User:IssaRice/Adherent point and limit point: Difference between revisions

Latest revision as of 05:02, 6 July 2019

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 $E ∖ {x_{0}}$

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 $ε > 0$ and every $N \geq 1$ there exists $n \geq N$ such that $d (x_{n}, x) \leq ε$

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

(replacing " $x_{0}$ is an adherent point of ${x_{n} : n \geq N}$ " with " $x_{0}$ is a limit point of ${x_{n} : n \geq N}$ " does not work. why not?)

$x_{0}$ is a limit point of $(x_{n})_{n = 1}^{\infty}$ iff there are infinitely many $N \geq 1$ for which $x_{0}$ is an adherent point of ${x_{n} : n \geq N}$

If $x_{0}$ is a limit point of ${x_{n} : n \geq 1}$ , then $x_{0}$ is a limit point of $(x_{n})_{n = 1}^{\infty}$ .

also limit point iff some subsequence converges to it

In the case of sequences, the notions of adherent and limit points collapses into the same idea, because isolated points are particularly useless. (the "image" of the sequence keeps shifting, so isolated points disappear after a finite amount of time. this is similar to how the starting index of a sequence is unimportant for convergence.)

In the case of sequences, a set like ${1}$ , where $1$ appears "isolated", might not be, since the sequence could be $(1, 1, 1, \dots)$ . In other words, there could be infinitely many points bunched up on top of each other.

Can a sequence have infinitely many limit points? I think so; just enumerate the rationals

@@ Line 1: / Line 1: @@
 Let <math>(X,d)</math> be a metric space, let <math>E</math> be a subset of <math>X</math>, and let <math>x_0\in X</math> be a point.
+==Adherent point==
 * there exists a sequence <math>(x_n)_{n=1}^\infty</math> of points in <math>E</math> which converges to <math>x_0</math>
@@ Line 5: / Line 7: @@
 * <math>x_0</math> is an interior point of <math>E</math> or is a boundary point of <math>E</math>
 * for every open set <math>U</math> such that <math>x \in U</math> one has <math>U\cap E \ne \emptyset</math>
+==Limit point==
 * for every open set <math>U</math> such that <math>x \in U</math> there is some <math>y \in U\cap E</math> such that <math>y \ne x</math>
 * for every open set <math>U</math> such that <math>x \in U</math>, the set <math>U \cap E</math> has infinitely many points
 * there exists a sequence <math>(x_n)_{n=1}^\infty</math> of distinct points in <math>E</math> (i.e. <math>x_n \in E</math> for all <math>n \geq 1</math> and <math>x_n \ne x_m</math> for all <math>n \ne m</math>) which converges to <math>x_0</math>
 * there exists a sequence <math>(x_n)_{n=1}^\infty</math> of points in <math>E</math>, none of which are equal to <math>x_0</math>, which converges to <math>x_0</math>
+==Relationship between adherent point and limit point==
+<math>x_0</math> is a limit point of <math>E</math> iff it is an adherent point of <math>E \setminus \{x_0\}</math>
+Every limit point of <math>E</math> is an adherent point of <math>E</math>, but the converse is false. Limit points which are not adherent points are called isolated points.
+==Limit point of a sequence==
+<math>x_0</math> is a limit point of <math>(x_n)_{n=1}^\infty</math> iff for every <math>\varepsilon > 0</math> and every <math>N\geq 1</math> there exists <math>n \geq N</math> such that <math>d(x_n,x) \leq \varepsilon</math>
+<math>x_0</math> is a limit point of <math>(x_n)_{n=1}^\infty</math> iff for every <math>N\geq 1</math>, <math>x_0</math> is an adherent point of <math>\{x_n : n \geq N\}</math>
+(replacing "<math>x_0</math> is an adherent point of <math>\{x_n : n \geq N\}</math>" with "<math>x_0</math> is a limit point of <math>\{x_n : n \geq N\}</math>" does not work. why not?)
+<math>x_0</math> is a limit point of <math>(x_n)_{n=1}^\infty</math> iff there are infinitely many <math>N\geq 1</math> for which <math>x_0</math> is an adherent point of <math>\{x_n : n \geq N\}</math>
+If <math>x_0</math> is a limit point of <math>\{x_n : n \geq 1\}</math>, then <math>x_0</math> is a limit point of <math>(x_n)_{n=1}^\infty</math>.
+also limit point iff some subsequence converges to it
+In the case of sequences, the notions of adherent and limit points collapses into the same idea, because isolated points are particularly useless. (the "image" of the sequence keeps shifting, so isolated points disappear after a finite amount of time. this is similar to how the starting index of a sequence is unimportant for convergence.)
+In the case of sequences, a set like <math>\{1\}</math>, where <math>1</math> appears "isolated", might not be, since the sequence could be <math>(1,1,1,\ldots)</math>. In other words, there could be infinitely many points bunched up on top of each other.
+Can a sequence have infinitely many limit points? I think so; just enumerate the rationals