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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"?

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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 $\max M$.

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  • $\begingroup$ I didn't know about ">!" -- cool feature(!) $\endgroup$ – BrianO Nov 11 '15 at 23:53
  • $\begingroup$ @Brian: It’s a bit limited – you can use it only once in an answer, and you can’t use displayed formulae within it – but it’s still a great feature. $\endgroup$ – Brian M. Scott Nov 11 '15 at 23:59
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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.

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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:

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

  2. $\mathscr{U}$ 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.

  3. $\mathscr{U}$ 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.

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  • $\begingroup$ @Jean-ClaudeArbaut 3 cases, I say... pulled the trigger too fast. $\endgroup$ – BrianO Nov 11 '15 at 23:51
  • $\begingroup$ I'd call that an indexed family. For me, & for many, a collection of open sets is just a set of open sets — for this topology, a subset of $\tau \subseteq \mathcal{P}(\Bbb N)$. As $\bigcup$ of an indexed family = $\bigcup$ of the range of the indexed family, it's splitting hairs over nothing. $\endgroup$ – BrianO Nov 11 '15 at 23:59
  • $\begingroup$ Ok, thank to have clarified. Yes, it's a minor question of vocabulary, but I really had indexed family in mind. Sorry to have insisted :-) $\endgroup$ – Jean-Claude Arbaut Nov 12 '15 at 0:03
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    $\begingroup$ @Jean-ClaudeArbaut No problem. If I had been talking about an indexed family, my language would have been different (and much wordier, in order to be precise: I wouldn't be able to say "if $\mathscr{U}$ contains $\emptyset$", and the like, as I'd have to mention some index set $I$ and talk about $U_i = \emptyset$ for some $i\in I$... and of course I would have written the unions differently.) My $\mathscr{U}$ is the range of the indexed family you had in mind :) $\endgroup$ – BrianO Nov 12 '15 at 0:08
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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 $X=\mathbb N$.
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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?

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