Exercise 2.4.5 (Subbase for a topology)

Question

Let $X$ be a set and let $\mathscr{S}$ be a family of subsets of $X$.
\begin{enumerate}[label=(\alph*)]
\item Show that there exists a unique smallest topology $\mathscr{T}$ on $X$ such that $\mathscr{S}$ $\subset \mathscr{J}$. The topology $\mathscr{T}$ is called the topology generated by $\mathscr{S}$, and $\mathscr{S}$ is called a subbase for the topology $\mathscr{T}$.
\item Let $\mathscr{B}$ be the family of subsets of $X$ consisting of $X, \varnothing$, and all finite intersections of sets in $\mathscr{S}$. Show that $\mathscr{B}$ is a base of open sets for the topology generated by $\mathscr{S}$.
\item Let $X$ have the topology generated by $\mathscr{S}$, let $Y$ be a topological space, and let $f: Y \rightarrow X$ be a function. Show that $f$ is continuous if and only if $f^{-1}(S)$ is an open subset of $Y$ for every set $S \in \mathscr{S}$.
\end{enumerate}

Elu Thingol · Accepted Answer

extbf{For any collection $\mathscr{S}$ of subsets of $X$ there is the smallest topology $\mathscr{T}$ on $X$ containing $\mathscr{S}$.}
\begin{proof}
We enlarge $\mathscr{S}$ minimally so that the topology axioms (1)-(3) are fulfilled for the enlarged set. For that purpose, we take three steps. First, add the empty set and the parent set to $\mathscr{S}$:
$$\mathscr{S}^{\prime} := \mathscr{S} \cup \{\emptyset, X\}.$$

Then consider the set $\mathscr{B}$ containing all finite intersections of sets in $\mathscr{S}^{\prime}$, i.e,
$$\mathscr{B} := \left\{U_1 \cap \ldots \cap U_n \subseteq X \mid U_1, \ldots, U_n \in \mathscr{S}^{\prime} ight\}.$$

If $\mathscr{B}$ would be a base for a topology, then the following set would become a topology
$$\mathscr{T} := \left\{\bigcup \mathscr{C} \mid \mathscr{C} \subseteq \mathscr{B} ight\}.$$

We argue that $\mathscr{T}$ is indeed the smallest topology containg $\mathscr{S}$ and thus $\mathscr{B}$ is a base for $\mathscr{T}$.\par

extit{First}, $\mathscr{S} \subseteq \mathscr{T}$ since $\mathscr{S} \subseteq \mathscr{S}^{\prime} \subseteq \mathscr{B} \subseteq \mathscr{T}$.\par

extit{Second}, $\mathscr{T}$ is indeed a topology over $X$. Since $\emptyset, X \in \mathscr{S}^{\prime} \subseteq \mathscr{T}$, we have (1.1). To demonstrate (1.2), consider $B,C \in \mathscr{T}$, i.e., $B = \bigcup \mathcal{B}$ and $C = \bigcup \mathcal{C}$ for $\mathcal{B},\mathcal{C} \subseteq \mathscr{B}$. Then,
\begin{align*}
B \cap C &= \bigcup \mathcal{B} \cap \bigcup \mathcal{C}\
         &= \bigcup \left\{ 	ilde{B} \cap 	ilde{C} \mid 	ilde B \in \mathcal{B} 	ext{ and } 	ilde{C} \in \mathcal{C}ight\}
\end{align*}
Since for each $	ilde{B}, 	ilde{C} \in \mathscr{B}$ we also have $	ilde{B} \cap 	ilde{C} \in \mathscr{B}$ and thus $\left\{ 	ilde{B} \cap 	ilde{C} \mid 	ilde B \in \mathcal{B} 	ext{ and } 	ilde{C} \in \mathcal{C}ight\} \in \mathscr{B}$ and therefore $B \cap C \in \mathscr{T}$. For (1.3), consider an arbitrary subset $\{B_i \mid i \in I \} \subseteq \mathscr{T}$. Each $B_i$ is of the form $B_i = \bigcup \mathcal{B}_i$ for $\mathcal{B}_i \in \mathscr{B}$. We have $\bigcup_{i \in I} \mathcal{B}_i \in \mathscr{T}$ and thus
\begin{align*}
\bigcup_{i \in I} B_i &= \bigcup \left\{ \bigcup \mathcal{B}_i \mid i \in I ight\}\
                      &= \left(\bigcup_{i \in I} \bigcup ight) \mathcal{B}_i 
\end{align*}
In other words, the above set is a union of basis sets in $\mathscr{B}$ and is thus contained in $\mathscr{T}$.\par

extit{Third}, the topology $\mathscr{T}$ is smaller than any other topology $\mathscr{T}^{\prime}$ containing $\mathscr{S}$. This follows from the fact that $\mathscr{T} \subseteq \mathscr{T}^{\prime}$ implies $\mathscr{T}^{\prime} \subseteq \mathscr{T}^{\prime}$ ab $\mathscr{T}^{\prime}$ satisfies axiom (1.1) as well; similarly $\mathscr{B} \subseteq \mathscr{T}^{\prime}$ since  $\mathscr{T}^{\prime}$ satisfies (1.2); finally, $\mathscr{T} \subseteq \mathscr{T}^{\prime}$ since $\mathscr{T}^{\prime}$ satisfies (1.3).\par

This proves (a) and (b).
\end{proof}

extbf{Let $\mathscr{S}$ be a subbasis of $Y$. Then the following are equivalent:
\begin{itemize}
\item $f: X 	o Y$ is continuous.
\item For all $S \in \mathscr{S}$ we have $f^{-1}(S)$ is open in $X$.
\end{itemize}}
\begin{proof}
One direction is obvious as the inverse image of any open set under a continuous function is open, in particular an image of an open set in the subbasis.\par
From the previous part of the exercise, we can recall that a topology of $Y$ can be constructed from $\mathscr{S}$ in three steps: by complementing $\mathscr{S}$ with $X$ and $\emptyset$, then constructing the basis $\mathcal{B}$ as the set of all finite intersections of the sets in $\mathscr{S}^{\prime}$ and finally getting the desired topology by $\mathscr{T}$ by taking all possible unions over $\mathscr{B}$. Notice that $f^{-1}(Y)=X$ and $f^{-1}(\emptyset)=\emptyset$ imply that for all $S \in \mathscr{S}^{\prime}$, $f^{-1}(S)$ is open. Furthermore, from $f^{-1}\left(\bigcap \mathcal{C}ight) = \bigcap f^{-1}\left(\mathcal{C}ight)$ and from the axiom (1.2) we deduce that for all $\mathcal{S} \in \mathscr{B}$, $f^{-1}(\mathcal{S})$ is open. Finally, from $f^{-1}\left(\bigcup Uight) = \bigcap f^{-1}\left(Uight)$ and from the axiom (1.3) we deduce that  all $U \in \mathscr{T}$, $f^{-1}(U)$ is open. In other words, $f$ is continuous.
\end{proof}

Exercise 2.4.5 (Subbase for a topology)

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Exercise 2.4.5 (Subbase for a topology)

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