Topology on $\mathbb N$ formed by taking the open sets to be $\emptyset, \mathbb{N}$ and $\{ 1, 2, 3, \ldots, n \} $ for each $n \in \mathbb{N}$ I am having some definition-wise problems.
Problem: Prove that we get a topology for $\mathbb{N} = \{ 1, 2, 3, \ldots \} $ by taking the open sets to be $\emptyset, \mathbb{N}$ and $\{ 1, 2, 3, \ldots, n \} $ for each $n \in \mathbb{N}$. 
My point of confusion: How do we show that any union of the open sets, as defined, is an open set "formally"?
 A: HINT: For each $n\in\Bbb N$ let $U_n=\{1,\ldots,n\}$, so that the open sets are $\varnothing,\Bbb N$, and the sets $U_n$ for $n\in\Bbb N$. Let $\mathscr{U}$ be a collection of open sets. If $\Bbb N\in\mathscr{U}$, then clearly $\bigcup\mathscr{U}=\Bbb N$, so $\bigcup\mathscr{U}$ is open. If $\mathscr{U}=\{\varnothing\}$, then $\bigcup\mathscr{U}=\varnothing$, which is open. The only remaining possibility is that $\Bbb N\notin\mathscr{U}$, and there is at least one $U_n\in\mathscr{U}$. Let $M=\{n\in\Bbb N:U_n\in\mathscr{U}\}$.


*

*What is $\bigcup\mathscr{U}$ if $M$ is infinite?  

*If $M$ is finite, what is $\bigcup\mathscr{U}$? If you get completely stuck on this part, mouse-over the spoiler protected further hint below.



 Consider $\max M$.

A: Let $N_n=\{1,2,\ldots, n\}$.  For $n_1<n_2<\ldots<n_k$, we have $N_{n_1}\subseteq N_{n_2}\subseteq\ldots\subseteq N_{n_k}$.  So,
$$\bigcup_{j=1}^k N_{n_j}=N_{n_k}.$$
The infinite case is similar.
A: Each open set $A$ of the topology that is not $\emptyset$ or $\Bbb N$ is uniquely determined by its greatest integer member $n = \max A$. Let's write $[n] = \{1,\dotsc,n\}$.
Suppose $\mathscr{U}$ is a collection of open sets. If $\mathscr{U}$ contains $\emptyset$ we can safely omit it without affecting the union. If $\mathscr{U}$ itself is empty then its union is $\emptyset$ and therefore open. There are then three cases:


*

*$\Bbb N \in \mathscr{U}$. Then clearly $\bigcup \mathscr{U} = \Bbb N$, which is open.

*$\mathscr{U}$ is finite, not containing $\Bbb N$. Then $\mathscr{U}$ contains one largest set $[n]$, which includes all other sets in $\mathscr{U}$: if $A\in \mathscr{U}$ then $A\subseteq [n]$. Then $[n] = \bigcup \mathscr{U}$, so the union is an open set.

*$\mathscr{U}$ is infinite, not containing $\Bbb N$. Then the set of $\max A$ for $A\in \mathscr{U}$ is unbounded, so $\bigcup \mathscr{U} = \Bbb N$, which is also an open set.
A: Call a subset $S$ of an ordered set $X$ 'downward closed' if $\forall a\in S, \forall b\in X, b<a \implies b\in S$. Now, it is an easy exercise to check that 


*

*The collection of all downward closed sets forms a topology on $X$. (It is infact closed under arbitrary intersections.)

*The collection of sets in the given problem is precisely the collection of downward closed sets in the case $X=\mathbb N$.

A: Take $\mathcal O$ to be a set of "open" sets as described. Then, proceed by cases:


*

*What is $\bigcup\mathcal O$ if $\mathcal O=\emptyset$?

*What is $\bigcup\mathcal O$ if $\mathcal O$ has a single element?

*What is $\bigcup\mathcal O$ if $\Bbb N\in\mathcal O$?

*What is $\bigcup\mathcal O$ if $\mathcal O$ is finite with more than one element, and $\Bbb N\notin\mathcal O$?

*What is $\bigcup\mathcal O$ if $\mathcal O$ is infinite?
Are there any other possibilities?
