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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?

zhw.
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Shamisen Expert
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    How is your set different from ${1,1/2,1/3,1/4,\ldots}$? Are you talking about limit points of a sequence? – Batominovski Jul 24 '15 at 04:42
  • As stated, the only limit point is $0$. (Limit points need not belong to the set.) However, if it was a sequence $(1, 1/2, 1, 1/3, 1, 1/4, ...)$, then 1 and 0 would be limit points, I believe. (Note the $()$ signs instead of ${}$ signs — $()$ means sequence, ${}$ means set.) – Akiva Weinberger Jul 24 '15 at 05:02
  • What @Batominovski said. It is about accumulation points of a sequence, I bet. – Daniel Fischer Jul 24 '15 at 07:55
  • Like Batominovsky and Daniel, I suspect that the question is actually about the sequence $$\left\langle 1,\frac12,1,\frac13,1,\frac14,1,\ldots\right\rangle;.$$ You may find the discussion in this answer helpful. – Brian M. Scott Jul 24 '15 at 08:26

2 Answers2

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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 $1$.

Daniel
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The point $0$ is only limit point of your set. there is no other limit point.

Ali
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