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Limit Point is defined as:

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

Proof Wiki : Some sources define a point $x\in S$ to be a limit point of $A$ if every neighborhood $U$ 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, 1/2, 1/3 ...} and states that 0 is the limit point and 0 is the ONLY limit point.

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

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The sequence he's defining is all fractions of the form $\frac 1 n$. Every neighbourhood of $\frac 1 2$ does not have such a non-empty intersection with this set; take $(\frac 5 {12}, \frac 7 {12})$ for example. –  Robert Mastragostino Jun 28 '12 at 13:22
    
@RobertMastragostino. Interesting, so, the only point which will have a neighbourhood which intersects in a non-zero fashion is 0? Seems hard to believe for some reason..... –  Topology_Man Jun 28 '12 at 13:29
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"Hard to believe" is why mathematicians have proofs, which are nothing more than careful, persuasive arguments as to why something is true. You might try to prove this one yourself; it is not too hard. I suggest considering three cases: a point $x < 0$, a point $x > 1$, or a point $x$ with $0<x<1$. –  MJD Jun 28 '12 at 13:40
    
@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. –  Robert Mastragostino Jun 28 '12 at 14:53
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up vote 1 down vote accepted

Interior points are indeed limit points. For example, consider the set $S$ which contains all numbers $x$ such that $1<x<2$. 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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In the first case, 1<x<2; If all points are limit points, what are 1 and 2? Aren't they limit points too? How do you differentiate them from the interior points (such as 1.24) –  Topology_Man Jun 28 '12 at 13:33
    
In that example, 1 and 2 are also limit points. Topology has several related notions here: a point p is in the interior of a set $S$ if there is a neighborhood of $p$ that is contained in $S$. A limit point of $S$ that is not in the interior is called a boundary point of $S$, and the set of boundary points of $S$ is called the boundary of $S$. For the set $1<x<2$, the set $\{1, 2\}$ is the boundary. For the set $1≤x≤2$, the boundary is again $\{1, 2\}$, but this time the set contains its boundary. Such a set is called closed. –  MJD Jun 28 '12 at 13:37
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Just a minor quibble: If $x$ is an isolated point of a topological space $X$ (i.e., if the singleton $\{ x \}$ is open), then $x$ cannot be a limit point of any subset of $X$, let alone of the open $\{ x \}$ (of which it is an interior point). For example, in a discrete space no subset has any limit points. –  Arthur Fischer Jun 28 '12 at 14:42
    
Thanks for the correction. I should have qualified my comment more carefully. –  MJD Jun 28 '12 at 15:03
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