# Topology - limit point of a sequence

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.

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

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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 $\,y\,$ ...?

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

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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?

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@DonAntonio Correct? ${}{}{}{}$ – saadtaame Jul 24 '13 at 23:32