Cluster point of set $\{a_n \,|\, n \in \mathbb N\}$ is limit point of sequence $(a_n)_n$

Question

Let $(a_n)_n$ be a sequence in $\mathbb R$ and $A := \{a_n \,|\, n \in \mathbb N\}$ the set of elements of the sequence. I want to show

If $(a_n)_n$ is injective, then a limit point of $(a_n)_n$ is also a cluster point of $A$.

My definitions are

$x_0$ is a cluster point of a set $A \iff$ for every $\varepsilon \gt 0$ there exists $a \in A$, such that $0 \lt |a - x_0| \lt \varepsilon$.

$x_0$ is a limit point of a sequence $(a_n)_n \iff$ there is a subsequence which converges to $x_0$ or equivalently if for all $\varepsilon \gt 0$ and all $N \in \mathbb N$ there exists $n \ge N$, such that $|a_n - x_0| \lt \varepsilon$.

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I don't know how to start. How should I use the injective property?

Answer 1

The injective property is needed because otherwise for example for $a_n=(-1)^n$ we have $A=\{-1,1\}$ and $-1$ is a limit point of $a_n$ but not a limit point of $A$...

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Let $a$ be a limit point of $a_n$. There is a subsequence $a_{n_k}$ converging to $a$. Let $\epsilon>0$ be given. There is $K_1$ such that for all $k>K_1$ we have: \begin{align} |a_{n_k}-a|<\epsilon \end{align} Not that we cannot have $a_{n_k}=a$ for all $k>K_1$, because $a_n$ is injective. Let $K_2$ be such that $a_{n_{K_2}}=a$ if there is one. If there is no $K_2$ with that property then we take $K_2=1$. Now take $K=\max\{K_1,K_2 \}$. So for all $k>K$ we have $0<|a_{n_k}-a|<\epsilon$. Since $\epsilon>0$ was arbitrary we can conclude $a$ is a limit point of $A$.