# Infimum

• Dec 14th 2008, 09:13 PM
anlys
Infimum
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, 09:28 PM
Mathstud28
Quote:

Originally Posted by anlys
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 $\xi$ to be a greatest lower bound of $E$ two criteria must be met

1. $\xi\leqslant x~~\forall x \in E$
2. If $\xi then $x$ is not a lower bound of $E$

Criterion 1. was given so we must just prove that if $x$ is greater than $\xi$ then it is not a lower bound of $E$

Since $\xi it follows that $d(x,\xi)=D>0$

But by the Archimidean property we may pick a $D_1\in\mathbb R$ such that $0. So now consider that since $\xi$ is a limit point of $E$ we may pick a $x_1\in E$ such that $x_1\in\mathcal{O}_{D_1}(\xi)$ (this is just the neighborhood of radius $D_1$). This implies that $d(x_1,\xi)\leqslant D_1 which in turn implies that $\xi with $x,x_1\in E$ which proves 2.
• Dec 14th 2008, 09:58 PM
anlys
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, 12:49 AM
anlys
Quote:

Originally Posted by anlys
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?

Can someone please check my work? Thanks...
• Dec 15th 2008, 05:27 AM
anlys
infimum (Urgent)
Quote:

Originally Posted by anlys
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?

Can someone please check my work? Thanks...
• Dec 15th 2008, 03:15 PM
Mathstud28
Quote:

Originally Posted by anlys
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?

What do you mean by "So, there exists a delta > 0 such that (a, a + delta) intersect with D = empty set."?
• Dec 16th 2008, 01:56 PM
Mathstud28
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 $\chi$ with $E\subset \chi$.

Assume $\xi$ is a limit point of a set $E$, a lower bound of $E$, and there exists a $x\in E$ such that $\xi is a lower bound of $E$.

So since $\xi it follows that $d_{\chi}(x,\xi)>0$. Now because $x$ is a lower bound of $E$ it follows that there does not exist a $x_1\in E$ such that $\xi but this is a contradiction to $\xi$ being a limit point because then there would be a neighborhood of $\xi$ not containing a point of $E$

EDIT: Someone informed me to the following typo: It should be $\chi=\mathbb R^n$...I use $\chi$ since it is easier to write than $\mathbb R^n$