Ad., Accum., Iso. Pts., Bound. Sets, Coverings, and Comp. Sets Review

# Adherent, Accumulation, and Isolated Points, Bounded Sets, Coverings, and Compact Sets Review

We will now review some of the recent content regarding adherent, accumulation, and isolated points alongside bounded sets, coverings, and compact sets in metric spaces.

Let $(M, d)$ be a metric space and let $S \subseteq M$.

• We said that a point $x \in M$ is an Adherent Point of $S$ if every ball centered at $x$ contains points of $S$, that is, for all positive $r > 0$, $B(x, r) \cap S \neq \emptyset$.
• We also said that a point $x \in M$ is an Accumulation Point (or Limit Point) of $S$ if every ball centered at $x$ contains points of $S$ that are different from $x$, that is, for all positive $r > 0$, $B(x, r) \cap S \setminus \{ x \} \neq \emptyset$.
• Furthermore, we aid a point $x \in S$ is an Isolated Point of $S$ if there exists a positive $r_0 > 0$ such that the ball centered at $x$ with radius $r_0$ contains no points of $S$ different from $x$, that is, $B(x, r_0) \cap S \setminus \{ x \} = \emptyset$.
• We then looked at some nice criteria for a set to be closed with respect to these definitions on the Criteria for a Set to be Closed in a Metric Space page. We saw that $S$ being closed, $S$ containing all of its adherent points, and $S$ contains all of its accumulation points were equivalent
(1)
\begin{align} \quad \bar{S} = S \cup \partial S \end{align}
• On The Derived Set of a Set in a Metric Space page we said that the Derived Set of $S$ denoted $S'$ is the set of all accumulation points of $S$ and we proved that another very important identity for the closure of a set:
(2)
\begin{align} \quad \bar{S} = S \cup S' \end{align}
(3)
\begin{align} \quad B(x, r) \cap S \neq \emptyset \end{align}
• We noted that a set $S$ is dense in $M$ if and only if $\bar{S} = M$.
• On the Separable Metric Spaces page we defined a special type of metric space. We said that $(M, d)$ is a Separable metric space if it contains a countable dense subset.
• On the Bounded Sets in a Metric Space page we said that $S$ is Bounded if there exists a positive real number $r > 0$ such that for some $x \in M$ we have that the ball centered at $x$ with radius $r$ fully contains $S$, i.e., $S \subseteq B(x, r)$. We said that $S$ is Unbounded if it is not bounded.
• On the Coverings of a Set in a Metric Space page we said that a Cover or Covering of $S$ is a collection of sets $\mathcal F$ from $M$ such $S$ is contained in the union of all sets in $\mathcal F$:
(4)
\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal F} A \end{align}
• We said that a subset $\mathcal S \subseteq \mathcal F$ is a Subcover or Subcovering of $S$ if $\mathcal S$ is also a cover of $S$, i.e., $S \subseteq \bigcup_{A \in \mathcal S} A$.
• Furthermore, an Open Cover or Open Covering of $S$ is a collection of open sets $\mathcal F$ that are a cover of $S$, and similarly, an Open Subcover or Open Subcovering of $S$ is a subset $\mathcal S \subseteq \mathcal F$ of open sets that also covers $S$.
• We then looked at two very nice properties of compact sets. The first important property we proved was on the Boundedness of Compact Sets in a Metric Space, and we proved that every compact set in a metric space is bounded.
• We summarized the results above on the Compact Sets in a Metric Space are Closed and Bounded page by noting that a subset of $\mathbb{R}^n$ or $\mathbb{C}^n$ (with the Euclidean metric) is compact if and only if it is closed and bounded - however, this is not true in general for general metric spaces. We exhibited a metric space which contains a set that is closed and bounded yet not compact.
• We also showed that the union of a finite collection of compact sets is compact and that the intersection of an arbitrary collection of compact sets is compact.

The following diagram shows some of the main connections between some of the important results we have proven so far.