I will award full points to whoever answers both part a and b with an actual pro
ID: 2987115 • Letter: I
Question
I will award full points to whoever answers both part a and b with an actual proof:
Use the reflection principle and analogous results about suprema to prove the following results:
a) [Approximation Property for Infima]: Prove that if a set E is a proper subst of R, real numbers, has a finite infimum and e,epsilon,>0 is any postive number, then there is a point a that exists in set E such that inf E + e>a>=inf E.
***for part a e is epsilon, >= reads greater than or equal.
b) [Completeness Property for Infima] If E is a subset of R, real numbers, is nonempty and bounded below, the E has a (finite) infimum.
Explanation / Answer
(a)
inf E is finite.
Consider inf E + epsilon.
Now this cannot be a lower bound for E since then inf E is defined to be the greatest lower bound, and since we have inf E+epsilon > inf E.
This means there exists some a in S such that a < inf E + epsilon.
Also since inf E is also a lower bound we have inf E <= a.
Thus we have an a in S such that inf E <= s < inf E + epsilon.
(b)
E is nonempty, so suppose e is in E.
Then inf E <= e and thereby inf E < infinity, since e < infinity.
Also E is bounded below, so there exists a lower bound B for set E.
Now inf E , being greatest lower bound >= B and hence > -infinity.
Thus -infinity < inf E < infinity.
Thus unf E is finite.