----- Let $(X, \tau)$ be a [[topological space]]. Here is a condition for obtaining a basis from the [[topological space|topology]] $\tau$. > [!proposition] Proposition. ([[condition for obtaining a basis from a topology]]) > Suppose $\mathcal{C} \subset \tau$ is a collection of [[open set]]s of $X$ that [[nestles in]] $\tau$. > \ > (That is, a collection such that for each [[open set]] $U \in \tau$ of $X$ and each $x \in U$, $\ex C \in \mathcal{C}$ s.t. $x \in C \subset U$.) > \ > Then $C$ is a [[basis for a topology|basis for the topology]] $\tau$ on $X$. ^805956 > [!NOTE] Remark. > > If $\tau$ is specified in terms of [[closed set|closed sets]], then the equivalent condition is to have that for each closed set $F$ of $X$ and each $x \notin F$, there exists a closed set $G \supset F$ with $x \notin G$ (draw picture). (Maybe this is not useful.) > [!proof]- Proof. ([[condition for obtaining a basis from a topology]]) > First we must show that $\mathcal{C}$ satisfies the two [[basis for a topology|basis]] conditions. > 1. Immediate. > 2. Suppose $x \in C_{1} \cap C_{2}$ for some $C_{1},C_{2} \in \mathcal{C}$. By third [[topological space]] condition, $C_{1} \cap C_{2}$ is [[open set|open in]] $X$, thus by assumption $\ex C \in \mathcal{C}$ s.t. $x \in C \subset C_{1} \cap C_{2}$. \ Next, we must show that $\tau$ indeed equals the [[topology generated by a basis|topology generated by]] $\mathcal{C}$ (call it $\tau'$) via two-way inclusion. First note that if $U \in \tau$ and $x \in U$, by hypothesis there exists $C \in \mathcal{C}$ such that $x \in C \subset U$. So $\tau \subset \tau'$. Conversely, if $W \in \tau'$ then $W$ equals a union of elements of $\mathcal{C}$ by [[open sets are unions of basis elements]]$\textcolor{Skyblue}{'}$. Since each element of $\mathcal{C}$ belongs to $\tau$ and $\tau$ is a [[topological space|topology]] (specifically, unions "stay in" $\tau$), it follows that $W \in \tau$. So $\tau' \subset \tau$. $\qedin$ ^933a96 > [!definition] Definition. ([[topology generated by a basis]]) > Recall: ![[basis for a topology#^c0c9e7]] > If $\mathscr{B}$ satisfies these two conditions, then we define the **topology $\tau$ generated by $\mathscr{B}$** to be the maximal collection subsets of $X$ that $\mathscr{B}$ [[condition for obtaining a basis from a topology#^933a96|nestles in]]. > \ > Note that each basis element is itself an element of $\tau$. > [!basicexample] **Proposition a.** The countable collection $\mathscr{B} := \{ (a,b) : a<b, a \text{ and } b \text{ rational} \}$ is a [[basis for a topology|basis]] that [[topology generated by a basis|generates]] the [[standard topology on the real line|standard topology]] $\tau_{\text{standard}}$ on $\mathbb{R}$. \ **Proof a.** We can demonstrate this by showing that it [[nestles in]] $\tau_{\text{standard}}$: \ Let $U \in \tau_{\text{standard}}$; note [[open sets are unions of basis elements|it is a union of]] [[open interval]]s in $\mathbb{R}$: $U=\bigcup_{\xi}^{}(\alpha,\beta)_\xi$. Let $x \in U$. Then $x \in (\alpha, \beta)$ for some $(\alpha,\beta) \in U$. Via density of $\mathbb{Q}$ in $\mathbb{R}$ obtain rational $a,b$ for which $\alpha < a < x < b < \beta.$ Then, we have $x \in (a,b) \subset (\alpha, \beta) \subset \bigcup_{\xi}^{}(\alpha, \beta)_\xi=U \text{ with } (a,b) \in \mathscr{B},$implying that $\mathscr{B}$ [[nestles in]] $\tau_{\text{standard}}$ as required. \ **Proposition b.** The collection $\mathcal{C}:= \{ [a, b) : a<b, a \text{ and } b \text{ rational} \}$ is a [[basis for a topology|basis]] that generates a [[topological space|topology]] *different* from the [[lower limit topology, upper limit topology|lower limit topology]] $\tau_{\ell}$ on $\mathbb{R}$. \ **Proof b.** If two [[topological space|topologies]] are equal, then each is [[comparable topologies|finer]] than the other. Therefore, we will show that the [[topological space|topology]] generated by $\mathcal{C}$ is not finer than $\tau_{\ell}$. By [[basis characterization of comparable topologies]], it suffices to show that $\mathcal{C}$ is [[basis for a topology|basis]] and it does not [[nestles in]] the set $\mathscr{B}$ of half-open [[interval]]s on $\mathbb{R}$ which (by definition) generates $\tau_{\ell}$. \ First, to show $\mathcal{C}$ is a [[basis for a topology]] on $X$. Covering is clear, as is nestling (since for any nontrivally intersecting $C_{1}, C_{2} \in \mathcal{C}$, we have $C_{1} \cap C_{2} \in \mathcal{C}$ too). \ Here is a (counter)example: let $x = \sqrt{ 2 }$. Then $x \in [\sqrt{ 2 }, 2) =: B \in \mathscr{B}$, meaning that there cannot exist $a,b \in \mathbb{Q}$ for which $x \in [a,b) \subset [\sqrt{ 2 }, 2)$. That is to say: there cannot exist $C \in \mathcal{C}$ for which $x \in C \subset B$. ^566ed2 ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```