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If $a_{n} = \frac{-1}{2}, \frac{1}{3}, \frac{-1}{4}, \frac{1}{5} ...$ It is clear to me by intuition that the limit point is $0$. But according to Rudin, "A point $p$ is a limit point of the set $E$ if every neighborhood of $p$ contains a point $q \neq p$ such that $q \in E$.

I'm having trouble understanding why $\frac{-1}{2}$ is not a limit point by the definition given in Rudin. Could someone explain?

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Jon, I would also suggest that you look at what it means for $p$ to NOT be a limit point by negating the definition of limit point. You need to show that "there is a neighborhood of $p$ such that..." (what?) –  The Chaz 2.0 Mar 23 '12 at 2:29
    
@TheChaz: The negation would be "A point $p$ is not a limit point of a set $E$ if there exists a neighborhood of $p$ that does not contain a $q \in E$ where $q \neq p$. –  Student Mar 23 '12 at 2:36
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(+1) for extra credit :) –  The Chaz 2.0 Mar 23 '12 at 2:37
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3 Answers

up vote 2 down vote accepted

Let $E=\{a_1,a_2,\ldots\}=\{\frac{-1}{2},\frac{1}{3},\ldots\}$. The point $\frac{-1}{2}$ is not a limit point of the set $E$ because the neighborhood $B_{1/4}(\frac{-1}{2})$ contains no point in $E$ other than $\frac{-1}{2}$. To see this, note that any element of $E$ is either positive, in which case it is at least $\frac{1}{2}$ away from $\frac{-1}{2}$, or negative and at least $\frac{-1}{4}$, which is $\frac{1}{4}$ away from $\frac{-1}{2}$ and so not in $B_{1/4}(\frac{-1}{2})$, which is defined as the set of points less than $\frac{1}{4}$ away from $\frac{-1}{2}$.

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So if I consider the interval $(\frac{-1}{2} - \frac{1}{4}, \frac{-1}{2} + \frac{1}{4}) = (\frac{-3}{4}, \frac{-1}{4})$, there is no element of $E$ in the open interval. So $\frac{-1}{2}$ cannot be a limit point. –  Student Mar 23 '12 at 2:17
    
@Jon Precisely. –  Alex Becker Mar 23 '12 at 2:17
    
Just a follow up question: If $E = \{1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ... \}$, then it is clear the limit points are 0 and 1. If I construct any interval around 1 no matter how small, it will always contain 1 and not 0. But 1 = 1, so doesn't that contradict the definition? –  Student Mar 23 '12 at 2:22
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@Jon, those look like isolated points to me... –  The Chaz 2.0 Mar 23 '12 at 2:23
    
@TheChaz: Thanks for pointing that out. It makes sense now. –  Student Mar 23 '12 at 2:27
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Every point in the set other than $-1/2$ itself is larger or equal than $-1/4$. Since $-1/2 < -3/8 < -1/4$, the open interval $(-1,-3/8)$ is an open neighborhood of $-1/2$ that contains no other point in the set.

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No one else explicitly said this so I will: The concept in definition of limit point that you seem to have misunderstood is the word 'every'. If you draw a picture it should be very clear that $\frac{-1}{2}$ cannot be a limit point.

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