General Topology: Topological Spaces
Bases
Topologies are defined by persnickety rules regarding unions and intersections, and these rules aren't always conducive to defining topologies directly. Thankfully, the concept of a basis allows us to define topologies in terms of other sets that can in turn be used to generate the open sets. Bases are recipes for topologies, if you will.
Let $X$ be a set. The collection $\mathcal{B}$ of subsets of $X$ is a basis for a topology on $X$ if the following two conditions hold:
- For each $x \in X$, there exists a $B \in \mathcal{B}$ such that $x \in B$.
- If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \bigcap B_2$, then there exists a third subset $B_3 \in \mathcal{B}$ containing $x$ such that $B_3 \subseteq B_1 \bigcap B_2$.
The elements of $\mathcal{B}$ are called the basis elements of $\mathcal{B}$.
Given a basis $\mathcal{B}$ for a set $X$, the topology generated by $\mathcal{B}$ is defined as follows: A subset $A \subseteq X$ is open if for each $x \in A$, there is a basis element $B$ such that $x \in B$ and $B \subseteq A$.
A subbasis $\mathcal{S}$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by $\mathcal{S}$ is the collection of all unions of finite intersections of $\mathcal{S}$.
Problems
Show that the union of two basis elements is not necessarily itself a basis element.
Let $X = \{a, b\}$, and let $\mathcal{B} = \{ \{a\}, \{b\} \}$. Then $\mathcal{B}$ is a basis of $X$, and the topology $\mathcal{T}$ generated by $\mathcal{B}$ is the discrete topology on $X$. Note that $\{a\}$ and $\{b\}$ are basis elements, but their union $\{a, b\}$ is not.
Prove that the topology generated by a basis is in fact a topology.
Let $X$ be a set and let $B$ be a topological basis of $X$. To prove that the topology $\mathcal{T}$ generated by $X$ is in fact a topology, we must show that it meets all four conditions of a topology.
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$\varnothing \in \mathcal{T}$ trivially. (That is, it's true that for all $x \in \varnothing$ there exists a basis element $B \in \mathcal{B}$ such that $x \in \varnothing$ because there are no $x \in \varnothing$.)
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For all $x \in X$, there exists by definition a basis element $B \in X$ such that $x \in B$. Therefore $X \in \mathcal{T}$.
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Proof by induction.
Base case: Let $A_1, A_2 \in \mathcal{T}$, and let $x \in (A_1 \bigcap A_2)$. There exist two basis elements $B_1 \subseteq A_1$ and $B_2 \subseteq A_2$ such that $x \in B_1$ and $x \in B_2$. Therefore $x \in (B_1 \bigcap B_2)$. By the second property of bases, there exists some third basis element $B_3$ such that $x \in B_3 \subseteq B_1 \bigcap B_2$. Therefore $B_3 \subseteq (A_1 \bigcap A_2)$, so $(A_1 \bigcap A_2) \in \mathcal{T}$.
Inductive step: Assume $\bigcap\limits_{i=1}^{n} A_i \in \mathcal{T}$. Consider a set $A_{n+1} \in \mathcal{T}$. Then $\bigcap\limits_{i=1}^{n+1} A_i= \left( \bigcap\limits_{i=1}^{n} A_i \right)\bigcap A_{n+1}$. Since the base case shows that the intersection of two open sets is in $\mathcal{T}$, $\bigcap\limits_{i=1}^{n+1} A_i \in \mathcal{T}$.
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Let $\{A_{\alpha}\}_{\alpha \in I}$ be an indexed collection of open subsets. Let $x \in \bigcup \{A_{\alpha}\}_{\alpha \in I}$. Then there exists some $\alpha \in I$ such that $x \in A_{\alpha}$. Therefore there exists a basis element $B \in \mathcal{B}$ such that $x \in B \subseteq A_{\alpha}$. Therefore $B \subseteq \bigcup \{A_{\alpha}\}_{\alpha \in I}$, so $\bigcup \{A_{\alpha}\}_{\alpha \in I} \in \mathcal{T}$.
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Let $\mathcal{B}$ be a basis for a set $X$. Prove that the topology $\mathcal{T}$ generated by $\mathcal{B}$ is the collection of all possible unions of elements of $\mathcal{B}$.
Hint: Show that all unions of basis elements are open sets and all open sets are unions of basis elements.
Let $\mathcal{B}_0$ be a subcollection of basis elements. Then for each $x \in \bigcup \mathcal{B}_0$ there exists a basis element $B_x \in \mathcal{B}_0$ such that $x \in B_x$. Therefore $\mathcal{B}_0$ is an open set. Conversely, let $A$ be an open set. Then for each $x \in A$ there exists a basis element $B$ such that $x \in B_x \subseteq A$. Therefore $A$ is the union of all such basis elements.
Let $\mathcal{B}$ be a basis for a topology on a set $X$. Show that the topology $\mathcal{T}$ generated by $\mathcal{B}$ is the intersection of all topologies on $X$ that contain $\mathcal{B}$.
Let $\{\mathcal{R}_{\alpha}\}$ be the indexed collection of topologies that each contain $\mathcal{B}$. Then $\mathcal{B} \subseteq \bigcap \{\mathcal{R}_{\alpha}\}$, and by the proof from the previous section, $\bigcap \{\mathcal{R}_{\alpha}\}$ is a topology. We must show that $\mathcal{T} = \bigcap \{\mathcal{R}_{\alpha}\}$.
Since $\bigcap \{\mathcal{R}_{\alpha}\}$ is a topology that contains $\mathcal{B}$, it contains all unions of elements of $\mathcal{B}$, thus it contains the topology $\mathcal{T}$ generated by $\mathcal{B}$. Conversely, let $A \in \mathcal{T}$. Then $A$ is a union of basis elements in $\mathcal{B}$. Because each $\mathcal{R}_{\alpha}$ contains $\mathcal{B}$, it contains all unions of basis elements in $\mathcal{B}$, therefore it contains $A$. Since $T \subseteq \mathcal{R}_{\alpha}$ for all $\mathcal{R}_{\alpha}$, $T \subseteq \bigcap \{\mathcal{R}_{\alpha}\}$. Therefore $T = \bigcap \{\mathcal{R}_{\alpha}\}$.
Let $\mathcal{T}$ be a topology on $X$, and let $\mathcal{R}$ be a collection of open sets in $\mathcal{T}$ such that for each open set $T \in \mathcal{T}$ and $x \in T$, there exists an open set $R_x \in \mathcal{R}$ such that $x \in R_{x} \in U$. Prove that $\mathcal{R}$ is a basis for $\mathcal{T}$.
We must show that $\mathcal{R}$ is a basis, so we check both the conditions. First, for all $x \in X$, there exists a basis element $R_x \in \mathcal{R}$ such that $x \in R_x$. Since $X \in \mathcal{T}$, $X \in \mathcal{R}$, so $R_x = X$ fulfills this requirement. Second, let $x \in X$, and assume $x \in R_1 \bigcap R_2$, where $R_1, R_2 \in \mathcal{R}$. Because the intersection of open sets is open, $R_1 \bigcap R_2$ is open and thus in $\mathcal{R}$. By hypothesis, there thus exists some $R_3 \in \mathcal{R}$ such that $x \in R_3 \subseteq R_1 \bigcap R_2$. Therefore $\mathcal{R}$ is a basis for a topology on $X$.
Next we must show that the topology on $X$ generated by $\mathcal{R}$, call it $\mathcal{T}'$, is in fact $\mathcal{T}$. First, let $T \in T$. Then for all $x \in T$, there exists an $R_x \in \mathcal{R}$ such that $x \in R_x \in T$. Therefore $T$ is a union of basis elements in $\mathcal{R}$, so $T \in \mathcal{T}'$. Therefore $T \subseteq \mathcal{T}'$. Conversely, let $T' \in \mathcal{T}'$. Then $\mathcal{T}'$ is a union of basis elements in $\mathcal{R}$. Because each basis element in $\mathcal{R}$ is open in $\mathcal{T}$, their union $T'$ is also open in $\mathcal{T}$. Therefore $\mathcal{T}' \subseteq \mathcal{T}$. Thus $\mathcal{T} = \mathcal{T}'$.
Let $X$ be a set, and let $\mathcal{B}_1$ be a basis for a topology $\mathcal{T}_1$ on $X$, and let $\mathcal{B}_2$ be a basis for a topology $\mathcal{T}_2$ on $X$.
Prove that $\mathcal{T}_2$ is finer than $\mathcal{T}_1$ if and only if for each $x \in X$ and $B_1 \in \mathcal{B}_1$, there is a basis element $B_2 \in \mathcal{B}_2$ such that $x \in \mathcal{B}_2 \in \mathcal{B}_1$.
Assume $\mathcal{T}_2$ is finer than $\mathcal{T}_1$. Then $\mathcal{T}_1 \subset \mathcal{T}_2$. Then each $B_1 \in \mathcal{B}_1$ is a union of basis elements in $\mathcal{B}_2$. Therefore for each $x \in B_1$, there exists some basis element $B_2 \in \mathcal{B}_2$ such that $x \in \mathcal{B}_2 \subseteq \mathcal{B}_1$.
Conversely, assume that for each $x \in X$ and $B_1 \in \mathcal{B}_1$ that contains that $x$ there exists some $B_2 \in \mathcal{B}_2$ such that $x \in \mathcal{B}_2 \in \mathcal{B}_1$. Then $B_1$ is a union of basis elements in $B_2$. Because each open set $T_1 \in \mathcal{T}_1$ is a union of basis elements in $\mathcal{B}_1$, it follows that it is also a union of basis elements in $\mathcal{B}_2$. Therefore $T_1$ is also open in $\mathcal{T}_2$, and so $\mathcal{T}_1 \subseteq \mathcal{T}_2$.
Show that the topology $\mathcal{T}$ generated by a subbasis $\mathcal{S}$ on a set $X$ is a topology.
Rather than show that $\mathcal{T}$ is a topology directly, we can show that the set of all finite intersections of $\mathcal{S}$ forms a basis. Since the set of all unions of basis elements forms a topology, we will have shown that $\mathcal{S}$ indeed generates a topology.
Let $x \in X$, and let $\mathcal{B}$ denote the set of all finite intersections of elements in $\mathcal{S}$. To check the first condition of a basis, note that by definition of subbasis, there is a subbasis element $S_x \in \mathcal{S}$ such that $x \in S_x$. Remember that the intersection of a single set is the set itself, so $S_x \in \mathcal{B}$. To check the second condition, consider two sets $B_1$ and $B_2$ that are finite intersections of subbasis elements of $\mathcal{S}$. Then $B_3 = B_1 \bigcap B_2$ is a finite intersection of subbasis elements, so it belongs to $\mathcal{B}$. Therefore $\mathcal{B}$ forms a basis for a topology on $X$, so we conclude that the topology $\mathcal{T}$ generated by the subbasis $\mathcal{S}$ is indeed a topology.
Consider the Euclidean plane $\mathbb{R}^2$. Define the open disk $D_{\mathbf{x}_0,r}$ in $\mathbb{R}^2$ as
$$D_{\mathbf{x}_0,r} = \{ \mathbf{x} \in \mathbb{R}^2 : ||\mathbf{x} - \mathbf{x}_0|| < r \}.$$
Show that the collection $\mathcal{B}$ of all open disks of all radii form a basis for a topology on $\mathbb{R}^2$.
The first condition is easily checked. Let $\mathbf{x}_0 \in \mathbb{R}^2$. Then $\mathbf{x}_0\in D_{\mathbf{x}_0, 1} \in \mathcal{B}$. For the second, let $\mathbf{x} \in D_{\mathbf{x}_1, r_1} \bigcap D_{\mathbf{x}_2, r_2}$. Denote the distance from $x$ to the perimeter of each disk as $d_1 = r_1 - ||\mathbf{x} - \mathbf{x}_1||$ and $d_2 = r_2 - ||\mathbf{x} - \mathbf{x}_2||$. Select $d = \frac{1}{2}\text{min}(d_1, d_2)$ as half the smaller of those two distances. Then $D_{\mathbf{x}, d} \in \mathcal{B}$ and $x \in D_{\mathbf{x}, d} \subset D_1 \bigcap D_2$. Therefore $\mathcal{B}$ is a topology.
Provide a counterexample to show that the union of a basis and another collection of subsets is not necessarily another basis.
Consider the basis $\mathcal{B}$ for the topology on $\mathbb{R}^2$ formed out of open disks from problem 8. Let $C= \{(0,0)\}\cup D_{(3,3),1}\}$, where $(0,0)$ denotes the origin rather than an empty interval, and let $\mathcal{C} = \{C\}$. Then for $D_{(0,0),1} \in \mathcal{B}$, $D_{(0,0),1} \cap C = \{(1,1)\}$, but there is no basis element in $\mathcal{B} \cup \mathcal{C}$ that is a subset of $\{(0,0)\}$ . Therefore $\mathcal{B} \cup \mathcal{C}$ is not a basis.