Prove it's topology First let's understand who these topologies are:
\begin{equation}
    \tau_1 = \{\{m \in \mathbb{N} : m < n\} : n \in \mathbb{N}\} \cup \{\mathbb{N}\}
\end{equation}
\begin{equation}
    \tau_1=\{\emptyset, \{0\},\{0,1\},\{0,1,2\}, \{0,1,2,3\}, \cdots, \mathbb{N}\}
\end{equation}
Let's verify that $\tau_1$ is a topology:

*

*In fact, $\emptyset$ is subnext to any set, so $\emptyset \in \tau_1$, plus $\mathbb{N}\in \tau_1$.

*We need to show that the finite intersection of the $\tau_1$ elements belongs to $\tau_1$. I can see that yes, for example, $\{0\}\cap \{0,1\}=\{0\}\in \tau_1$, $\{0\}\cap\{0,1\}\cap\{0,1,2\}=\{0\}\in \tau_1$ or $\{0,1\}\cap\{0,1,2\}\in \tau_1$. But I can't generalize. Can you help me?

*We need to show that the arbitrary union of $\tau_1$ elements belongs to $\tau_1$. I know it's worth that, for example, $\{0\}\cup \{0,1\}\cup \{0,1,2\} \cup \mathbb{N}=\mathbb{N} \in \tau_1$. But I can't generalize. Can you help me?

Thank you!
 A: Hints:

*

*Given finitely many sets of the form $\{0, \ldots, k_1\}, \ldots, \{0, \ldots, k_n\}$, can you see what the intersection must be? What if you put $k = \min\{k_1, \ldots, k_n\}$. What could it be?
Of course, intersecting further with $\varnothing$ or $\Bbb N$ is easy.

*Let $I$ be an arbitrary set and for each $i \in I$, suppose you are given $\{0, \ldots, k_i\}$. If the set $\{k_i\}$ is bounded, can you see deduce the union? (What if you consider $\max_{i \in I} k_i$?) What if the set is not bounded? Do you think the union is then $\Bbb N$?
As before, taking a further union with $\varnothing$ or $\Bbb N$ is easy.

A: Set $U_n=\{m\in\mathbb{N}:m<n\}$ and $U_\infty=\mathbb{N}$. Then $U_0=\emptyset$, so $\emptyset\in\tau_1$.
Consider a finite (nonempty) set of members of $\tau_1$ and write it as
$$
\{U_{n_1},U_{n_2},\dots,U_{n_k}\}
$$
Prove that $\bigcap_{i=1}^k U_{n_i}=U_n$, where $n=\min\{n_1,\dots,n_k\}$.
Suppose you have an arbitrary set of members of $\tau_1$. The set is $\{U_{n_i}:i\in I\}$. Now consider $\sup\{n_i:i\in I\}$.
