# 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:

$$\tau_{1}=\{\{m\in \mathbb{N}:m and $$\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 $$\tau_1$$.

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 $$\{m\in\Bbb N:m for $$n\in\Bbb N$$. Fix $$n\in\Bbb N$$, let $$U=\{m\in\Bbb N:m, 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 $$U_m=\{k\in\Bbb N:k for $$m. Thus, $$\mathscr{U}\subseteq\{U_m:m, and

$$\bigcup\mathscr{U}\subseteq\bigcup_{m

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 $$0$$.

• Do so a base for $\tau_{2}$ will be $\{ \{0 \}\}$? Dec 22 '20 at 12:20