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

Answer:

a) must be met

Step-by-step explanation:

We have two conditions:

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

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

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