# Test Review 2 - a property of the infimum

1. Oct 15, 2005

### cmurphy

Another question: Prove that if a set E in R has a finite infimum and e > 0 is any positive number, then there is a point a in E such that inf E <= a < inf E + e.

The first part, inf E <= a, is obvious from the definition of infimum.

I am having trouble showing that a < inf E + e, even though it seems obvious. My thought is to break it into two cases.

Case 1: inf E = a. Then, we know that since e is strictly greater than 0, that inf E < inf E + e. Thus a < inf E + e.

Case 2: inf E < a. Then what can I use to help me show that a < inf E + e?

Colleen

2. Oct 15, 2005

### HallsofIvy

Staff Emeritus
Suppose the statement were NOT true. Since a is an infimum, it is a lower bound so there are no members of the set less than a. If there were no member of the set between a and a+ epsilon, there would be no members of the set less than a+ epsilon. What does that make a+ epsilon? How does that contradict the hypotheses?

3. Oct 15, 2005

### cmurphy

HallsofIvy, what are you supposing is not true? Do you follow my proof and then suppose that Case 2 is not true?

Or are you supposing the original statement is not true?

I understand where you are going ... that your assumption would show that
a + epsilon would then become the infimum (but there is only one infimum for the set), but I'm not sure how you said that a was the infimum.

We have actually already shown that inf E <= a, so how do we know that a is exactly the infimum for the set?

4. Oct 15, 2005

### 1800bigk

whats wrong with just saying let m = inf E then since e>0 then m+e is not a lower bound so there is some a in E such that a<m+e so m<=a<m+e