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Math 344-1,2: Kursobjekt

Aaron Greicius

Section 1.2 Topological basis

When defining a topology \(\mathcal{T}\) for a set \(X\) it is often convenient to specify instead a more manageable subset of \(\mathcal{T}\) that we can use to generate the entire topology using arbitrary unions and/or finite intersections (our open set operators). A topological basis is a particularly nice example of this (1.2.1) as the open sets of the topology it generates can be expressed as (arbitrary) unions of the basis elements. Bases will also provide a convenient way of comparing different topologies and (later on) determining whether relevant topological properties are satisfied by a given space.

Definition 1.2.1. Topological basis.

Let \(X\) be a set. A topological basis on \(X\) is a collection \(\mathcal{B}\) of subsets of \(X\) satisfying the following axioms:
  1. Covers \(X\).
    For each \(x\in X\) there is a set \(B\in \mathcal{B}\) such that \(x\in B\text{.}\)
  2. Intersections covered by basis elements.
    If \(B_1, B_2\in \mathcal{B}\text{,}\) then for all \(x\in B_1\cap B_2\) there is a \(B\in \mathcal{B}\) such that \(x\in B\subseteq B_1\cap B_2\text{.}\)
When the context is clear, we usually shorten ‘topological basis’ to ‘basis’.

Remark 1.2.2. Equivalent alternative to axiom (2).

Axiom (2) in Definition 1.2.1 is equivalent to saying that \(B_1\cap B_2\) is equal to a union of elements of \(\mathcal{B}\text{.}\) In other words, we could replace (2) with the equivalent statement
\begin{equation*} (2') \ B_1, B_2\in \mathcal{B}\implies B_1\cap B_2=\bigcup_{\alpha\in I} B_\alpha \text{ for some } B_\alpha\in \mathcal{B}\text{.} \end{equation*}
This is a consequence of the more general fact expressed in Lemma 1.2.3.

Proof.

Implication: \((1)\implies (2)\).

Suppose \(Y=\bigcup_{i\in I}A_i\) for some sets \(A_i\in \mathcal{A}\text{.}\) Given any \(y\in Y\) we have \(y\in A_i\) for some \(i\in I\text{.}\) Since \(A_i\subseteq Y\text{,}\) we are done.

Implication: \((2)\implies (1)\).

By assumption, for all \(y\in Y\) we can find a set \(A_y\in \mathcal{A}\) such that \(y\in A_y\subseteq Y\text{.}\) It follows that \(Y=\bigcup_{y\in Y}A_y\text{,}\) showing that \(Y\) is a union of elements of \(\mathcal{A}\text{.}\)

Example 1.2.4. Some bases for \(\R\).

Show that the given collections \(\mathcal{B}\) are topological bases of \(\R\text{.}\)
  1. \(\displaystyle \mathcal{B}_1=\{(x-\epsilon, x+\epsilon)\colon x,\epsilon\in \R, \epsilon> 0\}=\{(a,b)\colon a,b\in\R, a< b\}\)
  2. \(\displaystyle \mathcal{B}_2=\{ (p,q)\colon p,q\in \Q, p< q\} \)
  3. \(\displaystyle \mathcal{B}_3=\{[a,b)\colon a,b\in \R, a< b\}\)
Solution.
It is clear that each collection of interval types covers \(\R\) in the sense of axiom (1). As for axiom (2), for each \(\mathcal{B}_i\text{,}\) \(1\leq i\leq 3\text{,}\) note that intersection of these types of intervals are either empty, or another interval of the same type. This is a result of the fact that
\begin{align*} (a,b)\cap (c,d)\amp = \left(\max\{a,c\},\min\{b,d\}\right) \\ [a,b)\cap [c,d) \amp = \left [\max\{a,c\}, \min\{b,d\}\right) \text{,} \end{align*}
where the interval described on the right is understood to be the empty set if \(\max(a,c)\geq \min(b,d)\text{.}\)

Proof.

Note: it is somewhat easier to show that \(\mathcal{T}\) is a topology using the description of \(\mathcal{T}\) in statement (1), and then to show that the description in statement (2) also holds. (This is how Munkres does it, and how I did it in class.) I’ve decided to go the hard way here as an alternative proof that puts us through some useful set theory calisthenics.
First we show that \(\mathcal{T}\) as defined in (2) is a topology. We treat each axiom separately.
  1. The empty set is the “empty union” of elements of \(\mathcal{B}\text{;}\) thus \(\emptyset\in \mathcal{T}\text{.}\) The covering axiom for bases implies that for all \(x\in X\) there is a \(B_x\in\mathcal{B}\) such that \(x\in B_x\text{.}\) It follows that \(X=\bigcup_{x\in X}B_x\) is a union of elements of \(\mathcal{B}\text{,}\) and thus that \(X\in \mathcal{T}\text{.}\)
  2. In plain English, a union of unions of elements of \(\mathcal{B}\) is a union of elements of \(\mathcal{B}\text{.}\) More rigorously, assume we have a collection of \(U_i\in \mathcal{T}\text{.}\) By definition, for each \(i\in I\) we have \(U_i=\bigcup_{j\in J_i} B_{i,j}\text{,}\) where \(B_{i,j}\in \mathcal{B}\) for each \(j\in J\text{.}\) Thus
    \begin{align*} \bigcup_{i\in I} U_i \amp = \bigcup_{i\in I}\bigcup_{j\in J_i} B_{i,j} \\ \amp \bigcup_{(i,j)\in K} B_{i,j} \text{,} \end{align*}
    where \(K\) is the set of all pairs of the form \((i,j)\) for \(i\in I\) and \(j\in I_j\text{.}\) Thus \(U=\bigcup_{i\in I} U_i\) is a union of basis elements, and hence an element of \(\mathcal{T}\text{.}\)
  3. It is enough to show that \(U, V\in \mathcal{T}\implies U\cap V\in \mathcal{T}\text{.}\) (See Remark 1.1.2.) Assume \(U=\bigcup_{i\in I}B_i\text{,}\) \(V=\bigcup_{j\in J}B_j\text{,}\) where \(B_i, B_j\in \mathcal{B}\) for all \(i\in I, j\in J\text{.}\) We have
    \begin{align*} U\cap V \amp= \left(\bigcup_{i\in I}B_i \right) \cap \left( \bigcup_{j\in J}B_j\right) \\ \amp = \bigcup_{(i,j)\in I\times J} B_i\cap B_j \text{.} \end{align*}
    Now by axiom (2’) for bases (see Remark 1.2.2), each \(B_i\cap B_j\) is a union of basis elements, and hence is an element \(U_{i,j}\in \mathcal{T}\text{.}\) We have already shown that \(\mathcal{T}\) is closed under arbitrary unions. Thus \(U\cap V\) is an element of \(\mathcal{T}\text{.}\)
This shows that \(\mathcal{T}\text{,}\) as defined in statement (2) of the theorem, is a topology. Statement (1) now follows from the general claim made in Remark 1.2.2.

Definition 1.2.6. Topology generated by basis.

Let \(\mathcal{B}\) be a topological basis of the set \(X\text{.}\) The topology
\begin{align} \mathcal{T} \amp = \{U\subseteq X\colon \text{for all } x\in U \text{ there exists } B\in \mathcal{B} \text{ s.t. } x\in B\subseteq U\}\tag{1.2.1}\\ \amp =\{U\subseteq X\colon U=\bigcup_{i\in I}B_i \text{ for some elements } B_i\in \mathcal{B}\} \tag{1.2.2} \end{align}
is called the topology generated by the basis \(\mathcal{B}\text{.}\) Similarly, we call \(\mathcal{B}\) a basis for \(\mathcal{T}\text{.}\)

Proof.

Observe first that basis elements are open in the topology \(\mathcal{T}\text{,}\) essentially by definition: i.e., we have \(\mathcal{B}\subseteq \mathcal{T}\text{.}\) The forward implication \((\implies)\) follows easily, since \(\mathcal{T}\subseteq \mathcal{T}'\) and \(\mathcal{B}\subseteq \mathcal{T}\) implies \(\mathcal{B}\subseteq \mathcal{T}'\text{.}\)
Now consider the reverse implication \((\impliedby)\text{.}\) Assume \(\mathcal{B}\subseteq \mathcal{T}'\text{.}\) Given an open set \(U\in \mathcal{T}\) we have \(U=\bigcup_{i\in I} B_i\) for some \(B_i\in \mathcal{B}\) by (1.2.2). Since \(B_i\in \mathcal{T}'\) for all \(i\in I\) (by assumption), and since \(\mathcal{T}'\) is closed under arbitrary unions, we conclude \(U\in \mathcal{T}'\text{.}\) This proves \(\mathcal{T}\subseteq \mathcal{T}'\text{,}\) as desired.

Example 1.2.8. A basis for the standard topology on \(\R\).

Show that the standard topology on \(\R\) is the topology generated by the basis \(\mathcal{B}_1\) in Example 1.2.4.
Solution.
Let \(\mathcal{T}\) be the topology generated by \(\mathcal{B}_1\) and let \(\mathcal{T}'\) be the standard topology on \(\R\). Since every open interval \((a,b)\) is open in \(\mathcal{T}'\) each element of \(\mathcal{B}_1\) is an element of \(\mathcal{T}'\text{.}\) By Theorem 1.2.7, we have \(\mathcal{T}\subseteq \mathcal{T}'\text{.}\) Next, take any open set \(U\in\mathcal{T}'\text{;}\) then by definition for all \(x\in U\) there exists \(\epsilon_x> 0\) such that \((x-\epsilon_x, x+\epsilon_x)\subseteq U\text{.}\) It follows that \(U\in \mathcal{T}\) by (1.2.1). This proves \(\mathcal{T}'\subseteq \mathcal{T}\text{,}\) and thus that \(\mathcal{T}=\mathcal{T}'\text{,}\) as desired.

Example 1.2.9. Multiple bases for the standard real topology.

Show that the bases \(\mathcal{B}_1, \mathcal{B}_2\) from Example 1.2.4 both generate the standard topology on \(\R\text{.}\)
Solution.
Let \(\mathcal{T}_i\) be the topology generated by \(\mathcal{B}_i\text{,}\) \(1\leq i\leq 2\text{.}\) We must show that \(\mathcal{T}_1\subseteq \mathcal{T}_2\) and \(\mathcal{T}_2\subseteq \mathcal{T}_1\text{.}\) Clearly any \((p,q)\in \mathcal{B}_2\) is an element of \(\mathcal{B}_1\text{,}\) and hence an element of \(\mathcal{T}_1\text{.}\) It follows from (1.2.3) that \(\mathcal{T}_2\subseteq \mathcal{T}_1\text{.}\) Next, take any \((a,b)\in \mathcal{B}_1\text{.}\) Since the rational numbers are dense in \(\R\text{,}\) for any \(x\in (a,b)\) we can find \(p,q\in\R\) satisfying \(x\in (p,q)\subseteq (a,b)\text{.}\) From (1.2.1) it follows that \((a,b)\in \mathcal{T}_2\) for all \((a,b)\in \mathcal{B}_1\) and hence that \(\mathcal{T}_1\subseteq \mathcal{T}_2\) by Theorem 1.2.7.
As it turns out, the basis \(B_3\) of Example 1.2.4 generates a nonstandard topology of \(\R\text{.}\) Accordingly, we give this basis (and its corresponding topology) an official name.

Topological specimen 4. Lower limit topology.

The collection \(\mathcal{B}_l=\{([a,b)\colon a,b\in \R, a< b\}\) is a topological basis of \(\R\) called the lower limit basis, and the topology \(\mathcal{T}_l\) it generates is called the lower limit topology on \(\R\text{.}\) We denote the topological space \((\R, \mathcal{T}_l)\) as \(\R_l\text{.}\)

Example 1.2.10. Lower limit is finer than standard.

Show that the lower limit topology on \(\R\) is strictly finer than the standard one.
Solution.
Let \(\mathcal{T}, \mathcal{T}'\) be the Euclidean and lower limit topologies, respectively. According to Theorem 1.2.7, to show \(\mathcal{T}\subseteq \mathcal{T}'\) we must show that for any basis element \(B=(a,b)\in \mathcal{T}\) and any \(x\in B\text{,}\) there is a basis element \([c,d)\in \mathcal{T}'\) satisfying \(x\in [c,d)\subseteq (a,b)\text{.}\) Since \(x\in (a,b)\text{,}\) we have \(a < x < b\text{,}\) and thus we have \(x\in [x,b)\subseteq (a,b)\text{,}\) as desired.
To show \(\mathcal{T}'\) is strictly finer than \(\mathcal{T}\) it suffices to find a basis element \([a,b)\in \mathcal{T}'\) that is not an element of \(\mathcal{T}\text{.}\) I claim \([0,1)\notin\mathcal{T}\text{.}\) Indeed, consider \(0\in [0,1)\text{.}\) Any open set \((a,b)\) containing \(0\) must contain negative numbers, and hence does not satisfy \((a,b)\subseteq [0,1)\text{.}\) It follows that \([0,1)\) is not an element of
\begin{equation*} \mathcal{T}=\{U\subseteq X\colon \text{for all } x\in U \text{ there exists } (a,b) \text{ s.t. } x\in (a,b)\subseteq U \}\text{.} \end{equation*}

Topological specimen 5. \(K\)-topology.

Let \(K=\{1/n\colon n\in \Z_+\}=\{1, 1/2, 1/3,\dots\}\text{,}\) and define
\begin{equation*} \mathcal{B}_K=\{(a,b)\colon a,b\in \R, a< b\}\cup \{ (a,b)-K\colon a,b\in \R, a< b\}\text{.} \end{equation*}
The collection \(\mathcal{B}_K\) is a basis in \(\R\text{.}\) The topology \(\mathcal{T}_K\) it generates is called the \(K\)-topology on \(\R\text{.}\) We denote the topological space \((\R, \mathcal{T}_K)\) as \(\R_K\text{.}\)

Proof.

We show that \(\mathcal{B}_K\) is a basis, treating each axiom separately.
  1. Since by definition \(\mathcal{B}_K\) contains all finite open intervals, it clearly covers \(\R\text{.}\)
  2. We claim that in fact given \(B_1, B_2\in \mathcal{B}_K\text{,}\) their intersection \(B_1\cap B_2\) is either empty or another element of \(\mathcal{B}_K\text{.}\) Axiom (2) follows directly from this claim.
    Why is the claim true? An element of \(\mathcal{B}\) is either of the form \(I\) or \(I-K\text{,}\) where \(I=(a,b)\) is an open interval. There are then three different cases describing intersections of two elements, as shown below. Here \(I_1, I_2\) denote finite open intervals.
    \begin{align*} I_1\cap I_2 \amp \\ (I_1-K)\cap I_2\amp =I_1\cap I_2-K\cap I_2=I_1\cap I_2-K \\ (I_1-K)\cap (I_2-K) \amp= (I_1-K)\cap I_2 - K\cap (I_2-K)\\ \amp = I_1\cap I_2-K\text{.} \end{align*}
    As we have already remarked, \(I_1\cap I_2\) is either empty or a finite open interval \(I\text{.}\) The claim now follows easily.

Example 1.2.11. Comparing standard, lower limit, and \(K\)-topology.

  1. Show that the \(K\)-topology is strictly finer than the standard topology on \(\R\text{.}\)
  2. Show that the lower limit topology and \(K\)-topology are incomparable.
Solution.
Let \(\mathcal{T}, \mathcal{T}', \mathcal{T}''\) be the Euclidean, lower limit, and \(K\)-topologies, respectively.
  1. By definition, the basis for the \(K\)-topology contains all finite open intervals. It follows from (1.2.3) that \(\mathcal{T}\subseteq \mathcal{T}''\text{.}\) Next, using an argument similar to the one in Example 1.2.10, we show that the basis element \((-1,1)-K\in \mathcal{T}''\) is not open in the standard topology. Indeed, consider the element \(0\in (-1,1)-K\text{.}\) For any \(\epsilon > 0\) the open interval \((-\epsilon,\epsilon)\) containing \(0\) must contain an element of the form \(1/N\) for some positive integer \(N > 1\text{.}\) Since \(1/N\notin (-1,1)-K\text{,}\) it follows that \((-\epsilon, \epsilon)\not\subseteq (-1,1)-K\text{.}\) We conclude that \((-1,1)-K\) is not open in the standard topology.
  2. This is left as an exercise.

Proof.

We prove both implications of the given “if and only if” statement separately.

Case: \((\implies)\).

Assume \(\mathcal{T}\) is generated by \(\mathcal{B}\text{.}\) According to the description of \(\mathcal{T}\) in (1.2.1), for all \(U\in \mathcal{T}\) and \(x\in U\) there is an element \(B\in \mathcal{B}\) such that \(x\in B\subseteq U\text{.}\)

Case: \((\impliedby)\).

Assume for that for all \(U\in \mathcal{T}\) and \(x\in U\) there is an element \(B\in \mathcal{B}\) such that \(x\in B\subseteq U\text{.}\) Taking \(U=X\) we see in particular that for all \(x\in X\) there is a \(B\in \mathcal{B}\) such that \(x\in B\text{.}\) Thus \(\mathcal{B}\) satisfies axiom (1) for a basis. Next, let \(B_1, B_2\in \mathcal{B}\) and take \(x\in B_1\cap B_2\text{.}\) Since \(B_1, B_2\in \mathcal{T}\) are open, so is \(U=B_1\cap B_2\text{.}\) Our assumption now implies that there is a \(B\in \mathcal{B}\) such that \(x\in B\subseteq U=B_1\cap B_2\text{.}\) Thus axiom (2) of Definition 1.2.1 holds, and we condlude \(\mathcal{B}\) is a basis.

Example 1.2.13. Bases for discrete topology.

Let \(X\) be a set, and let \(\mathcal{T}\) be the discrete topology on \(X\text{.}\) Show the following:
  1. \(\mathcal{B}=\{ \{x\}\colon x\in X\}\) is a basis for \(\mathcal{T}\text{.}\)
  2. If \(\mathcal{B}'\) is a basis for \(\mathcal{T}\text{,}\) then \(\mathcal{B}\subseteq \mathcal{B}'\text{.}\) In other words, \(\mathcal{B}\) is the minimal basis of the discrete topology.
Solution.
This is left as an exercise.

Topological specimen 6. Order topology.

Let \(X\) be a set equipped with a total ordering: i.e., a binary relation \(\leq\) satisfying the following four axioms:
  • Reflexive.
    For all \(x\in X\text{,}\) we have \(x\leq x\text{.}\)
  • Transitive.
    For all \(x,y,z\in X\text{,}\) if \(x\leq y\) and \(y\leq z\text{,}\) then \(x\leq z\text{.}\)
  • Antisymmetric.
    For all \(x,y\in X\text{,}\) if \(x\leq y\) and \(y\leq x\text{,}\) then \(x=y\text{.}\)
  • Comparability.
    For all \(x,y\in X\text{,}\) we have \(x\leq y\) or \(y\leq x\text{.}\)
Given the total ordering \(\leq\) on \(X\text{,}\) we define the relation \(<\) as \(x< y\) if \(x\leq y \) and \(x\ne y\text{.}\) A smallest element of \(X\) is an element \(x_0\in X\) satisfying \(x_0\leq y\) for all \(y\in X\text{;}\) a largest element of \(X\) is an element \(y_0\in X\) satisfying \(x\leq y_0\) for all \(x\in X\text{.}\) Lastly, given elements \(x,y\in X\) we define the following subsets of \(X\text{:}\)
\begin{align*} (x,y) \amp =\{z\in X\colon x< z \text{ and } z< y\} \\ [x,y] \amp =\{z\in X\colon x\leq z \text{ and } z\leq y\} \\ [x,y) \amp =\{z\in X\colon x\leq z \text{ and } z< y\} \\ (x,y] \amp =\{z\in X\colon x< z \text{ and } z\leq y\} \text{.} \end{align*}
An set of the form \((x,y)\) is called an open interval and a set of the form \([x,y]\) is called a closed interval; sets of the form \([x,y)\) or \((x,y]\) are called half-open intervals.
With this notation in place, we define \(\mathcal{B}\) to be the collection of all subsets \(B\subseteq X\) satisfying one of the following properties:
  1. \(B=(x,y)\) for some \(x,y\in X\text{.}\)
  2. \(B=[x_0, y)\) for some \(y\in X\text{,}\) and some smallest element \(x_0\ in X\text{.}\)
  3. \(B=(x,y_0]\) for some \(x\in X\) and some largest element \(y_0\in X\text{.}\)
The collection \(\mathcal{B}\) is a topological basis and the topology \(\mathcal{T}\) it generates is called the order topology of \(X\) with respect to the total ordering \(\leq\text{.}\)

Proof.

The proof is left as an exercise.
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