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, 09:13 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, 09:28 PMMathstud28
For $\displaystyle \xi$ to be a greatest lower bound of $\displaystyle E$ two criteria must be met

1. $\displaystyle \xi\leqslant x~~\forall x \in E$

2. If $\displaystyle \xi<x$ then $\displaystyle x$ is not a lower bound of $\displaystyle E$

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

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

But by the Archimidean property we may pick a $\displaystyle D_1\in\mathbb R$ such that $\displaystyle 0<D_1<D$. So now consider that since $\displaystyle \xi$ is a limit point of $\displaystyle E$ we may pick a $\displaystyle x_1\in E$ such that $\displaystyle x_1\in\mathcal{O}_{D_1}(\xi)$ (this is just the neighborhood of radius $\displaystyle D_1$). This implies that $\displaystyle d(x_1,\xi)\leqslant D_1<D$ which in turn implies that $\displaystyle \xi<x_1<x$ with $\displaystyle x,x_1\in E$ which proves 2. - Dec 14th 2008, 09:58 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, 12:49 AManlys
- Dec 15th 2008, 05:27 AManlysinfimum (Urgent)
- Dec 15th 2008, 03:15 PMMathstud28
- Dec 16th 2008, 01:56 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 $\displaystyle \chi$ with $\displaystyle E\subset \chi$.

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

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

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