Does $\{(-1)^n\mid n\in\mathbb N\}=\{-1,1\}$

Question

In my course of topology, we distinguish sequence s.t. $\#\{x_n\mid n\in\mathbb N\}<\infty $ and sequence where $\#\{x_n\mid n\in\mathbb N\}=\infty $, and in particular, we have at a moment that $$\{x_n\mid n\in\mathbb N\}=\{(-1)^n\mid n\in\mathbb N\}=\{-1,1\}.$$ In a way I agree with the last equality, but I'm also very disturbed by it. Indeed, if we have two sets $A$ and $B$, then $$A=B\implies A'=B',$$ where $A'$ and $B'$ denote the limits point of $A$ and $B$ respectively. Now, if I set $$A=\{(-1)^n\mid n\in\mathbb N\}\quad \text{and}\quad B=\{-1,1\},$$ then $A'=\{-1,1\}$, whereas $B'=\emptyset$. So, how can be interpreted $$\{(-1)^n\mid n\in\mathbb N\}=\{-1,1\} \ \ \ ?$$

Answer 1

$1$ is not a limit point of $A=\left\{(-1)^n \;|\; n \in \mathbb{N}\right\}$ (however you write it) because for $1$ to be a limit point of $A$, every neighborhood of $1$ would have to contain a point of $A$ distinct from $1$. This extra condition of the definition is important here. (Alternately, it's necessary for there to be a sequence in $A \setminus \{1\}$ whose limit is $1$.)