Let D be a set of real numbers. Prove that if y is a lower bound of D and y also a limit point of D, then y = inf D.
Appreciate someone's help on this?
For to be a greatest lower bound of two criteria must be met
1.
2. If then is not a lower bound of
Criterion 1. was given so we must just prove that if is greater than then it is not a lower bound of
Since it follows that
But by the Archimidean property we may pick a such that . So now consider that since is a limit point of we may pick a such that (this is just the neighborhood of radius ). This implies that which in turn implies that with which proves 2.
Can you also show the proof by contradiction?
So, supposed that a is not an infimum of D. So, there must exists a larger lower bound of D, let denote it as b.
i.e. a < b <= x for all x in D
So, there exists a delta > 0 such that (a, a + delta) intersect with D = empty set.
Also, a does not belong to the set D because all elements of D are bounded below by b which is greater than a.
Now, let delta = b-a > 0. This implies that for any sequence {X_n} in D\{a}, |X_n - a| >= b-a
So, this means that {X_n} does not converge to a.
So, we get that for any sequence {X_n} in D\{a}, it is not convergent to a. This implies that a is not a limit point. Contradiction!
Is this correct?
Sorry to take so long to get back to you. You can do this proof by contradiction. In the following we are talking about a metric space with .
Assume is a limit point of a set , a lower bound of , and there exists a such that is a lower bound of .
So since it follows that . Now because is a lower bound of it follows that there does not exist a such that but this is a contradiction to being a limit point because then there would be a neighborhood of not containing a point of
EDIT: Someone informed me to the following typo: It should be ...I use since it is easier to write than