Question

A and B are bounded non-empty subsets of R. For inf(A) to be less than or equal to inf(B), which of the following conditions must be met?

a) For every b in B and epsilon > 0, there exists a in A, such that a < b + epsilon.

b) There exists a in A, and b in B such that a < b.

If neither of these conditions are appropriate, what would be appropriate conditions for inf(A) to be less than or equal to inf(B)?

Answer 1

a) must be met

Explanation:

We have two conditions:

a) For every
`b\in B` and
`\epsilon>0`, there exists
`a\in A`, such that
`a<b+\epsilon`.

b) There exists
`a\in A` and
`b\in B` such that
`a<b`.

We will prove that conditon a) is equivalent to
`inf(A)\leq inf(B)`

If a) is not satisfied, then it would exist
`b\in B` and
`\epsilon >0` such that, for every
`a\in A`,
`a\geq b+\epsilon`. This implies that
`b+\epsilon` is a lower bound for A and in consequence

`inf(A)\geq b+\epsilon > b\geq inf(B)`

Then,
`inf(A) \leq inf(B)` implies a).

If
`inf(A) \leq inf(B)` is not satisfied then,
`inf(A) > inf(B)` and in consequence exists
`b\inB` such that
`b-inf(A)=\epsilon >0`. Then
`b-\epsilon=inf(A)` and, for every
`a\in A`,

`b-\epsilon =inf(A)\leq a`.

So, a) is not satisfied.

In conclusion, a) is equivalent to
`inf(A)\leq inf(B)`

Finally, observe that condition b) is not an appropiate condition to determine if
`inf(A)\leq inf(B)` or not. For example:

• A={0}, B={1}. b) is satisfied and
  `inf(A)=0<1=inf(B)`
• A={0}. B={-1,1}. b) is satisfied and
  `inf(A)=0>-1=inf(B)`