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Section 2.3 Specifying topologies efficiently

A common way of specifying a topology is to speak of the coarsest topology such that blah or the finest topology such that blah. Specifying a topology this way has its pluses and minuses: on one hand, these properties often make it easy to verify features of these topologies, but on the other hand, one has to do the work to actually confirm that a finest or coarsest topology with the given property exists. Let's do some of that work now.

Let \(X\) be a topological space, and let \(Y\subseteq X\) is a subspace, then \(Y\) has the coarsest topology such that the inclusion map \(\into{Y}{X}\) is continuous.

Here's a handy fact of set theory that we'll be using a lot in this section. Let \(X\) be a set, and let \(A \subseteq \PP(X)\) be a collection of subsets of \(X\text{.}\) Then the following are equivalent for a subset \(U \subseteq X\text{:}\)

  • The set \(U\) can be expressed as the union of elements of \(A\text{.}\)

  • For every point \(u \in U\text{,}\) there exists an element \(V \in A\) such that \(u \in V \subseteq U\text{.}\)

Let's define \(O\subseteq\PP(X)\) as the collection of all those subsets \(U\subseteq X\) such that for any point \(x\in U\text{,}\) there exist finitely many elements \(V_1,\dots,V_n\in B\) such that \(x\in V_1\cap\cdots\cap V_n\subseteq U\text{.}\) It is easy to see that \(O\) is a system of open sets for a topology on \(X\text{.}\) We claim that this is the desired topology \(\tau_B\text{.}\) Indeed, every element of \(B\) is an element of \(O\text{,}\) and at the same time, every element of \(O\) is a union of finite intersections of elements of \(B\text{,}\) and hence must be open in any topology in which the elements of \(B\) are open.

Definition 2.3.3.

In the situation of the previous exercise, one says that \(B\) generates the topology on \(X\) or that \(B\) is a subbase

for the topology on \(X\text{.}\)

The desired topology on \(X\) is the one generated by the set

\begin{equation*} \{f_{\alpha}^{-1}(U) : (\alpha\in A)\wedge(U\subseteq Y_{\alpha}\text{ is open } ) \}\text{.} \end{equation*}

If \(g\) is continuous, then since the composition of continuous maps is continuous, it follows that each \(f_{\alpha} \circ g\) is continuous too.

Conversely, let us assume that each \(f_{\alpha} \circ g\) is continuous. Now let \(U \subseteq X\) be an open set. Since \(X\) has the initial topology with respect to \(F\text{,}\) it follows that \(U\) is a union of finite intersections of sets of the form \(f_{\alpha}^{-1}(V)\) for \(\alpha \in A\) and \(V \subseteq Y_{\alpha}\) open. Consequently, \(g^{-1}(U)\) is a union of finite intersections of sets of the form \(g^{-1}(f_{\alpha}^{-1}(V))\) for \(\alpha \in A\) and \(V \subseteq Y_{\alpha}\text{.}\) Since \(f_{\alpha} \circ g\) is continuous, it follows that \(g^{-1}(U)\) is open.

Dually, but simpler, suppose \(X\) a set, and suppose \(F \coloneq \{ f_{\alpha} \colon \fromto{Y_{\alpha}}{X} \}_{\alpha\in A}\) an indexed family of maps. Suppose that each source \(Y_{\alpha}\) is endowed with a topology \(\tau_{\alpha}\text{.}\)

  • There exists a unique finest topology on \(X\) relative to which every map of \(F\) is continuous. This is called the final topology with respect to \(F\text{.}\) A subset \(U \subseteq X\) is open with respect to the final topology if and only if, for every \(\alpha \in A\text{,}\) the inverse image \(f_{\alpha}^{-1}(U) \subseteq Y_{\alpha}\) is open.

  • A map \(g\colon\fromto{X}{X'}\) is continuous if and only if, for each \(\alpha\in A\text{,}\) the composite \(g\circ f_{\alpha}\) is continuous.

Each of the following sets form a subbase for the standard topology on \(\RR\text{;}\)

  • the set of all open subsets of \(\RR\text{;}\)

  • the set \(\left\{\left]a,b\right[ : a,b\in\RR\right\}\) of open intervals; and

  • the set \(\left\{\left]a,+\infty\right[ : a\in\RR\}\cup\{\left]-\infty,b\right[ : b\in\RR\right\}\) of open rays.

Let \(X \subseteq \RR^n\) be a subspace of Euclidean space. The set \(\{B(x,\varepsilon) : x\in X,\ \varepsilon>0\}\) is a subbase for the induced topology on \(X\text{.}\)

If \(B\) is a subbase for a topology, then any open set in that topology is a union of finite intersections of subbase elements. One might wish for a better subbase with enough sets therein to ensure that every open set can be written as a union of elements of that subbase.

Definition 2.3.8.

Let \(X\) be a topological space. Let \(x\in X\) be a point. Then a local base at \(x\in X\) is a set \(N_x\) of open neighbourhoods of \(x\) with the following property: For any open neighbourhood \(W\) of \(x\text{,}\) there is an open neighbourhood \(V\in N_x\) with \(V\subseteq W\text{.}\)

A base for \(X\) is a collection \(B\) of open sets such that for any point \(x\in X\text{,}\) the set

\begin{equation*} B_x \coloneq \{ U\in B : x\in U \} \end{equation*}

is a local base at \(x\in X\text{.}\)

Let \(X \subseteq \RR^n\) be a subspace of Euclidean space. Then the set

\begin{equation*} \{ B(x,\varepsilon) : (x\in X) \wedge (\epsilon>0) \} \end{equation*}

is a base for the induced topology on \(X\text{.}\)

Let \(X\) be a set with two topologies \(\tau_1\) and \(\tau_2\text{;}\) let \(B_1\) be a base for \((X,\tau_1)\text{,}\) and let \(B_2\) be a base for \((X,\tau_2)\text{.}\) Show that \(\tau_1\) is finer than \(\tau_2\) if and only if, for every point \(x\in X\) and every element \(U\in(B_2)_x\text{,}\) there exists an element \(V\in(B_1)_x\) such that \(V\subseteq U\text{.}\)