Show that x is also a cluster point of E...

**tn11631** · Feb 28th 2010, 06:02 PM

I'm trying to study for an upcoming exam for a class in which the book title is Intro to analysis, so i think it belongs in this category. However, this question is not in the book and I am completely lost. I hope someone could help so I can understand and learn the answer. Thanks!

Let E be a set of real numbers. Suppose that x = lim_{n to infinity} x_n and that each x_n is a cluster point of E. Show that x is also a cluster point of E. (Warning: the numbers x_n may not belong to E.)

**Drexel28** · Feb 28th 2010, 06:08 PM

Originally Posted by tn11631

I'm trying to study for an upcoming exam for a class in which the book title is Intro to analysis, so i think it belongs in this category. However, this question is not in the book and I am completely lost. I hope someone could help so I can understand and learn the answer. Thanks!

Let E be a set of real numbers. Suppose that x = lim_{n to infinity} x_n and that each x_n is a cluster point of E. Show that x is also a cluster point of E. (Warning: the numbers x_n may not belong to E.)

What have you tried? Do you know what a limit point, or any of the other relevant definitions are?

**tn11631** · Feb 28th 2010, 06:15 PM

Originally Posted by Drexel28

What have you tried? Do you know what a limit point, or any of the other relevant definitions are?

Well I thought to show that there exists a sequence and that there is an e(epsilon)>0 such that |x_n-x|<e and let n>=N. But honestly I'm completely lost and confused with this question. I have a tendency to just make things up when I write proofs. So I guess i'll say I haven't tried too much.

**Drexel28** · Feb 28th 2010, 06:25 PM

Originally Posted by tn11631

So I guess i'll say I haven't tried too much.

Haha, true dat.

Ok, so we call $x$ a limit (clutser) point of $E$ if given any arbitrary $\varepsilon>0$ there exists some $e\in B_{\varepsilon}(x)$ such that $e\in E-\{x\}$ .

Ok, so let $B_{\varepsilon}(x)$ be arbitrary.

We have two cases: either $\{x_n\}$ contains infinitely many points or it is eventually $x$ . If it is the latter we may conclude. So, assume not.

Since $x_n\to x$ and $\{x_n\}$ has infinitely many points we know that given any $\varepsilon>0$ we have that $B_{\varepsilon}(x)$ contains infinitely many distinct points of $\{x_n\}$ . In particular, there's at least two, call them $x_{n_1},x_{n_2}$ . Since $x_{n_1}\ne x_{n_2}$ we cannot have that they both equal $x$ . So, let's assume WLOG that $x_{n_1}\ne x$ . Since $B_{\varepsilon}(x)$ is open we know there exists some $\delta>0$ such that $B_{\delta}(x_{n_1})\subseteq B_{\varepsilon}(x)$ . But, since $x_{n_1}$ is a limit point of $E$ it must contain infinitely many distinct points of $E$ . Namely there has to be at least two distinct points of $E$ , call them $e_1,e_2$ . Since they are distinct we know they both can't be equal to $x$ . So, we may assume WLOG that $e_1\ne x$ . But, $e_1\in B_{\delta}(x_{n_1})$ and since $B_{\delta}(x_{n_1})\subseteq B_{\varepsilon}(x)$ it follows that $e_1\in B_{\varepsilon}(x)$ .

Since $\varepsilon>0$ was arbitrary the conclusion follows.

**tn11631** · Feb 28th 2010, 06:38 PM

Oh man, thanks so much. Following it along while reading the def given in the book, its making sense. As you can tell I'm not fond of theory

Everything else I'm good with. However, I'm not sure I'm completely following the lingo (if it is lingo), what is WLOG? Thanks again

**Drexel28** · Feb 28th 2010, 06:42 PM

Originally Posted by tn11631

Oh man, thanks so much. Following it along while reading the def given in the book, its making sense. As you can tell I'm not fond of theory

Everything else I'm good with. However, I'm not sure I'm completely following the lingo (if it is lingo), what is WLOG? Thanks again

Haha, I feel for you man. For the longest time I didn't know what W.R.T., WLOG, or N.B. meant.

W.R.T.- With Respect To

WLOG- Without Loss Of Generality

N.B.- Nota Benne (or something...it means note well in latin I think )

**HallsofIvy** · Mar 1st 2010, 03:19 AM

"well" is "bene" in Latin- only one "n".

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