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?
Best Answer
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 $\frac{-1}{2}$.