# When is $\mathbb{N}\setminus A$ finite?

I have the topological space $$(\mathbb{N},\tau_{\text{kof}})$$ where $$\tau_{\text{kof}}=\{U\subseteq\mathbb{N}|\mathbb{N}\setminus U\,\,\text{finite}\} \cup \{\emptyset\}$$

Now I want to show, that for $$U$$ non-empty and open (with regards to $$\tau_{\text{kof}}$$), there exists an element $$u\in U$$ such that every successor of $$u$$, is an element of $$\mathbb{N}\setminus U$$. So for $$v\in\mathbb{N}$$ with $$v>u$$ we have $$v\in \mathbb{N}$$.

Proof:

Since $$|\mathbb{N}\setminus U|=m<\infty$$ is finite, we can note $$\mathbb{N}\setminus U=\{a_1,\dotso, a_m\}$$ with $$a_i\neq a_j$$ for $$i\neq j$$. Without loss of generality it is $$a_1<\dotso.

Now we have $$a_m\notin U$$ and since $$a_m$$ is the maximal element of $$\mathbb{N}\setminus U$$ it is $$a\in U$$ for every $$a>a_m$$. So it exists $$u=a_m+1$$ such that every successor is an element of $$U$$, which gives a contradiction.

I feel like this could be proven much more elegant and precisely. Somehow I am not satisfied with what I did.

What do you think? Thanks in advance.

• Fixed the topology definition and the problem statement. The empty set is also open and would be a counterexample... – Henno Brandsma Jul 14 '19 at 5:37
• The statement should have “there exists $u\in U$ such that every successor of $u$ is an element of $U$” (not $\mathbb{N}\setminus U$). – egreg Jul 14 '19 at 9:40

Fact: A finite subset of a linearly ordered set has a maximum.

Now, if $$U$$ is non-empty and open it is of the cofinite form, so $$\mathbb{N} \setminus U$$ is finite.

Define $$M=\max(\mathbb{N} \setminus U)$$ which exists by the first fact.

If $$u > M$$, then $$u \in U$$ (or else $$u \in \mathbb{N} \setminus U$$ which would contradict the maximality of $$M$$). So take $$u=M+1$$ and then $$v > u$$ implies $$v > M$$ so $$v \in U$$.

Yes, it's the same proof, more compactly, with less notation. It's a matter of opinion which is better. I think mine is more readable.

• I agree. Thank you. – Cornman Jul 14 '19 at 11:54
• @Cornman Glad I could help. – Henno Brandsma Jul 14 '19 at 11:55

If $$A \subseteq \mathbb{N}$$ is finite, then $$A$$ is bounded above.
If $$A$$ is not bounded above then
for all $$n$$, exists $$a_n$$ in A with $$n < a_n$$.
Show $$\{ a_n : n \in \mathbb{N} \}$$ is an infinite subset of $$A$$.
Thus any upper bound of $$\mathbb{N} - U$$ will suffice.