Prove that if $B1$ is a base for $\tau_{1}\{\emptyset, \mathbb{N}\}$ then $\tau_{1}\{\emptyset, \mathbb{N}\}\subseteq B_{1}$ Be $\tau_{1}$ and $\tau_{2}$ topologies over $\mathbb{N}$ defined by:
$\tau_{1}=\{\{m\in \mathbb{N}:m<n\}:n\in \mathbb{N}\}\cup \{\mathbb{N}\}\}$ and $\tau_{2}=\{A\subseteq \mathbb{N}: 0\in A\}\cup \{\emptyset\}$.
Prove that if $B_1$ is a base for $\tau_{1}\{\emptyset, \mathbb{N}\}$ then $\tau_{1}\{\emptyset, \mathbb{N}\}\subseteq B_{1}$.
(i) If $B_{1}$ is an open base for $\tau_1$ then $\tau_{1}-\{\emptyset, \mathbb{N}\} \subseteq B_1$. If $B_1$ is a basis for the Topology $\tau_1$ then any element of $\tau_1$ can be written as a union of the elements of $B_1$. We need to show that $\tau_1-\{\emptyset, \mathbb{N}\} \subseteq B_1$.
Let $V \in \tau_1- \{\emptyset, \mathbb{N}\}$ so as $B_1$ is a base of $\tau_1$, we have that $V = \bigcup_ {\lambda \in I} B_\lambda$.
Is the union of base elements at the base?
(ii) How is the $B_1$ base? A $B_2$ base for $\tau_2$ I would kick $\tau_1$.
 A: I believe that the question actually asks you to show that $\tau_1\color{red}{\setminus}\{\varnothing,\Bbb N\}\subseteq B_1$. In other words, you’re to show that $B_1$ must contain all of the sets $\{m\in\Bbb N:m<n\}$ for $n\in\Bbb N$. Fix $n\in\Bbb N$, let $U=\{m\in\Bbb N:m<n\}$, and suppose that $U\notin B_1$. Since $B_1$ is a base for $\tau_1$, and $U\in\tau_1$, there must be some $\mathscr{U}\subseteq B_1$ such that $U=\bigcup\mathscr{U}$. What sets could belong to $\mathscr{U}$? Clearly they must all be subsets of $U$, as otherwise $\bigcup\mathscr{U}$ could not be equal to $U$. The only members of $\tau_1\setminus\{U\}$ that are subsets of $U$ are the sets $U_m=\{k\in\Bbb N:k<m\}$ for $m<n$. Thus, $\mathscr{U}\subseteq\{U_m:m<n\}$, and
$$\bigcup\mathscr{U}\subseteq\bigcup_{m<n}U_m=U_{n-1}=\{k\in\Bbb N:k<n-1\}\,.$$
But $n-1\in U\setminus U_{n-1}$, so $\mathscr{U}\ne U$. That is, $U$ is not the union of members of $B_1$, contradicting the assumption that $B_1$ is a base for $\tau_1$. This show that $U$ must in fact belong to $B_1$, and since $U$ was an arbitrary member of $\tau_1\setminus\{\varnothing,\Bbb N\}$, we’ve proved that $B_1\supseteq\tau_1\setminus\{\varnothing,\Bbb N\}$, as desired.
I can’t tell what (ii) actually is; I suspect that it contains some significant typos. However, I can tell you that $\tau_2$ has a base that contains just one open set: it is the smallest open set containing $0$.
