Can someone please explain to me why the set {1, 1/2, 1, 1/3, 1, 1/4, ...} has two points of accumulation?

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Those are not the neighborhoods of $1$, moreover, they're not even neighborhoods since those doesn't contain any open set (a neighborhood is either an open set or a set which contains an open set).

$1$ is not an accumulation point of that set since, for example, the neighborhood $\left(\frac{1}{2}, \frac{3}{2}\right)$ doesn't contain any point from the set besides $1$.

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Shamisen Expert
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Updated on August 01, 2022

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  • Shamisen Expert
    Shamisen Expert over 1 year

    A point of accumulation in $X$ is a point $c$ where every neighborhood of $c$ contains at least one point of $X$ distinct from $c$

    For example, the neighborhood of $1$ is $\{1, 1/2\}, \{1,1/2,1\}, \{1, 1/2, 1, 1/3\}...$, since it contains itself, therefore $1$ is definitely not a point of accumulation

    But the solution states that the two points are accumulation are $1$ and $0$, and neither of them belongs to the set.

    I am very confused, can someone help?