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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.

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

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

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  • $\begingroup$ Is this not simply a short version of my proof, where the details are left to the reader? This does not really answer my question then, but there might be no answer in the first place. $\endgroup$ – Cornman Jul 14 '19 at 1:58
  • $\begingroup$ See my edit. @Cornman $\endgroup$ – William Elliot Jul 14 '19 at 2:00
  • $\begingroup$ I still think, that this is basically the same proof, but I am fine with that. Or do you disagree? $\endgroup$ – Cornman Jul 14 '19 at 2:02

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