Topology without Tears - 2.2 Definitions and Propositions

Proposition:

A subset \(S\) of \(\mathbb{R}\) is open if and only if it is a union of open intervals.

Proof:

• We shall first prove that a union of open intervals is an open set in \(\mathbb{R}\). Let \(S = \bigcup_{\alpha \in \Gamma} (a_{\alpha},b_{\alpha})\) where \((a_{\alpha},b_{\alpha})\) is an open interval in \(\mathbb{R}\). We proved in the previous section that any open interval is an open set. From the definition of a topology, we have any arbitrary union of open sets to be an open set. Hence, \(S = \bigcup_{\alpha \in \Gamma} (a_{\alpha},b_{\alpha})\) is an open set.
• Now we shall now prove that any open set in \(\mathbb{R}\) is a union of open intervals. From the definition of an open set in \(\mathbb{R}\), we have that for every \(x \in S\), we can find an open interval \((a_x,b_x)\) containing \(x\) such that \((a_x,b_x) \subseteq S\). Now consider \(\cup_{x \in S} (a_x,b_x)\). We shall now prove that \(S = \cup_{x \in S} (a_x,b_x)\).

• We shall prove first that \(S \subseteq \cup_{x \in S} (a_x,b_x)\). Consider any \(y \in S\). We hence have \(y \in (a_y,b_y)\). Hence, \(y \in \cup_{x \in S} (a_x,b_x)\). Hence, \(S \subseteq \cup_{x \in S} (a_x,b_x)\).
• We shall prove next that \(S \supseteq \cup_{x \in S} (a_x,b_x)\). Consider any \(y \in \cup_{x \in S} (a_x,b_x)\). This means that \(y \in (a_x,b_x)\) for some \(x \in S\). But \((a_x,b_x) \subseteq S\). Hence, we get \(y \in S\). Hence, \(S \supseteq \cup_{x \in S} (a_x,b_x)\).

This gives us that a subset \(S\) of \(\mathbb{R}\) is open if and only if it is a union of open intervals.

Definition:

Let \((X, \tau)\) be a topological space. A collection \(\mathcal{B}\) of open subsets of \(X\) is said to be a basis for the topology \(\tau\) if every open set is a union of members of \(\mathcal{B}\).

Proposition:

Let \((X,\tau)\) be a discrete space and \(\mathcal{B}\) the family of all singleton subsets of \(X\); that is, \(\mathcal{B} = \{\{x\}: x \in X\}\). Then \(\mathcal{B}\) is a basis for \(\tau\).

Proof:

The discrete topology contains all the subsets of \(X\). Note that any subset, \(A \subseteq X\), can be written as \(A = \cup_{x \in A} \{x\}\). Hence, the singleton set generate the topology.

Proposition:

There are many different bases for the same topology. If \(\mathcal{B}\) is a basis for a topology \(\tau\) on a set \(X\) and \(\mathcal{B}_1\) is a collection of subsets of \(X\) such that \(\mathcal{B} \subseteq \mathcal{B}_1 \subseteq \tau\), then \(\mathcal{B}_1\) is also a basis for \(\tau\).

Proof:

The topology generated by \(\mathcal{B}\) is \(\tau\). Let \(\tau(\mathcal{B}_1)\) be the topology generated by \(\mathcal{B}_1\). We need to show that \(\tau(\mathcal{B}_1) = \tau\).

• We shall first show that \(\tau(\mathcal{B}_1) \supseteq \tau\). We are given that \(\mathcal{B} \subseteq \mathcal{B}_1\) and \(\tau(\mathcal{B}) = \tau\). This means that any element in \(\tau\) can be written as \(\bigcup_{A_{\alpha} \in \mathcal{B}} A_{\alpha}\). Since any element in \(\mathcal{B}\) is also an element of \(\mathcal{B}_1\), we have that any element in \(\tau\) can be written as \(\bigcup_{A_{\alpha} \in \mathcal{B}_1} A_{\alpha}\). Hence, we have \(\tau(\mathcal{B}_1) \supseteq \tau\).
• Now we shall show that \(\tau(\mathcal{B}_1) \subseteq \tau\). We are given that \(\mathcal{B}_1 \subseteq \tau\). Let us look at the topology generated by \(\mathcal{B}_1\). Any element in the \(\tau(\mathcal{B}_1)\) is an arbitrary union of elements in \(\mathcal{B}_1\). Consider any arbitrary union of elements in \(\mathcal{B}_1\) i.e. \(\bigcup_{A_{\alpha} \in \mathcal{B}_1} A_{\alpha}\). Since \(\mathcal{B}_1 \subseteq \tau\), we have that \(A_{\alpha} \in \tau\) for all \(A_{\alpha} \in \mathcal{B}_1\). Since \(\tau\) is a topology, it is closed under arbitrary unions and hence we get \(\bigcup_{A_{\alpha} \in \mathcal{B}_1} A_{\alpha} \in \tau\). Hence any element in \(\tau(\mathcal{B}_1)\) is also an element in \(\tau\). Hence, we get \(\tau(\mathcal{B}_1) \subseteq \tau\). 

Hence, we get that \(\mathcal{B}_1\) is also a basis for \(\tau\).

Proposition:

Let \(X\) be a non-empty set and let \(\mathcal{B}\) be a collection of subsets of \(X\). Then \(\mathcal{B}\) is a basis for a topology on \(X\) if and only if \(\mathcal{B}\) has the following properties:

1. \(X = \bigcup_{B \in \mathcal{B}} B\)
2. For any \(B_1,B_2 \in \mathcal{B}\), then the set \(B_1 \cap B_2\) is a union of members of \(B\).

Proof:

If \(\mathcal{B}\) is a basis for a topology \(\tau\), then any element in \(\tau\) can be written as an arbitrary union of elements in \(\mathcal{B}\). Since, \(X \in \tau\), we have that \(X = \bigcup_{\alpha \in \Gamma} B_{\alpha}\) where \(B_{\alpha} \in \mathcal{B}\). But \(\bigcup_{\alpha \in \Gamma} B_{\alpha} \subseteq \bigcup_{B \in \mathcal{B}} B\). Hence, we get \(X \subseteq \bigcup_{B \in \mathcal{B}} B\). Further, for any \(B \in \mathcal{B}\), we have \(B \subseteq X\). Hence, we get \(\bigcup_{B \in \mathcal{B}} \subseteq X\). Hence, we have \(X = \bigcup_{B \in \mathcal{B}} B\). Also, for any \(B_1,B_2 \in \mathcal{B}\), we also have \(B_1,B_2 \in \tau\). Since \(\tau\) is closed under intersections, we have \(B_1 \cap B_2 \in \tau\). Since \(\mathcal{B}\) generates \(\tau\), we have \(B_1 \cap B_2\) is a union of members of \(\mathcal{B}\).

To prove the other way around, that is, if \(X = \bigcup_{B \in \mathcal{B}} B\) and for any \(B_1,B_2 \in \mathcal{B}\), the set \(B_1 \cap B_2\) is a union of members of \(\mathcal{B}\), then \(\mathcal{B}\) is a basis for the topology on \(X\). Let \(\tau(\mathcal{B}) = \{\bigcup_{\alpha} B_{\alpha}: B_{\alpha} \in \mathcal{B}\}\). We need to prove that \(\tau\) is a topology.

1. Clearly, we have \(X, \emptyset \in \tau\). (This follows from the condition in the problem and from the fact that the empty set is a union of no sets.)
2. Any arbitrary union of elements in \(\tau\) is nothing but an arbitrary union of elements in \(\mathcal{B}\) which is again in \(\tau\).
3. Consider two elements in \(\tau\). These two elements can be written as \(\bigcup_{\alpha} B^1_{\alpha}\) and \(\bigcup_{\beta} B^2_{\beta}\) where \(B^1_{\alpha},B^2_{\beta} \in \mathcal{B}\) forall \(\alpha,\beta\). The intersection of these two elements is given by \(\left(\bigcup_{\alpha} B^1_{\alpha} \right) \cap \left(\bigcup_{\beta} B^2_{\beta} \right) = \bigcup_{\alpha,\beta} \left(B^1_{\alpha} \cap B^2_{\beta} \right)\). But we are given that intersection of two elements in \(\mathcal{B}\) can be written as union of members of \(\mathcal{B}\) i.e. \(B^1_{\alpha} \cap B^2_{\beta} = \bigcup_{\gamma} B_{\gamma,\alpha,\beta}\). Hence, the intersection of any two elements in \(\tau\) is given by \(\bigcup_{\alpha,\beta,\gamma} B_{\gamma,\alpha,\beta}\). Hence, closed under finite intersections.

Hence, \(\tau(\mathcal{B})\) is a topology.