Distinguish between accumulation points and adherent points of a set with examples. Show that if x is an accumulation point of S, where S is a subset of R^n, then every open n-ball... Distinguish between accumulation points and adherent points of a set with examples. Show that if x is an accumulation point of S, where S is a subset of R^n, then every open n-ball B(x) contains infinitely many points of S. Also, prove that a set S in R^n is closed if, and only if, it contains all its adherent points. Read more

Steps to Solve

1. Distinguish between accumulation points and adherent points with examples.

   • Accumulation Point (or Limit Point): A point $x$ is an accumulation point of a set $S$ if every open neighborhood of $x$ contains at least one point of $S$ different from $x$ itself. Formally, for every $\epsilon > 0$, the set $(B(x, \epsilon) \setminus {x}) \cap S$ is non-empty, where $B(x, \epsilon)$ is the open ball centered at $x$ with radius $\epsilon$.

   • Adherent Point: A point $x$ is an adherent point of a set $S$ if every open neighborhood of $x$ contains at least one point of $S$. Formally, for every $\epsilon > 0$, the set $B(x, \epsilon) \cap S$ is non-empty. This means either $x$ is in $S$ or $x$ is an accumulation point of $S$.

   • Example: Let $S = {1/n : n \in \mathbb{N}} = {1, 1/2, 1/3, 1/4, ...}$.

     • The point $0$ is an accumulation point of $S$ because any open interval containing $0$ will contain infinitely many points of $S$.
     • The set of adherent points of $S$ is $S \cup {0} = {1, 1/2, 1/3, 1/4, ..., 0}$.
     • The points in $S$ are isolated points, meaning there exists a neighborhood around each point that contains no other points of $S$. For instance, $1/2$ is an isolated point of $S$.

2. Show that if $x$ is an accumulation point of $S \subset \mathbb{R}^n$, then every open n-ball $B(x)$ contains infinitely many points of $S$.

   Assume $x$ is an accumulation point of $S$. Suppose, for contradiction, that there are finitely many points of $S$ in $B(x)$, say $x_1, x_2, \dots, x_k$ are the points of $S$ in $B(x)$ other than $x$. Let $r = \min{|x - x_i| : i = 1, 2, \dots, k}$. Since each $x_i$ is distinct from $x$, $r > 0$. Now consider the open n-ball $B(x, r/2)$. This ball contains no point $x_i$ because $|x - x_i| \ge r > r/2$ for all $i$. Furthermore, $B(x, r/2)$ cannot contain any other point of $S$ besides $x$, because we assumed $x_1, x_2, \dots, x_k$ were all the points of $S$ in $B(x)$ other than $x$. This contradicts the assumption that $x$ is an accumulation point of $S$, because $B(x, r/2)$ contains no point of $S$ other than possibly $x$. Therefore, $B(x)$ must contain infinitely many points of $S$.

3. Also prove that a set $S$ in $\mathbb{R}$ is closed if, and only if, it contains all its adherent points.

   • ($\Rightarrow$) Suppose $S$ is closed. We want to show that $S$ contains all its adherent points. Let $x$ be an adherent point of $S$. If $x \in S$, we are done. Suppose $x \notin S$. Since $S$ is closed, $S^c = \mathbb{R} \setminus S$ is open. Because $x \in S^c$, there exists an $\epsilon > 0$ such that $B(x, \epsilon) \subset S^c$. But this means $B(x, \epsilon) \cap S = \emptyset$, which contradicts the assumption that $x$ is an adherent point of $S$ (since an adherent point must have every open neighborhood intersect the set). Thus, $x$ must be in $S$.

   • ($\Leftarrow$) Suppose $S$ contains all its adherent points. We want to show that $S$ is closed. We will show that $S^c$ is open. Let $x \in S^c$. Since $S$ contains all its adherent points, $x$ is not an adherent point of $S$. This means there exists some $\epsilon > 0$ such that $B(x, \epsilon) \cap S = \emptyset$. Thus, $B(x, \epsilon) \subset S^c$. Since for every $x \in S^c$, there exists an open ball $B(x, \epsilon) \subset S^c$, then $S^c$ is open. Therefore, $S$ is closed.