0 an accumulation point for sin(1/n)

Statement

Show that $0$ is an accumulation point for $D = \{\sin(1/n) : n\ge 1\}$ and state whether $0$ belongs to $D$.

Solution

$0$ is an accumulation point. As $n$ grows we have $1/n\to 0$, and hence $\sin(1/n)\to \sin 0 = 0$. Given any $\delta>0$, there exists $n$ large enough that $0<\sin(1/n)<\delta$; indeed there are infinitely many, all distinct and non-zero. So in every punctured neighbourhood of $0$ there fall infinitely many points of $D$: $0$ is an accumulation point.

$0\notin D$. For $0$ to belong to $D$ we would need $\sin(1/n)=0$ for some $n\ge 1$, i.e. $1/n = k\pi$ for some integer $k$. But $1/n\in(0,1]$ is never a multiple of $\pi$: hence $\sin(1/n)\ne 0$ for every $n\ge 1$, and $0\notin D$.

It is the typical example of an accumulation point that does not belong to the set.

Links

Topics: Limits
Concepts: Accumulation point
Skills: Proving
Exercise type: Proof