Properties:: *[[Properties]]* Sufficiencies::[[open sets are unions of basis elements]] Equivalences:: *[[Equivalences]]* ---- > [!definition] Definition. ([[topology generated by a basis]]) > Recall: ![[basis for a topology#^8d8ba4]] > If $\mathscr{B}$ satisfies these two conditions, then we define the **topology $\tau$ generated by $\mathscr{B}$** as follows: a subset $U \subset X$ is declared [[open set|open in]] $X$ (i.e., is a member of $\tau$) if for each $x \in U$, there is a basis element $B \in \mathscr{B}$ such that $x \in B \subset U$. > \ > That is: If $\mathscr{B}$ satisfies these two conditions, then we define the **topology $\tau$ generated by $\mathscr{B}$** to be the maximal collection of 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 $\tau$. > [!justification] (Proof that this is really a topology) > We have to show [[topological space|the three properties]]. > 1. If $U=\emptyset$, it belongs to $\tau$ vacuously. And trivially we have that for each $x \in X$, there exists $B \in \mathscr{B}$ with $x \in B \subset X$, so $X \in \tau$. > 2. Take an indexed family $\{ U_\alpha \}_{\alpha \in J} \subset \tau$. For each $x \in \bigcup_{\alpha \in J} U_{\alpha}$, we have $x \in U_{\alpha}$ for some $\alpha \in J$. Then since $U_{\alpha} \in \tau$, we have that $\ex B \in \mathscr{B}$ s.t. $x \in B \subset U_{\alpha} \subset \bigcup_{\alpha \in J}^{} U_{\alpha}$. So $\bigcup_{\alpha \in J}^{} U_{\alpha} \in \tau$. >3. **Lemma.** First we'll take *two* elements $U_{1}, U_{2} \in \tau$ and show $U_{1}\cap U_{2} \in \tau$. Let $x \in U_{1} \cap U_{2}$. Since $x \in U_{1} \in \tau$, $\ex B_{1} \in \mathscr{B}$ s.t. $x \in B_{1} \in U_{1}$. Since $x \in U_{2} \in \tau$, $\ex B_{2} \in \mathscr{B}$ s.t. $x \in B_{2} \in U_{2}$. $x \in B_{1} \cap B_{2}$, so by the second condition in [[basis for a topology|basis]] definition $\ex B_{3} \in \mathscr{B}$ s.t. $x \in B_{3} \subset B_{1} \cap B_{2} \subset U_{1} \cap U_{2}$. So $U_{1} \cap U_{2} \in \tau$. \ Proceed inductively. The base case $n=1$ is trivial. Suppose the (finite) intersection of elements $U_{1},\dots,U_{n-1}$ of $\tau$ is again in $\tau$. Let $U_{n} \in \tau$ and consider $\bigcap_{i \in [n]}U_{i} = (U_{1} \cap\dots \cap U_{n-1})\cap U_{n}.$ That $\bigcap_{i \in [n] U_{i}} \in \tau$ now follows from the induction hypothesis and **Lemma**. ---- #### ---- #### 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 > ```