User:IssaRice/Adherent point and limit point

Let $(X,d)$ be a metric space, let $E$ be a subset of $X$, and let $x_{0}\in X$ be a point.

Adherent point

• there exists a sequence $(x_{n})_{n=1}^{\infty }$ of points in $E$ which converges to $x_{0}$
• for every radius $r>0$ the ball $B(x_{0},r)$ has nonempty intersection with $E$
• $x_{0}$ is an interior point of $E$ or is a boundary point of $E$
• for every open set $U$ such that $x\in U$ one has $U\cap E\neq \emptyset$

Limit point

• for every open set $U$ such that $x\in U$ there is some $y\in U\cap E$ such that $y\neq x$
• for every open set $U$ such that $x\in U$, the set $U\cap E$ has infinitely many points
• there exists a sequence $(x_{n})_{n=1}^{\infty }$ of distinct points in $E$ (i.e. $x_{n}\in E$ for all $n\geq 1$ and $x_{n}\neq x_{m}$ for all $n\neq m$) which converges to $x_{0}$
• there exists a sequence $(x_{n})_{n=1}^{\infty }$ of points in $E$, none of which are equal to $x_{0}$, which converges to $x_{0}$

Relationship between adherent point and limit point

$x_{0}$ is a limit point of $E$ iff it is an adherent point of $E\setminus \{x_{0}\}$

Every limit point of $E$ is an adherent point of $E$, but the converse is false. Limit points which are not adherent points are called isolated points.

Limit point of a sequence

$x_{0}$ is a limit point of $(x_{n})_{n=1}^{\infty }$ iff for every $\varepsilon >0$ and every $N\geq 1$ there exists $n\geq N$ such that $d(x_{n},x)\leq \varepsilon$

$x_{0}$ is a limit point of $(x_{n})_{n=1}^{\infty }$ iff for every $N\geq 1$, $x_{0}$ is an adherent point of $\{a_{n}:n\geq N\}$