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

Question

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 $a_1<\dotso<a_m$.

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.

Answer 1

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.

Answer 2

If $A \subseteq \mathbb{N}$ is finite, then $A$ is bounded above.
Proof by contradiction.
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.