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Let $\lbrace \tau_n:n\in \mathbb{N}\rbrace$ be an ascending chain of topologies on a nonempty set $X$. Then is $\bigcup\limits^{\infty}_{n=0}\tau_n$ a topology on $X$?

I have a strong notion that it may not be a topology. But I am not getting the proper counter example. Please help. I know this must be a topology if the collection is finite.

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3 Answers

up vote 4 down vote accepted

This is an addendum to Asaf’s answer and Pete’s comment.

Suppose that $\tau=\bigcup_{n\in\Bbb N}\tau_n$ is not a topology on $X$. Then, as noted in Asaf’s answer, there must be some family $\mathscr{U}\subseteq\tau$ such that $\bigcup\mathscr{U}\notin\tau$, since the other requirements for $\tau$ to be a topology are certainly met.

For each $n\in\Bbb N$ let $\mathscr{U}_n=\mathscr{U}\cap\tau_n$, and let $U_n=\bigcup\mathscr{U}_n$; then $U_n\in\tau_n$, and $\bigcup\mathscr{U}=\bigcup_{n\in\Bbb N}U_n\notin\tau$. Note that $\mathscr{U}_n\subseteq\mathscr{U}_{n+1}$, since $\tau_n\subseteq\tau_{n+1}$, so $U_n\subseteq U_{n+1}$ for each $n\in\Bbb N$. Thus, a counterexample to the conjecture that $\tau$ is a topology on $X$ can always be taken to consist of a non-decreasing nest $\{U_n:n\in\Bbb N\}\subseteq\tau$ such that $U_n\in\tau_n$ for each $n\in\Bbb N$, but $\bigcup_{n\in\Bbb N}U_n\notin\tau$.

Of course one very simple way to construct such an example is to let $U_n=\{k\in\Bbb N:k<n\}$, take $X=\Bbb N\cup\{p\}$, where $p\notin\Bbb N$, and let $\tau_n=\{\varnothing,X\}\cup\{U_k:k<n\}$ for each $n\in\Bbb N$: since $\Bbb N\notin\bigcup_{n\in\Bbb N}\tau_n$, $\tau$ is not a topology on $X$. This is in some sense a minimal counterexample.

It can evidently be adapted along the lines of Asaf’s construction to build an example starting with any space $\langle X,\sigma\rangle$ containing a countably infinite set $A$ that is not open: just enumerate $A=\{a_k:k\in\Bbb N\}$, and for $n\in\Bbb N$ let $U_n=\{a_k:k<n\}$ and $\tau_n=\sigma\lor\{U_k:k<n\}$, the topology generated by the subbase $\sigma\cup\{U_k:k<n\}$. If $\sigma$ is $T_1$, $\tau_n$ will be the discrete topology on $U_n$, and this will be precisely Asaf’s construction in the case $X=\Bbb R$ with $\sigma$ the Euclidean topology and $A=\Bbb N$.

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Non-decreasing isn't strong enough. It needs to be increasing. –  dfeuer Aug 15 '13 at 17:53
    
@dfeuer: It needs to be non-decreasing with union not in $\tau$, which is exactly what I said. And I prefer the term non-decreasing to weakly increasing. –  Brian M. Scott Aug 15 '13 at 17:56
    
Non-decreasing may mean either $\lnot(\tau_i \supseteq \tau_{i+1})$ or $\lnot (\tau_i \supsetneq \tau_{i+1})$, neither of which is intended. We're not working in the context of a totally ordered set here, so non-decreasing and weakly increasing have completely different senses. –  dfeuer Aug 15 '13 at 18:03
    
@dfeuer: I thought that I’d written nest; fixed now. –  Brian M. Scott Aug 15 '13 at 18:04
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(Instead of just giving a counterexample, let's walk through the approach you should take in a situation like that.)

This requires simply to verify the three axioms of topology. Let's take any chain of topologies, not just countably infinite. So we have $\tau=\bigcup_{i\in I}\tau_i$ which are an increasing chain of topologies on $X$.

Recall the axioms.

  1. $\varnothing$ and $X$ are open;
  2. finite intersections are open; and
  3. arbitrary unions are open.

The first is obviously true, and the second is also fairly easy, given finitely many $U_i$ there are $\tau_i$ such that $U_i\in\tau_i$. Therefore in the maximal of the $\tau_i$'s all the sets appear, and so does their intersection.

But what about the third? Given an arbitrary collection of open sets, if we could have bound it in the chain, i.e. find a topology which contains them all, then we were done. But what if we choose one from each topology (which didn't appear before)? Then we can get in trouble. Let's use this to construct a counterexample!

Consider $X=\Bbb R$, and let $\tau_i$ for $i\in\Bbb N$ be the topology which is generated from declaring that $\{j\}$ for $j<i$ are open sets. So $\tau_0$ is the trivial topology, and $\tau_1$ contains only $\{0\}$ as a nontrivial open set, and so on.

Let $\tau$ be the union of $\tau_n$, and consider the collection $\{\{n\}\mid n\in\Bbb N\}$. Every $\{n\}\in\tau_n$, so this entire collection is a collection of elements of $\tau$. But its union is $\Bbb N$. It's not hard, at all, to show that all the open sets in $\tau_n$ are finite, for every $n$ (except $X$ of course). Therefore in the union there is no infinite open set (again, except $X$). So $\Bbb N$ is not in $\tau$, so $\tau$ is not a topology.

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Drat. Beat me by 4 minutes and came up with (basically) the same counterexample I did...and explained more thoroughly. (+1) Incidentally, you mean "indiscrete" in the second-to-last paragraph. –  Cameron Buie Aug 15 '13 at 15:20
    
@Cameron, you should buy a Unicomp Classic with blank keys. You'll not believe that fun and speed that one can get on a buckling spring mechanism when the distractions of letters are gone! –  Asaf Karagila Aug 15 '13 at 15:21
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Guys: in fact I was thinking of the same counterexample (with $\mathbb{N}$ in place of $\mathbb{R}$). I'm glad I didn't type it up because my answer would have come out a few minutes later than Asaf's, but I think it's fair to say that this is somehow a prototypical counterexample. It might even be useful to try to explain/formalize this. –  Pete L. Clark Aug 15 '13 at 15:27
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Here's something kind of similar that I'm pretty sure works.

Let $\alpha$ be an infinite limit ordinal.

For each $\beta<\alpha$, let $\tau_\beta$ be the topology on $\alpha^+$ generated by the indiscrete topology on $\beta$.

Then $\alpha = \bigcup \{\beta\mid \beta<\alpha\}$, but $\alpha$ is not open in any $\tau_\beta$.

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