Real Analysis: Local Metric Topology

Problems

1. 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 $\mathbb{R}.$

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2. 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 $\mathbb{R}.$

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3. 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 $(a, b).$

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4. 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 $B.$

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5. 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.

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6. 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 $A.$

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7. 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.

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8. 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.

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9. 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.

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10. 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.

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11. 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 $x \in \overline{A}.$

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12. Give examples of the following using any metric space:

    1. An isolated point that is not an interior point.

    2. An isolated point that is not a boundary point.

    3. A limit point that is not an interior point.

    4. A limit point that is not a boundary point.

    5. An interior point that is not an isolated point.

    6. An interior point that is not a limit point.

    7. A boundary point that is not an isolated point.

    8. A boundary point that is not a limit point.

    Let $(X, d)$ be a metric space where $d$ is the discrete metric.

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

    2. 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.

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

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

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

    6. 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 $\{x\}.$

    7. 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 $(0, 1).$

    8. 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 $A.$

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13. Give an example of the following using any metric space:

    1. An isolated point that is an interior point.

    2. An isolated point that is a boundary point.

    3. A limit point that is an interior point.

    4. A limit point that is a boundary point.

    1. 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 $\{x\}.$

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

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

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

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14. 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 $A.$

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15. 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 $(2, 4).$

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16. 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.

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