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 $(x_n{)}_{n=1}^{\mathrm{\infty }}$ 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 $U\cap E\ne \mathrm{\varnothing }$
• for every open set $U$ such that $x\in U$ there is some $y\in U\cap E$ such that $y\ne x$
• for every open set $U$ such that $x\in U$, the set $U\cap E$ has infinitely many points
• there exists a sequence $(x_n{)}_{n=1}^{\mathrm{\infty }}$ of distinct points in $E$ (i.e. $x_n\in E$ for all $n\ge 1$ and $x_n\ne x_m$ for all $n\ne m$) which converges to $x_0$
• there exists a sequence $(x_n{)}_{n=1}^{\mathrm{\infty }}$ of points in $E$, none of which are equal to $x_0$, which converges to $x_0$

Limit point of a sequence

$x_0$ is a limit point of $(x_n{)}_{n=1}^{\mathrm{\infty }}$ iff for every $\epsilon >0$ and every $N\ge 1$ there exists $n\ge N$ such that $d(x_n,x)\le \epsilon$

$x_0$ is a limit point of $(x_n{)}_{n=1}^{\mathrm{\infty }}$ iff for every $N\ge 1$, $x_0$ is an adherent point of $\{a_n:n\ge N\}$