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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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  • $\begingroup$ By the way, you tagged this "general-topology", but speak of radii of nhoods. Are you working in metric spaces? $\endgroup$
    – David Mitra
    Aug 1, 2012 at 22:06
  • $\begingroup$ 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 $\endgroup$
    – Tengu
    Aug 1, 2012 at 22:09

2 Answers 2

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

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  • $\begingroup$ Brilliant answer, Ink! $\endgroup$
    – WishingFish
    Jul 28, 2013 at 3:13
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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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  • $\begingroup$ 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... $\endgroup$
    – Tengu
    Aug 1, 2012 at 22:16
  • $\begingroup$ 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. $\endgroup$
    – William
    Aug 1, 2012 at 22:18
  • $\begingroup$ Thank you very much. I think I should study more in topology. $\endgroup$
    – Tengu
    Aug 1, 2012 at 22:23
  • $\begingroup$ @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). $\endgroup$
    – MartianInvader
    Aug 1, 2012 at 22:23
  • $\begingroup$ 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? $\endgroup$
    – Tengu
    Aug 2, 2012 at 23:09

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