Can someone please explain to me why the set {1, 1/2, 1, 1/3, 1, 1/4, ...} has two points of accumulation?
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 $1$.
Related videos on Youtube
Shamisen Expert
Updated on August 01, 2022Comments
-
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?