Real Analysis: Local Metric Topology
Isolated Points and Limit Points
Isolated Points
Consider a subset $A$ of a metric space $(X, d).$ A point $x \in A$ is an isolated point of $A$ if there exists an open set $S$ such that $(S  \{x\}) \cap A = \varnothing.$ Plainly, $x$ is an isolated point if there exits some neighborhood of $x$ that contains no other points of $A$ other than $x$ itself. Equivalently, $x$ is an isolated point of $A$ if it has positive distance from the rest of $A.$
Limit Points
A point $x \in X$ is a limit point of $A$ if for every open set $S$ containing $x,$ $\left(S  \{x\}\right) \cap A \neq \varnothing.$ Plainly, $x$ is a limit point of $A$ if every open set containing $x$ also contains another point in $A$. In contrast to isolated points, which have a positive distance from the rest of $A,$ limit points are those that are zero distance from the rest of $A.$ As a result, a point can either be an isolated point or a limit point, but not both. Note that a point does not have to be an element of a set in order to be a limit point of that set, but an isolated point does.
Alternate Construction of the Closure
The closure was previously defined as the union of a set with its boundary. However, an equivalent definition defines the closure as the union of the interior of a set and the boundary of the set. This alternate definition is particularly useful because the interior and the boundary are disjoint, and thus form a partition, or disjoint union, of the closure.
Similarly, a set's isolated points and limit points also form a disjoint union of the closure. However, upon checking our definitions, we find that the set of isolated points and limit points are not alternate definitions of the interior or boundary. Thus they form a wholly independent way of constructing, and thus conceptualizing, "the set and all the points it touches."
Problems
Show that every point is a limit point of $\mathbb{R}.$
Let $x \in \mathbb{R}.$ Pick $r > 0.$ Then $B_{r}(x)$ contains the point $x + \frac{r}{2}.$ It follows that $\left(B_{r}(x)  \{x\}\right) \neq \varnothing.$ Therefore $x$ is a limit point of $\mathbb{R}.$
Show that $\mathbb{R}$ has no isolated points.
Consider $x \in \mathbb{R}$ and pick $r > 0.$ The point $x + \frac{r}{2}$ is in $B_{r}(x)  \{x\}.$ Therefore $\left(B_{r}(x)  \{x\}\right)$ is nonempty, and so $x$ is not an isolated point of $\mathbb{R}.$
Consider the interval $(a, b) \subset \mathbb{R}.$ Show that $a$ and $b$ are limit points of $(a, b).$
Set $l = b  a,$ pick $r > 0,$ and select $p = \text{min}\{l, r\}.$ Then $a + \frac{p}{2} \in (a, b)$ and $a + \frac{p}{2} \in \left(B_{r}(a)  \{a\}\right),$ so $a$ is a limit point of $(a, b).$ Likewise, $b  \frac{p}{2} \in (a, b)$ and $b  \frac{p}{2} \in \left(B_{r}(b)  \{b\}\right),$ so $b$ is also a limit point of $(a, b).$
Show that if $x$ is a limit point of $A,$ and $A \subseteq B,$ then $x$ is a limit point of $B.$
Since $x$ is a limit point of $A,$ for every $r > 0$ it follows that $B_{r}(x)  \{x\}$ contains another point of $A,$ call it $y.$ Since $A \subseteq B,$ it follows that $y \in B.$ Therefore $x$ is a limit point of $B.$
Show that $a$ and $b$ are limit points of the real intervals $[a, b),$ $(a, b],$ and $[a, b].$
Recall that $a$ and $b$ are limit points of the real interval $(a, b).$ Since $[a, b),$ $(a, b],$ and $[a, b]$ are supersets of $(a, b),$ it follows that $a$ and $b$ are limit points of each of them as well.
Equivalent definition: Show that a point $x$ in a subset $A$ of a metric space $X$ is an isolated point if and only if it is of positive distance from the rest of $A.$
First, let $x$ be an isolated point of $A$. Then there exists a $r> 0$ such that $(B_{r}(x)  \{x\}) \cap A = B_{r}(x) \cap (A  \{x\}) = \varnothing.$ Therefore $d(x, y) \geq r$ for all $y \in (A  \{x\}).$ Therefore $\text{Dist}(x, A  \{x\}) \geq r > 0.$
Conversely, assume $\text{Dist}(x, A  \{x\}) = r> 0.$ Then $d(x, y) \geq r$ for all $y \in A  \{x\}.$ Since $B_{r}(x)  \{x\}$ contains all points $y \neq x$ such that $d(x, y) < r,$ it follows that $(B_{r}(x)  \{x\}) \cap A = B_{r}(x) \cap (A  \{x\}) = \varnothing.$ Therefore $x$ is an isolated point of $A.$
Show that every open set containing at a limit point of $A$ contains an infinite number of points in $A.$
Let $A$ be a subset of a metric space $(X, d),$ and let $x$ be a limit point of $A.$
Proof by contradiction. Assume there is an $r > 0$ such that $B_{r}(x) \subseteq A$ has a finite number of points in $A.$ Pick $q$ such that $q = \text{min}\{ d(x, t) : t \in (B_{r}(x) \cap A) \}$. It follows that $B_{q/2}(x) \cap A = \{x\},$ which is a contradiction, since every open ball centered at $x$ must contain another point in $A$ other than $x.$ Therefore $B_{r}(x)$ must have had an infinite number of points in it after all.
Equivalent definition: Show that a set is closed if and only if it contains all of its limit points.
Let $A$ be a subset of a metric space $(X, d)$, and let $A'$ be the set of all limit points of $A.$
First, assume $A$ is closed. Then $A^c$ is open, which means that all of its points are interior points. In turn, this means that for all $x \in A^c,$ there exists an $r > 0$ such that $B_{r}(x) \subseteq A^c.$ It follows that $x$ is not a limit point of $A.$ Since none of the points in $A^c$ are limit points, it follows that $A' \subseteq A.$
Conversely, assume $A' \subseteq A.$ Therefore if $x \in A^c,$ then $x$ is not a limit point of $A.$ It follows that there is some $r > 0$ such that $(B_{r}(x)  \{x\}) \cap A = \varnothing.$ This implies that $(B_{r}(x)  \{x\}) \subseteq A^c.$ Since $x \in A^c$, it follows that $(B_{r}(x)  \{x\}) \subset B_{r}(x) \subseteq A^c.$ It follows that $x$ is an interior point of $A^c.$ Therefore $A^c$ is open, which means that $A$ is closed.
Equivalent definition: Show that $x$ is a limit point of $A$ if and only if it is zero distance from the rest of $A.$
First, let $x$ be a limit point of $A.$ Proof by contradiction. Assume $\text{dist}(x, A  \{x\}) > 0.$ Then there exists some $r > 0$ such that $\text{dist}(x, A  \{x\}) > r > 0.$ It follows that $B_{r}(x)$ contains no other points of $A$ aside from $x.$ But this is a contradiction. Therefore $\text{dist}(x, A  \{x\}) = 0$ after all.
Conversely, assume $\text{dist}(x, A  \{x\}) = 0.$ Proof by contradiction. Assume $x$ is not a limit point. Then there exists some $r > 0$ such that $B_{r}(x)$ contains no other points in $A$ aside from $x.$ It follows that $\text{dist}(x, A  \{x\}) \geq r.$ But this is a contradiction. Therefore $x$ is a limit point after all.
Let $A$ be a subset of a metric space $(X, d).$ Show that the set of limit points of $A$ and the set of isolated points of $A$ are disjoint.
Let $x$ be an isolated point of $A.$ Then there is an $r > 0$ such that $(B_{r}(x)  \{x\}) \cap A = \varnothing.$ Since $B_{r}(x)$ contains no point in $A$ other than $x,$ it follows that $x$ is not a limit point of $A.$
Alternatively, let $x$ be a limit point of $A.$ Then for every $r > 0,$ it follows that $(B_{r}(x)  \{x\}) \cap A \neq \varnothing.$ It follows that there is no open set containing $x$ that contains no other points in $A,$ and so $x$ is not an isolated point.
Equivalent definition: Show that the closure of a set is equal to the union of its limit points and its isolated points.
Consider a subset $A$ of a metric space $(X, d).$
Let $x \in \left(\overline{A}\right)^c = \text{Ext}(A).$ Then $x \in A^c,$ and so $x$ is not an isolated point of $A$. Likewise, $x \in \left(A^c\right)^{\circ}$, so there exists an $r > 0$ such that $B_{r}(x) \cap A = \varnothing.$ Therefore $x$ is not a limit point of $A$. Therefore, if $x$ is an isolated point or a limit point of $A,$ then $x \in \text{Ext}(A)^c = \overline{A}.$
Conversely, let $x$ be either an isolated point or a limit point of $A.$ If $x$ is an isolated point, then $x \in A,$ and so $x \in \overline{A}.$ If instead $x$ is a limit point, then for every $r > 0$ there is an open ball $B_{r}(x)$ such that $(B_{r}(x)  \{x\}) \cap A \neq \varnothing.$ If $x \in A,$ then $x \in \overline{A}.$ If instead $x \in A^c,$ then $x \in \partial A,$ and therefore $x \in \overline{A}.$
Give examples of the following using any metric space:

An isolated point that is not an interior point.

An isolated point that is not a boundary point.

A limit point that is not an interior point.

A limit point that is not a boundary point.

An interior point that is not an isolated point.

An interior point that is not a limit point.

A boundary point that is not an isolated point.

A boundary point that is not a limit point.
Let $(X, d)$ be a metric space where $d$ is the discrete metric.

Consider $\{3\} \subset \mathbb{R}.$ The point $3$ is an isolated point and boundary point of $\{3\}$ and therefore not an interior point of $\{3\}.$

Every point $x \in A \subseteq X$ is an isolated point, but $B_{\frac{1}{2}}(x) = \{x\}$ and contains no point outside of $A,$ and so is not a boundary point.

Consider $(0,1) \subset \mathbb{R}.$ The point $0$ is a limit point of $(0,1)$ but is not an interior point.

Consider $(0,2) \subset \mathbb{R}.$ The point $1$ is a limit point of $(0,2)$ but is not a boundary point.

Consider $(0,2) \subset \mathbb{R}.$ The point $1$ is an interior point of $(0,2)$ but is not an isolated point.

Consider a singleton set $\{x\}$ in $X.$ Then $B_{\frac{1}{2}}(x) = \{x\} \subseteq \{x\},$ and so $x$ is an interior point of $\{x\}.$ However, $B_{\frac{1}{2}}(x)$ contains no other points in $\{x\},$ and so $x$ is not a limit point of $\{x\}.$

Consider $(0,1) \subset \mathbb{R}.$ The point $1$ is a boundary point of $(0, 1)$ but is not a member of $(0, 1)$ and therefore not an isolated point of $(0, 1).$

Consider the set $A = (0, 1) \cup \{3\} \subset \mathbb{R}.$ The point $3$ is a boundary point and an isolated point of $A$, and therefore not a limit point of $A.$

Give an example of the following using any metric space:

An isolated point that is an interior point.

An isolated point that is a boundary point.

A limit point that is an interior point.

A limit point that is a boundary point.

Let $(X, d)$ be a metric space where $d$ is the discrete metric. Consider the singleton set $\{x\} \subseteq X.$ It follows that $B_{\frac{1}{2}}(x) = \{x\},$ so $x$ is an isolated point of $\{x\}.$ Since $\{x\} \subseteq \{x\},$ it follows that $x$ is also an interior point of $\{x\}.$

Consider the singleton set $\{2\} \subset \mathbb{R}.$ The point $2$ is an isolated point and a boundary point of $\{2\}.$

Consider the set $(0, 2) \subset \mathbb{R}.$ The point $1$ is a limit point and an interior point of $(0, 2).$

Consider the set $(0, 1) \subset \mathbb{R}.$ The point $1$ is a limit point and a boundary point of $(0, 1).$

Show that every point of a finite set is an isolated point.
Let $A$ be a finite set and let $x \in A.$ Pick $y$ such that $d(x, y) = \text{min}( \{ d(x, z) : z \in A \} ).$ Set $r = \frac{1}{2}d(x, y).$ Then $B_{r}(x)$ contains no other points of $A$ than $x.$ Therefore $x$ is an isolated point of $A.$
Counterexample: Show that if $A \subset B$ and $x$ is a limit point of $B,$ then $x$ is not necessarily a limit point of $A.$
Consider the real interval $(1, 5).$ The point $1$ is a limit point of $(1, 5),$ but $(2, 4) \subset (1, 5)$ and $1$ is not a limit point of $(2, 4).$
Give an example of sets $A$ and $B$ where $x$ is a limit point of each but $x$ is not a limit point of $A \cap B.$
Let $A$ and $B$ be real intervals where $A = [1, 2]$ and $B = [2, 3].$ Then $2$ is a limit point of $A$ and $B.$ However, $A \cap B = \{2\}.$ Because $A \cap B$ is finite, it follows that $2$ is an isolated point of $A \cap B$ and therefore not a limit point.