A \cap B = \varnothing$?

 

[Bolz]

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?

 

[verty]

Have you heard of Zeno's paradox (the well-known one I mean)?

 

[Bolz]

Have you heard of Zeno's paradox (the well-known one I mean)?

Yes. Why?

 

[HallsofIvy]

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.

 

[Bolz]

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?

 

[STEMucator]

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 $sup(ø) = -∞$.

 

[Bolz]

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.