2.6Closed Sets and Limit Points

Proof.

1. Note $\emptyset^C = X$ and $X^C = \emptyset$ are both open, so they're both closed.

2. Let $A_1, \dots, A_n$ be finitely many closed sets. Then

   $$\left( \bigcup_{k = 1}^n A_k \right)^C = X \setminus \left( \bigcup_{k = 1}^n A_k \right) = \bigcap_{k = 1}^n (X \setminus A_k)$$

   is the finite intersection of open sets, which is open. Thus the finite union is closed.

3. Let $\{A_\lambda\}_{\lambda \in \Lambda}$ be a collection of closed sets. Then

   $$\left( \bigcap_{\lambda \in \Lambda} A_k \right)^C = X \setminus \left( \bigcap_{\lambda \in \Lambda} A_k \right) = \bigcup_{\lambda \in \Lambda} (X \setminus A_k)$$

   is the arbitrary union of open sets, which is open. Thus the arbitrary intersection is closed.

❦

Proof.

$\Longrightarrow$: Suppose $A$ is closed in $Y$. By definition, its complement $Y \setminus A$ is open in $Y$. By the definition of the subspace topology, there exists an open set $U$ in $X$ such that $Y \setminus A = U \cap Y$. Then,

Since $U$ is open in $X$, its complement $X \setminus U$ is closed in $X$. Letting $C = X \setminus U$, we have $A = C \cap Y$, where $C$ is a closed set in $X$.

$\Longleftarrow$: Suppose $A = C \cap Y$, where $C$ is a closed set in $X$. We wish to show that $A$ is closed in $Y$. Consider the complement of $A$ in $Y$:

Since $C$ is closed in $X$, its complement $X \setminus C$ is open in $X$. By the definition of the subspace topology, the intersection of an open set in $X$ with $Y$ is open in $Y$. Therefore, $Y \setminus A$ is open in $Y$, which implies that $A$ is closed in $Y$, completing the backward direction. ❦

Proof.

$A$ is the intersection of some closed subset of $X$ with $Y$, which is also a closed subset of $X$—thus $A$ is closed in $X$. ❦

Proof.

We show $\Cl_Y(A) = \Cl_X(A) \cap Y$ by proving a forward and backward inclusion.

$\subseteq$: Suppose $x \in \Cl_Y(A)$. Then note $x$ is in every closed superset of $A$ contained in $Y$, and of course $\Cl_X(A) \cap Y$ is one example, so $x \in \Cl_X(A) \cap Y$.

$\supseteq$: Suppose $x \in \Cl_X(A) \cap Y$. Then let $C$ be any arbitrary closed superset of $A$ such that $A \subseteq C \subseteq Y$. If $x \in C$, then it follows that $x \in \Cl_Y(A)$. Since $C$ is closed in $Y$, and $Y$ is a subspace of $X$ (and thus endowed with the subspace topology), there is some closed subset $C_X \subseteq X$ such that $C = C_X \cap Y$. But $A \subseteq (C = C_X \cap Y)$, implying $A \subseteq C_X$. But $\Cl_X(A)$ is the minimal superset of $A$ that is closed, so $A \subseteq \Cl_X(A) \subseteq C_X$. Thus $x \in C$, completing the proof. ❦

Proof.

1. We proceed by contrapositive. That is, we show that $x \notin \bar{A}$ if and only if there exists some open neighborhood $U \subseteq X$ of $x$. First, suppose $x \notin \bar{A}$. Note $X \setminus \bar{A}$ is an open subset of $X$ that must contain $x$, but does not intersect $A$ by construction. Conversely, suppose some open neighborhood $U \subseteq X$ of $x$ exists that doesn't intersect $A$. Then $X \setminus U$ is a closed superset of $A$ that does not contain $x$, and so $x \notin \bar{A}$, completing the proof.

2. Basis elements that contain $x$ are open neighborhoods about $x$, and open neighborhoods about $x$ that intersect $A$ contain basis elements.

❦

Proof.

We prove equality by forward and backward inclusion.

$\subseteq$: Suppose $x \in \bar{A}$. If $x \in A$, then $x \in A \cup A'$. Otherwise, if $x \notin A$, then $A = A \setminus \{x\}$. But note that, since $x \in \bar{A}$, every open neighborhood $U \subseteq X$ of $x$ intersects $A$ nontrivially, so $U \cap (A \setminus \{x\}) \ne \emptyset$. Thus $x \in A' \subseteq A \cup A'$.

$\supseteq$: Suppose $x \in A \cup A'$. If $x \in A$, $x \in \bar{A}$, obviously. Now suppose $x \notin A'$, and so $x \in A'$. Analogous to before, this means $A \setminus \{x\} = A$. Since $x \in A'$, for every open neighborhood $U \subseteq X$ of $x$, $U \cap (A \setminus \{x\}) = U \cap A \ne 0$. Thus $x \in \bar{A}$, completing the proof. ❦

Proof.

Let $A$ be a subset of topological space $X$, and let $A'$ be the set of limit points. First, suppose $A$ is closed. Then $A = \bar{A} = A \cup A'$, meaning $A' \subseteq A$. Now suppose $A$ contains all its limit points, and so $A' \subseteq A$. Then $A' \cup A \subseteq A$ and $A \subseteq A' \cup A$, meaning $A = A \cup A' = \bar{A}$, completing the proof. ❦

Proof.

Suppose $\{x_0\} \subseteq X$. Let $x \in X$ be such that $x \ne x_0$. By Hausdorffness of $X$, there are disjoint open neighborhoods $U_1, U_2 \subseteq X$ about $x$ and $x_0$. Thus $x \notin \bar{A}$. This implies $\bar{A} = \{x_0\}$, meaning the singleton is closed. ❦

Proof.

$(\Longrightarrow)$: Suppose $x$ is a limit point of $A$. Then there is some open neighborhood $U \subseteq X$ of $x$ such that $(U \cap (A \setminus \{x\}) = S) \ne \emptyset$. Suppose $S$ is finite, so $S = \{x_1, \dots, x_n\}$. Note that $S$ is a finite union of singleton subsets of $X$, and since singletons are closed in $T_1$ spaces, $S$ is closed. Thus $X \setminus S$ is open. Let $U' = U \cap (X \setminus S)$. Then $U'$ is open, $x \in U$, but $U \cap (A \setminus \{x\}) = \emptyset$. We have constructed an open neighborhood about $x$ that does not intersect $A \setminus \{x\}$, meaning $x$ is not a limit point, forcing a contradiction. Thus $U \cap (A \setminus \{x\})$ is infinite.

$(\Longleftarrow)$: Every open neighborhood about $x$ intersects $A$ infinitely amount of times, so every open neighborhood about $x$ intersects $A \setminus \{x\}$ infinitely many times, meaning $x$ is a limit point. ❦

Proof.

Suppose $x_n$ converges, so $x_n \to x$. Let $x_0 \ne x$: we show $x_n \not \to x_0$, confirming $x$ is the only limit for the sequence. Since $x \ne x_0 \in X$, a Hausdorff space, we can construct disjoint open neighborhoods $U$ and $U_0$ about $x$ and $x_0$ respectively such that $U \cap U_0 = \emptyset$. Recall that since $x_n \to x$, there exists some $N \in \bZ_+$ such that $x_n \in U$ for every $n \ge N$. But since $U_0$ is disjoint from $U$, it follows that the only terms of the sequence $U_0$ can possibly contain are $x_1, \dots, x_{N - 1}$. In other words, the tail for $(x_n)$ is not contained in $U_0$, so $x_n \not \to x_0$, completing the proof. ❦

Solution.

Note $\bar{(a, b)}$, by definition, is the smallest closed superset of $(a, b)$. Thus, if $[a, b]$ is a closed superset of $(a, b)$, then $\bar{(a, b)} \subseteq [a, b]$ follows immediately. Note that $[a, b]^c = (\leftarrow, a) \cup (b, \rightarrow)$ is a union of two open sets (so it's open), making $[a, b]$ closed, completing the proof.

Equality is achieved when $[a, b] \subseteq \bar{(a, b)}$, since we have forward and backward inclusion. Since $[a, b] = (a, b) \cup \{a\} \cup \{b\}$, simply knowing that $a, b$ are limit points of $(a, b)$ forces equality. We show $a$ (resp. $b$) is a limit point of $(a, b)$ if $a$ (resp. $b$) has no immediate successor (resp. immediate predecessor).

Suppose $a$ has no immediate successor. Let $U \subseteq X$ be an open neighborhood about $a$. Then note that we can locally refine $U$ by choosing some basis element of the order topology, $V$, such that $a \in V \subseteq U$. Note a basis element is of the form $V = (a, c)$, where $c: a < c \le b$. In any case, since $a$ has no immediate successor, there is some $x < c$ for which $x \in V \subseteq U$. This means $x \in U \cap ((a, b) \setminus \{a\}) = U \cap (a, b)$, meaning $a$ is a limit point for $(a, b)$.

Solution.

1. Suppose $A \subseteq B$. Let $x \in \bar{A}$. If $x \in A$, then $x \in B \subseteq \bar{B}$ trivially, so suppose $x \notin A$. Then $x$ is a limit point, so every open neighborhood $U \subseteq X$ about $x$ is such that $U \cap (A \setminus \{x\}) \ne \emptyset$. But $A \subseteq B$, so this implies $U \cap (B \setminus \{x\}) \ne \emptyset$, meaning $x$ is a limit point of $B$ too, so $x \in \bar{B}$.

Solution.

Suppose an ordered set $X$ is endowed with the order topology. Let $x \ne y \in X$, and without loss of generality, suppose $x < y$. Suppose $y$ is $x$'s immediate successor: then there exist open neighborhoods $U = (-\infty, y) \ni x$ and $V = (x, \infty) \ni y$ where $U \cap V = \emptyset$. If $y$ is not $x$'s immediate successor, there exists $t: x < t < y$. Then $U = (-\infty, t) \ni x$ and $V = (t, \infty) \ni y$ such that $U \cap V = \emptyset$, completing the proof.

Solution.

Let $(x_1, y_1) \ne (x_2, y_2) \in X \times Y$, where $X$ and $Y$ are Hausdorff topological spaces. Without loss of generality, suppose $x_1 \ne x_2$. By Hausdorffness of $X$, there exists open neighborhoods $U_X \ni x_1$ and $V_X \ni x_2$ such that $U_X \cap V_X = \emptyset$. There also exists some open neighborhood $V \ni y_1, y_2$. Then $(U_X \times V) \ni (x_1, y_1)$ and $(V_X \times V) \ni (x_2, y_2)$ are disjoint open neighborhoods of $X \times Y$, completing the proof.

Solution.

Suppose $X$ is a Hausdorff space and $Y \subseteq X$ is a subspace. Let $x \ne y \in Y$. Note there exists open neighborhoods $U \subseteq X$ of $x$ and $V \subseteq X$ of $y$ such that $U \cap V = \emptyset$. The open analogue of these neighborhoods in the subspace are sets of the form $U \cap Y$ and $V \cap Y$. Of course, $x \in U \cap Y$ and $y \in V \cap Y$, and note $(U \cap Y) \cap (V \cap Y) = (U \cap V) \cap Y = \emptyset \cap Y = \emptyset$, completing the proof.

Proof.

$(\Longrightarrow)$: Suppose $X$ is Hausdorff. Suppose $(x, y) \in X \times X$ is a limit point for $\Delta$. Then, for every open neighborhood $U \subseteq X \times X$ of $(x, y)$, it follows that $U \cap (\Delta \setminus (x, y)) \ne \emptyset$. For the sake of contradiction, suppose $x \ne y$. Then, by Hausdorffness of $X$, there exist disjoint open neighborhoods $U \subseteq X$ and $V \subseteq X$ such that $x \in U$ and $y \in V$. Note then that $U \times V$ is open in $X \times X$. Since $U$ and $V$ are disjoint, there are no tuples with the same entry in each slot when taking their Cartesian product—that is, $(U \times V) \cap (\Delta \setminus (x, y)) = \emptyset$, which is a contradiction. Thus $x = y$, implying $\Delta$ contains all its limit points and thus is closed.

$(\Longleftarrow)$: Suppose $\Delta$ is closed, and so $\Delta^c$ is open. Let $x \ne y \in X$, then observe $(x, y) \in \Delta^c$. Since $\Delta^c$ is an open subset of $X \times X$, there is some basis element $U \times V$ (where $U, V$ are open subsets of $X$) such that $(x, y) \in U \times V \subseteq \Delta^c$. For $U \times V$ to be a subset of $\Delta^c$, this implies no elements of $U \times V$ may be in $\Delta$: thus $U \cap V = \emptyset$. Thus we have constructed two disjoint open neighborhoods that contain $x$ and $y$ respectively, completing the proof. ❦

Solution.

Note a set $O \subseteq \bR$ is open if $O = \emptyset$, or $\bR \setminus O$ is finite (i.e., it misses finitely many points of $\bR$). Let $x \in \bR$ be arbitrary, and let $O \subseteq \bR$ be an arbitrary open neighborhood such that $x \in O$. Note $O$ can only miss finitely many elements of $\bR$, so $O$ must contain infinitely many elements of $x_n$. That is, there is some $N \in \bZ_+$ such that $x_n \in O$ for every $n \ge N$. Thus $x_n \to x$, but $x$ is arbitrary, so $(x_n)$ converges to every real number.

Solution.

Suppose $x \in \Int(A)$. Since $\Int(A) \subseteq A$, it follows that $x \in A$. For the sake of contradiction, suppose $x \in \partial A$, implying $x \in \bar{X \setminus A}$. Note $X \setminus A = (X \setminus A) \cap (X \setminus A)'$ (where $S'$ refers to the limit points of $S$). Since $x \in A$, $x \notin (X \setminus A)$, so it is only possible that $x \in (X \setminus A)'$. This implies that every open subset $U \subseteq X$ must be such that $U \cap (A^c \setminus \{x\}) \ne \emptyset$. But note $x \in \Int(A) \subseteq A$, meaning every open neighborhood of $x$ must be fully contained in $A$, which is a contradiction. Thus $\Int(A) \cap \partial A = \emptyset$.

Solution.

We prove this by forward and backward inclusion. First, suppose $x \in \bar{A}$, and suppose $x \notin \Int(A)$. This implies there are no $X$-open neighborhoods of $x$ that are subsets of $A$. Indeed, this implies the only $X$-open neighborhoods of $x$ must intersect $X \setminus A$. Thus $x \in \bar{X \setminus A}$, so $x \in \partial A$. Thus every $x \in \bar{A}$ must be either in $\Int(A)$ or $\partial A$, and so $\bar{A} \subseteq \Int(A) \cup \partial A$. Conversely, suppose $x \in \Int(A) \cup \partial A$. If $x \in \Int(A)$, then $x \in A \subseteq \bar{A}$. If $x \in (\partial A = \bar{A} \cap \bar{X \setminus A})$, then $x \in \bar{A}$, completing the proof.

Solution.

Suppose $A$ is $X$-clopen. Since $A$ is closed, $A = \bar{A}$. Let $x \in A = \bar{A}$ be a limit point of $A$. Then note every $X$-open neighborhood $O \ni x$ is such that $O \subseteq A$, so no open neighborhoods intersect $X \setminus A$. Thus $x \notin \bar{X \setminus A}$, implying $\partial A = \emptyset$. Conversely, suppose $\partial A = \bar{A} \cap \bar{X \setminus A} = \emptyset$. If $x$ is a limit point of $A$, then $x \in \bar{A}$, then it cannot be in $X \setminus A$ and so $x \in A$, meaning $A$ is closed. The same argument can be made to show $X \setminus A$ is also closed, meaning $A$ is open as well, completing the proof.