Let $\beta$ be the collection of all open intervals $(a,b) \subset \mathbb{R}$, with $a,b \in \mathbb{R}$. Prove that the topology $\tau_\beta$ generated by $\beta$ is in fact a topology. Observation:
$\tau_\beta = \{U \subset \mathbb{R} \ \vert \ \forall x \in U ,\exists B \in \beta \text{ such that } x \in B \subset U\}$
My proof: $\emptyset \in \tau_\beta$ vacuously. Since for every $x \in \mathbb{R}$ and $\varepsilon > 0$, we have that $x \in (x-\varepsilon, x+\varepsilon) \subset \mathbb{R}$, then $\mathbb{R} \in \tau_\beta$. Now, given any arbitrary $x \in A = \displaystyle{\bigcup_{j \in J} U_j} $, where $U_j$ is an open set for each $j$, then $x \in U_i$ for some $i \in J$. Since $U_i $ is open, then there exists an interval $B$ such that $x \in B \subset U_i$. Finally, consider $W = U_1 \bigcap U_2$, where $U_1$ and $U_2$ are open sets. Given any $x \in W$, since $x \in U_1$ and $x \in U_2$, then there exist $B_1, B_2 \in \beta$ such that $x \in B_1 \subset U_1$ and $x \in B_2 \subset U_2$. Then $x \in B_1 \bigcap B_2 \subset U_1 \bigcap U_2$ and we're done (the rest is trivial by induction).
Are all my steps correct?
Best Answer
That is a vacuous problem because the topology generated by any subset K of P(X) is by definition a topology for X, the smallest topology for X containing K, the intersection of all topologies for X containing K.
Exercise. If T is a not empty collection of topologies for X,
show $\cap$T is a topology for X.
Problem. Is there a topology for X that contains any subset of P(X)?
Why is this important?