Showing the GLB of the union is the same as the min of the GLBs

[k3k3]

Homework Statement

Let A and B be non-empty bounded sets of real numbers.
Show the infimum of A union B is equal to the min{infA,infB}

Homework Equations

If a set is bounded below, a set called S for example, there exists a number N such that x≥N for all x in S

And if S is bounded below, then there exists a number α such that α is a lower bound of S and there is no number greater than it that is a lower bound.

The Attempt at a Solution

Let A and B be non-empty bounded sets of real numbers.

Show that the inf(A∪B) = min{infA,infB}

Since A and B are subets of A∪B, then the set, A or B, that has the smallest greatest upper bound is clearly the one that is the the result of the minimum function.

Since A union B is the entirety of both sets, the greatest upper bound will be the one from the set that contains the smallest value.

I am thinking that I need to break it up into cases when infA ≤ infB and infA > infB and some how link that to the inf(A union B).

If infA ≤ infB
Then A has the smallest greatest upper bound. When you find the infimum of A union B, it will have to be A because A is a subset of A union B.

If infA = infB then the set A=B, then A union B = A = B

If infA > infB, then use the same argument but replace A with B and vice versa.


Showing this properly is the issue I am running into. Any tips on which direction I should go in proving this?

 

[Deveno]

Homework Statement

Let A and B be non-empty bounded sets of real numbers.
Show the infimum of A union B is equal to the min{infA,infB}

Homework Equations

If a set is bounded below, a set called S for example, there exists a number N such that x≥N for all x in S

And if S is bounded below, then there exists a number α such that α is a lower bound of S and there is no number greater than it that is a lower bound.

The Attempt at a Solution

Let A and B be non-empty bounded sets of real numbers.

Show that the inf(A∪B) = min{infA,infB}

Since A and B are subets of A∪B, then the set, A or B, that has the smallest greatest upper bound is clearly the one that is the the result of the minimum function.

Since A union B is the entirety of both sets, the greatest upper bound will be the one from the set that contains the smallest value.

I am thinking that I need to break it up into cases when infA ≤ infB and infA > infB and some how link that to the inf(A union B).

If infA ≤ infB
Then A has the smallest greatest upper bound. When you find the infimum of A union B, it will have to be A because A is a subset of A union B.

If infA = infB then the set A=B, then A union B = A = B

If infA > infB, then use the same argument but replace A with B and vice versa.


Showing this properly is the issue I am running into. Any tips on which direction I should go in proving this?

let d = min{inf(A),inf(B)}.

step 1 should involve showing that d is "a" lower bound for AUB. so given x in AUB, you need to show that d ≤ x. i would "use cases" like so:

1) if inf(A) < inf(B), then for any b in B, inf(A) < b, and for any a in A, inf(A) ≤ a (since inf(A) is a lower bound for A). now if x is in AUB, either:

a) x is in B
b) x is not in B, so x is in A. draw appropriate conclusions accordingly.

2) otherwise, inf(B) ≤ inf(A) and...this should be similar to (1) above.

step 2 should involve you choosing some y such that d < y, and y is a lower bound for AUB. derive a contradiction (hint: show y cannot be less than both inf(A) and inf(B)).

 

[k3k3]

Sorry for the late response. Thank you for the help! I think I got it now.