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Limit Point is defined as:

Wolfram MathWorld: A number $x$ such that for all $\epsilon \gt 0$, there exists a member of the set $y$ different from $x$ such that $|y-x| \lt \epsilon$.

Proof Wiki: Some sources define a point $ x \in S$ to be a limit point of $A$ iff every open neighbourhood $U$ of $x$ satisfies: $ A \cap (U \smallsetminus \{x\}) \neq \emptyset $

What I don't understand is what prevents the above definitions from calling interior points (points which lie in the interior of the boundaries?) as limit points?

For instance, George Simmons defines the sequence $\{1, \frac{1}{2}, \frac{1}{3} \cdots \}$ and states that $0$ is the limit point and $0$ is the ONLY limit point.

If I select $\frac{1}{2}$, every neighbourhood of $\frac{1}{2}$ (minus the point $\frac{1}{2}$) has a non-zero intersection with the set $A$. Why not call $\frac{1}{2}$ the limit point?

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    @Topology_Man yes. If you give me a minimum distance $\epsilon$, I can just take $n=\lceil \frac 1 {\epsilon} \rceil+1$, and then $1/n$ is within the boundary. So $0$ is a limit point. Any other one is not. Think about it this way: These fractions can all be listed in order. So for a given fraction $\frac 1 n$, the closest fraction is $\frac 1 {n+1}$. So the minimum distance to any $\frac 1 n$ is $\frac 1 {n(n+1)}$. Pick an interval smaller than that and you're good.2012-06-28

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Interior points are indeed limit points. For example, consider the set $S$ which contains all numbers $x$ such that $1. Then every element of $S$ is a limit point of $S$. (As you noted below, 1 and 2 are also limit points of $S$.)

In your example, $T=\{1, \frac12, \frac13\ldots\}$. Here $\frac12$ is not a limit point of $T$ because we need, for every positive $\epsilon$, there is a point $y$ of $T$ different from $\frac12$ with $|y-\frac12| < \epsilon$. But when $\epsilon < \frac16$, there is no such point $y$.

Similarly, the ProofWiki definition says that $\frac12$ will be a limit point of $T$ if every neighborhood of $\frac12$ intersects $T\setminus\left\{\frac12\right\}$. But as before, a sufficiently small neighborhood of $\frac12$, say $\left(\frac5{12}, \frac7{12}\right)$ as suggested by Mr. Mastragostino, does not intersect $T\setminus\left\{\frac12\right\}$.

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    Thanks for the correction. I should have qualified my comment more carefully.2012-06-28