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?

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- Dec 14th 2008, 10:13 PManlysInfimum
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? - Dec 14th 2008, 10:28 PMMathstud28
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. - Dec 14th 2008, 10:58 PManlys
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? - Dec 15th 2008, 01:49 AManlys
- Dec 15th 2008, 06:27 AManlysinfimum (Urgent)
- Dec 15th 2008, 04:15 PMMathstud28
- Dec 16th 2008, 02:56 PMMathstud28
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