Given $A_n = \{n, n+1, \dots\} , n \in \mathbb{N} $ and $\tau = \{ \emptyset, A_1, A_2, \dots \} $, I would like to determine which sets are open in $(\mathbb{N}, \tau) $

By definition: $A_n$ is open and any intersection/union of finitely/infinitely many $A_n$ is also open. Furthermore: $A_n$ is infinite, ${A_n}^C$ is finite.

However this does not tell or really help me deciding, what are any other open/closed sets regarding this topology? For example: Is set of odd/even numbers open? What about an arbitrary set? etc.

It seems I am missing a very basic property of topologies. Therefore how does one approach above problem? What am I missing?


EDIT: This answer is based on a misreading of the question. I read the question as asking about the topology generated by $\{A_n: n\in\mathbb{N}\}\cup\{\emptyset\}$ - equivalently, asking to prove that that that set is already a topology.

As noted above, given that the OP already is told that $\{A_n: n\in\mathbb{N}\}\cup\{\emptyset\}$ is a topology, their question does boil down to just understanding how a topology describes open sets (namely, that the topology is exactly the set of open sets).

However, I suspect that the OP may be confused about why this family is a topology on $\mathbb{N}$. Since I think my (misinformed) answer may be useful to them with respect to this, I've left it up (with minor changes).

A set with the property that its complement is finite is called cofinite. For example, all the $A_i$s above are cofinite.

Here are a couple hints towards showing that the set of even numbers is not open in the above sense:

  • Show that the union of aritrarily many cofinite sets is cofinite. (HINT: any set containing a cofinite set is cofinite ...)

  • Show that the intersection of finitely many cofinite sets is cofinite. (HINT: think about the complement of a finite intersection of cofinite sets - remember that the complement of an intersection is the union of the complements ...)

  • Conclude that every open set (in the sense above) is cofinite.

Note that the above does not fully characterize the open sets (again, in the sense of your question). For example, the set of all numbers except $3$ and $17$ is cofinite; is it open?

  • Show that if $B$ is open in the sense above, and $b\in B$ and $c>b$, then $c\in B$.

  • Use the above to give a complete characterization of the open sets in the sense above.


$\tau$ is already a topology on $\mathbb{N}$. The open sets are precisely $\emptyset$ and $A_n$ for $n \in \mathbb{N}$.

Perhaps you meant to ask how to verify that $\tau$ is a topology. Notice that $A_{n+1} \subseteq A_n$ for all $n \in \mathbb{N}$.

Therefore, $$\bigcup_{i \in S} A_i = A_{\min S} \in \tau$$

for every $S \subseteq \mathbb{N}$.

Also, $$\bigcap_{i=1}^n A_{i} = A_n \in \tau$$ for every $n \in \mathbb{N}$.

We have $\phi \in \tau$ and $\mathbb{N} = A_1 \in \tau$. We conclude that $\tau$ is a topology.


The meaning of the word 'topology' is the collection of all open sets. Therefore, if someone tells you that the topology is $$\tau=\{\emptyset, A_1,A_2,\dots\}$$ then this is precisely the collection of all open sets. There are no others. You can check that this is indeed a topology, i.e., the union open sets is open (that is, a member of the set $\tau$), etc.


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