1. 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?

2. 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.

3. 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?

4. 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...

5. infimum (Urgent)

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...

6. 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."?

7. 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 $\mathbb R^n$