# finest non-discrete topology

Suppose we choose a topology which is not properly contained in any topology which is not discrete, appealing to Zorn's lemma. Does this yield anything interesting? I do not know a lot of topology, so I don't know how to approach this, or if it gives anything useful.

Searching for maximal topologies always gives results where extra conditions are added, such as being compact. I could not find any examples of the above choice. Perhaps there is an elementary reason it is uninteresting?

Edit: since it seems that boring examples exist, is it possible to consider a topology as above which can only be obtained via ZL? I.e., analogous to a non principal ultrafilter... Thank you for the quick response.

Further edit: I wonder then if there are any 'strange' examples of such topologies at all... But since I assumed Zorn held, this is a rather different question. I'll accept the current answer a few days if there are no more responses.

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I can't see any way to apply Zorn here. How are you doing so? –  Chris Eagle Jan 13 '13 at 17:11
Hmm. I naively thought it would fall through in the obvious way, but it seems not to. –  user10193 Jan 13 '13 at 17:49

Here’s a characterization of maximal non-discrete topologies.

Lemma. Let $\tau$ be a non-discrete topology on a set $X$, and let $N=\big\{x\in X:\{x\}\notin\tau\big\}\ne\varnothing$. Then $\tau$ is maximal non-discrete iff

1. $\tau$ induces the discrete topology on each $A\in\wp(X)\setminus\tau$, and
2. $A\supseteq N$ whenever $A\in\wp(X)\setminus\tau$.

Proof. Assume first that $\tau$ is maximal non-discrete. Let $A\in\wp(X)\setminus\tau$. If $\tau_A$ is the topology generated by the subbase $\tau\cup\{A\}$, it’s not hard to check that $$\tau_A=\big\{U\cup(A\cap V):U,V\in\tau\big\}\;,$$ so every subset of $X$ must be expressible in the form $U\cup(A\cap V)$ for some $U,V\in\tau$. In particular, for each $x\in X$ there must be $U_x,V_x\in\tau$ such that $\{x\}=U_x\cup(A\cap V_x)$. If $\{x\}\in\tau$ we may take $U_x=\{x\}$ and $V_x=\varnothing$; if, however, $\{x\}\notin\tau$, we must have $U_x=\varnothing$ and $A\cap V_x=\{x\}$. Thus, $\tau$ induces the discrete topology on $A$.

Suppose that $A\nsupseteq N$ for some $A\in\wp(X)\setminus\tau$, and fix $x\in N\setminus A$. Then $\tau_A$ is strictly finer than $\tau$, but $\{x\}\notin\tau_X$, contradicting the maximality of $\tau$.

Now suppose that $\tau$ satisfies (1) and (2), let $\tau'$ be a topology strictly finer than $\tau$, and let $U\in\tau'\setminus\tau$. By hypothesis $U\supseteq N$ and $\tau$ induces the discrete topology on $U$, so $\{x\}\in\tau'$ for each $x\in N$, and $\tau'$ is the discrete topology on $X$. Thus, $\tau$ is maximal non-discrete. $\dashv$

But now we have an easy

Theorem. Let $\tau$ be a topology on a set $X$. Then $\tau$ is maximal non-discrete iff $X$ has a unique non-isolated point.

Proof. In the notation of the lemma this just says that $\tau$ is maximal non-discrete iff $N$ is a singleton. If $N$ is a singleton, then (1) and (2) are clearly satisfied, so $\tau$ is maximal non-discrete. Suppose now that there are distinct $x,y\in N$; then $\{x\}\in\wp(X)\setminus\tau$, but $\{x\}\nsupseteq N$, so $\tau$ is not maximal non-discrete. $\dashv$

In Michael Greinecker’s example the unique non-isolated point is of course $y$, and its nbhd filter is the fixed filter generated by the set $\{x,y\}$. Among the nicer examples are $\omega+1$ (i.e., a simple sequence with its limit point) and $\{p\}\cup\omega$ for any $p\in\beta\omega\setminus\omega$.

(I could probably have done this more simply, but this is how I actually discovered the result, so I thought that I’d let it stand as is.)

Added: I realized somewhat belatedly that these spaces can be described even more precisely. Let $X$ be a set with more than one element, and fix $p\in X$. Let $Y=X\setminus\{p\}$, let $\mathscr{F}$ be any filter on $Y$, and let $\tau=\big\{\{p\}\cup F:F\in\mathscr{F}\big\}\cup\wp(Y)$. Then $\tau$ is a maximal non-discrete topology on $X$, and all such topologies are obtained in this way.

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I particularly like your final addendum. Thank you for this. –  user10193 Jan 15 '13 at 18:59
@user10193: You’re very welcome; that question was fun. –  Brian M. Scott Jan 15 '13 at 19:04

Let $X$ be a set with at leas two elements $x$ and $y$. Let the topology consist of arbitrary unions of singletons from $X\backslash \{y\}$ and the doubleton $\{x,y\}$. The result is a fairly boring topology such that any finer topology is discrete.

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