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Good morning, i'm working in this exercise and i solve this, but, i don't know it's fine, please how you can find the limit point?

1) $\left\{ 1-\frac{1}{n}\::\:n=1,2,3...\right\}$ Well, i say the limit point is 1. for the Archimedean property.

2) $\left\{ \left(1+\frac{1}{n}\right)^{n}\::\:n=1,2,3,...\right\} $ Limit point = 1.

3) $\sqrt[n]{n}\:n=1,2...$ Limit point:1.

Now,how i can prove the limit point is 1?

Edit: Definition of limit point: Let $A\in\mathbb{R}$, and $p\in\mathbb{R}$, exist $r>0$ then $A\cap\left\{ \left(p-r,p+r\right)-\left\{ p\right\} \right\} \neq\phi $

rcoder
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1 Answers1

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For the first one:

Consider the open neighborhood $B(1,\epsilon)$. By the archimedean property, there exists some $n \in \mathbb{N}$ so that $\frac{1}{n}<\epsilon$. Then $B(1, \frac{1}{n}) \subseteq B(1,\epsilon)$. However, $1-\frac{1}{n+1} \in B(1,\frac{1}{n}$), and $1-\frac{1}{n+1} \neq 1$, so the result follows.

However, how do you prove that $1$ is the limit point as worded in the question? Assume there is a limit point greater than $1$ and derive a contradiction. Assume there is $a<1$ that is a limit point show that there is some $N$ so that $1-\frac{1}{N} \leq a<\frac{1}{N-1}$ (or something like this) and find some neighborhood that has trivial intersection with $\{1-\frac{1}{n} \mid n \in \mathbb{N}\}-\{a\}$.

For the second one: $\lim_{n \to \infty} (1+\frac{1}{n})^n=e$. See here

To show you why the limit point is not $1$: We first note that that $a_n=(1+\frac{1}{n})^n$ is strictly increasing. See here

Yet $a_1=(1+1)^1=2$. But then just just pick $\epsilon=1$ the ball $B(1,\epsilon)$ to derive a contradiction.

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Andres Mejia
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