Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading through Kinsey L. Christine. Topology of Surfaces. The author proves the following theorem.

Therem. If $x$ is a limit point of a set $A\subseteq \mathbb{R^n}$, then there is a sequence of points $\{x_i\}_{i=1}^\infty$ where $x_i\in A$ so that $x$ is a limit point of the sequence $\{x_i\}_{i=1}^\infty$.

Proof. If $x$ is a limit point of the set $A$, then every disc about $x$ contains points of $A$. We wish to choose a sequence of points $x_k$ in $A$ with limit $x$. Consider the family of discs $B(x, 1/k)$ for $k=1,2,3,\dots$, These are discs centered at $x$ with radii $1/2, 1/3, \dots$, and must contain points of $A$. Choose $x_k\in B(x, 1/k)\cap A$ for each $k$. This gives us a sequence of points from $A$. For any disc $B(x, r)$, an integer $N$ may be chosen so that $1/n\lt r$. It follows that the discs are nested inside each other: $B(x,r)\supseteq B(x,\frac{1}{N})\supseteq B(x, \frac{1}{N+1})\supseteq \dots$, and so $$x_N, x_{n+1}, \dots \in B(x, \frac{1}{N})\subseteq B(x,r)$$ Therefore, $x$ is a limit point of

I need a hint for answering the question: show that $x$ is the only limit point of the sequence constructed in the theorem.

The author defines a limit point of a set $A$ to be any point whose neighborhoods (all) $N$ satisfy $N\cap A\neq \varnothing$

A limit point of a sequence is a point whose neighborhoods (all) contain infinitely many points of the sequence.

I think I should prove that if $y$ is another limit point of the sequence than it must be equal to $x$ but I need a clue to get myself started.

share|improve this question

4 Answers 4

Hint: Use the triangle inequality to show $d(x,y)$ is arbitrarily small.

share|improve this answer
1  
and the symmetry axiom for the metric (crucially). –  Ittay Weiss Jul 21 '13 at 23:04

Hint:

If $\,y\,$ is another limit point of the constructed sequence, then

$$\forall\,n\in\Bbb N\;,\;\;y\in B\left(x,\frac1n\right)$$

The question is: what's the distance between $\,x\,$ and $\,y\,$ ...?

share|improve this answer

Use the fact that the disks described form a neighborhood basis at $x$ and the fact that $\mathbb R^n$ is a Hausdorff space.

share|improve this answer

Suppose that $y$ is a limit point of the above sequence, then $d(x_n, y)\lt \frac{1}{n}$ and $d(x_n, x)\lt \frac{1}{n}$. So $d(x,y)\le d(x,x_n)+d(y,x_n)\lt \frac{2}{n}$ (or $0\le n\times d(x,y)\lt 2$) $\forall n$.

Suppose that $d(x,y)\gt 0$. For $n=2$, $2\times d(x,y)<2$ which is a contradiction. So $d(x,y)\le 0\implies d(x,y)=0$

Correct?

share|improve this answer
    
@DonAntonio Correct? ${}{}{}{}$ –  saadtaame Jul 24 '13 at 23:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.