📂MetricSpace

Neighborhood, Limit Point, Open, Closed in Metric Space

Definition

Let’s say $(X,d)$ is a metric space. Suppose $p \in X$ and $E \subset X$.

• The set that includes all $q$s satisfying $d(q,p)<r$ is defined as the neighborhood of point $p$ and is denoted as $N_{r}(p)$. Here, $r$ is called the radius of $N_{r}(p)$. If the distance can be omitted, it may also be denoted as $N_{p}$.

• If all neighborhoods of $p$ contain $q$, which is $q\ne p$ and $q\in E$, then $p$ is called a limit point of $E$.

• If $p\in E$ and $p$ is not a limit point of $E$, then $p$ is called an isolated point of $E$.

• If all limit points of $E$ are included in $E$, then $E$ is said to be closed.

• If there exists a neighborhood $N$ satisfying $N\subset E$, then $p$ is called an interior point of $E$.

• If every point of $E$ is an interior point of $E$, then $E$ is said to be open.

• The set that includes all $p$s that are $p \in X$ and $p \notin E$ is called the complement of $E$ and is denoted as $E^{c}$.

• If $E$ is closed and every point of $E$ is a limit point of $E$, then $E$ is said to be perfect.

• If there exists a point $q\in X$ and a real number $M$ satisfying $\forall p\in E,\ d(p,q)<M$, then $E$ is said to be bounded.

• If every point of $X$ is either a limit point of $E$ or a point of $E$, then $E$ is said to be dense in $X$.

• The set of all limit points of $E$ is called the derived set of $E$ and is denoted as $E^{\prime}$.

• The union of $E$ and $E^{\prime}$ is called the closure and is denoted as $\overline{E}=E\cup E^{\prime}$.

Explanation

The concepts of openness, limit points, denseness, interior points, etc., mentioned above can be defined through different statements but are essentially the same. Why each concept is defined and named as such can be easily grasped by directly drawing them in one or two dimensions. An isolated point is defined as a point that is not a limit point, so it cannot be both an isolated and a limit point at the same time. Conversely, open and closed sets are defined based on separate conditions, so contrary to the intuitive feeling their names might convey, there can exist sets that are both open and closed, or neither open nor closed. An example of the former is $\mathbb{R}^{2}$, and an example of the latter is $\left\{ {\textstyle \frac{1}{n}}\ |\ n\in \mathbb{N} \right\}$. Considering the definitions of interior points and neighborhoods, the condition for $x$ to be an interior point of $E$ is the same as the existence of some positive number $\varepsilon>0$ such that

$$d(x,p) <\varepsilon \implies x \in E$$

is satisfied. Several theorems and proofs related to the above concepts are introduced, following the notation from the definitions.

Theorem 1

All neighborhoods are open sets.

Proof

Let’s say $E=N_{r}(p)$. Also, consider any $q \in E$. Then, by the definition of a neighborhood, there must exist a positive real number $h$ that satisfies the following equation:

$$d(p,q)=r-h<r$$

Then, by the definition of distance, for all $s$ that satisfy $d(q,s)<h$, the following equation holds:

$$d(p,s)\le d(p,q)+d(q,s)<(r-h)+h=r$$

Therefore, by the definition of a neighborhood, $s \in E$ is true. This shows that any point $s$ within the neighborhood $N_{h}(q)$ of $q$ is an element of $E$. Hence, $N_{h}(q) \subset E$, meaning $q$ is an interior point of $E$. Since we initially considered $q$ to be any point of $E$, all points of $E$ are interior points. Therefore, $E$ is an open set.

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Theorem 2

A set $E$ being an open set is equivalent to $E^c$ being a closed set.

Proof

• $(\impliedby)$

  Assume $E^c$ is closed. Now, for any $p\in E$, since $p \notin E^c$ and by the definition of being closed, $p$ is not a limit point of $E^c$. Thus, there exists a neighborhood $N$ satisfying $N \cap E^c=\varnothing$. This implies $N \subset E$ and, by the definition of an interior point, $p$ is an interior point of $E$. Since any $p\in E$ is an interior point of $E$, by definition, $E$ is an open set.

• $(\implies)$

  Assume $E$ is open. And let $p$ be a limit point of $E^{c}$. Then, by the definition of a limit point, every neighborhood of $p$ contains at least one point of $E^{c}$. Thus, every neighborhood of $p$ does not include $E$, which means $p$ is not an interior point of $E$. Since we assumed $E$ is open, $p\notin E$ is true. Therefore, since all limit points $p$ of $E^{c}$ are included in $E^{c}$, $E^{c}$ is closed.

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Theorem 3

Let’s say $p$ is a limit point of $E$. Then, the neighborhood of $p$ contains infinitely many points of $E$.

This can be expressed differently as ‘A finite set does not have a limit point’, ‘A set with a limit point is an infinite set’.

Proof

Assume the neighborhood $N$ of $p$ includes only a finite number of elements of $E$. And let $q_{1},q_{2},\cdots,q_{n}$ be points of $N\cap E$ that are not $p$. Also, let $r$ be the minimum of the distances between $p$ and $q_{i}$.

$$r= \min \limits _{1\le i \le n}d(p,q_{i})$$

Since each $q_{i}$ is different from $p$, all distances are positive, and the minimum of positive numbers is also positive, thus $r>0$ is true. Now, consider another neighborhood $N_{r}(p)$ of $p$. Then, by the definitions of neighborhood and distance, $N_{r}(p)$ contains no $q_{i}$. Thus, by the definition of a limit point, $p$ is not a limit point of $E$. This contradicts the fact that $p$ is a limit point of $E$. Therefore, by reductio ad absurdum, the assumption is incorrect, proving the theorem.

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Corollary

A set with only a finite number of points does not have a limit point.

Theorem 4

For a metric space $(X,d)$ and $E \subset X$, the following facts hold: $(a)$ $\overline{E}$ is closed.$(b)$ Being $E=\overline{E}$ is equivalent to $E$ being closed.$(c)$ For all closed sets $F\subset X$ satisfying $E\subset F$, $\overline{E} \subset F$ holds.

$(a)$ and $(c)$ imply that $\overline{E}$ is the smallest closed subset of $X$ that contains $E$.