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Is any interior point also a limit point?

Judging from the definition, I believe every interior point is a limit point, but I'm not sure about it. If this is wrong, could you give me a counterexample?

(Since an interior point $p$ of a set $E$ has a neighborhood $N$ with radius $r$ such that $N$ is a subset of $E$. Obviously any neighborhood of $p$ with radius less than $r$ is a subset of $E$. Also, any neighborhood with radius greater than $r$ contains $N$ as a subset, so (I think) it is a limit point.)

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By the way, you tagged this "general-topology", but speak of radii of nhoods. Are you working in metric spaces? –  David Mitra Aug 1 '12 at 22:06
    
Yes. I just started analysis (Walter Rudin The principle of mathematical analysis) and I came up with this question when reading the chapter named the basic topology –  Tengu Aug 1 '12 at 22:09

2 Answers 2

up vote 6 down vote accepted

A discrete space has no limit points, but every point is an interior point.

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Thank you very much. –  Tengu Aug 1 '12 at 22:23
    
Brilliant answer, Ink! –  WishingFish Jul 28 '13 at 3:13

No an interior point is not a limit point in general. Suppose you have $\mathbb{N}$ with the discrete metric. The point $1$ is a interior point since $\{1\}$ is an open set. However, $1$ is not a limit point of $\mathbb{N}$ because $\{1\}$ is an open neighborhood of $1$ which does not intersect $\mathbb{N}$ in any point beside $1$.

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Is a discrete space a space all of whose points are isolated points? (I could not find it in my textbook) Anyway I don't quite understand why 1 is a interior point of N... –  Tengu Aug 1 '12 at 22:16
    
A discrete topology is a space where every subset is open. Topologically equivalent, it is a space endowed with the discrete metric. Since every subset is open, the set $\{1\}$ is open. –  William Aug 1 '12 at 22:18
    
Thank you very much. I think I should study more in topology. –  Tengu Aug 1 '12 at 22:23
    
@Tengu in the language of Rudin, the set of points in $\mathbb{N}$ whose distance from 1 is less than, say, $\frac{1}{2}$, is exactly $\{1\}$, thus this single-element set is an open ball (and thus open). –  MartianInvader Aug 1 '12 at 22:23
    
Thank you for your reply. Since the neighborhood of 1 with radius 1/2 does not include any point except 1 on N, 1 is an open subset of N? –  Tengu Aug 2 '12 at 23:09

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