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This is Exercise I.2.6b of Bourbaki's General Topology.

In a set $X$, a topology $\tau$ is called quasimaximal iff it is maximal in the set of all topologies $\sigma$ in $X$ such that $X$ does not have isolated points (i.e. $X=X^{d_\sigma}$, where $X^{d_\sigma}$ is the derived set in respect to $\sigma$).

If $X$ is Kolmogoroff and every non empty open set is infinite, show that there is a quasimaximal topology finer than the given topology.

May somebody verify whether I did not make any mistakes? I think I used only a much weaker hypothesis (namely, that $X=X^{d_\tau}$) than the one given in the exercise.

Here is my attempt:

Let $\tau$ be the given topology.

For every $x\in X$, than for $U\in\tau$ such that $x\in U$, then $U\neq\emptyset$, then $U$ is infinite, so in particular $U\cap\{x\}^c\neq\emptyset$. Therefore, $X=X^{d_\tau}$.

Let $\Gamma$ be the set of all topologies $\sigma$ in $X$ such that $\tau\subseteq\sigma$ and $X=X^{d_\sigma}$.

Then $\tau\in\Gamma$. Moreover, for every non empty chain $\Delta$ of $\Gamma$, let $\alpha$ be the topology generated by $\bigcup\Delta$, then for $x\in X$, for every $U\in\alpha$ such that $x\in U$, then there is a non empty finite subset $\mathcal{B}$ of $\bigcup\Delta$ such that $x\in\bigcap\mathcal{B}\subseteq U$, so there is a $\sigma\in\Delta$ such that $\mathcal{B}\subseteq\sigma$, so $\bigcap\mathcal{B}\in\sigma$, but $X=X^{d_\sigma}$, so $\bigcap\mathcal{B}\cap\{x\}^c\neq\emptyset$, so $U\cap\{x\}^c\neq\emptyset$; therefore $X=X^{d_\alpha}$, and $\tau\subseteq\alpha$, so $\alpha\in\Gamma$ and $\forall\sigma\in\Delta:\sigma\subseteq\alpha$; therefore, by Zorn's lemma, $\Gamma$ has a maximal element $\alpha$, so $\tau\subseteq\alpha$, and for every topology $\sigma$ such that $X=X^{d_\sigma}$ and $\alpha\subseteq\delta$, then $\tau\subseteq\delta$, so $\sigma\in\Gamma$, so $\alpha=\sigma$; therefore $\alpha$ is quasimaximal and $\tau\subseteq\alpha$.

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By an earlier exercise (par 1, 2b?), the condition that $X$ is Kolmogorov (a.k.a $T_0$) and every non-empty open set is infinite is equivalent to $X$ Kolmogorov and $X' = X$ (or $X^{d_\sigma} = X$) (which is alo called $X$ is dense in itself or $X$ is "crowded").

So the only extra thing is demanding $T_0$, and a refinement of a $T_0$ space is still $T_0$ (the quasi- in quasi-maximal is probably the absence of any separation axiom like in quasi-compact).

This essential common "sublemma" is implicit in your proof:

If $\mathcal{C}$ is a chain of topologies then any basic element in the topology $\mathcal{T}_{\mathcal{C}}$ generated by $\bigcup \mathcal{C}$ (so using that set as a subbase) is already in one of the topologies from $\mathcal{C}$.

So if some $\{p\}$ were in $\mathcal{T}_{\mathcal{C}}$, it must be in any of its bases, and so it must have been an open set in one of the members, which cannot be by construction. Obviously $\mathcal{T}_{\mathcal{C}}$ is an upperbound for $\mathcal{C}$, so the set of crowded topologies refining the initial $\tau$ is "inductive" (in Bourbaki terminology; it satisfies the Zorn lemma requirements) and so has a maximal element, and we're done. The same argument applies if we asume the stronger fact that all non-empty open sets are infinite, and it's indeed not an essential difference.

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  • $\begingroup$ Ah, I have already solved this. I think it is exercise I.1.2d. $\endgroup$ Dec 25 '17 at 3:42
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I think you are right. Starting with any crowded space you can refine the topology to get a quasimaximal one. Sometimes this property is called just maximal (or maybe maximal is reserved for $T_1$ such spaces).

Note that for $T_0$ spaces the property of being crowded is equivalent to nonempty open subsets being infinite, so the assumption is stronger only in additionally supposing $T_0$. This way you obtain a $T_0$ quasimaximal topology, and such topology is necessarily $T_1$, so it is maximal.

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This is a long comment on the conditions of the exercise.

Let $\tau$ be a $T_0$ topology on $X\ne \emptyset$ with no isolated points. Suppose, by contradiction, that $X$ has a non-empty finite open subset . Let $Y$ be such a set with the least possible number, $n,$, of members. Then $n>1$ as there are no isolated points. Now $\overline {\{p\}}\supset Y$ for each $p\in Y.$ Otherwise, for some $p\in Y$ the set $Y\setminus \overline {\{p\}}$ is a non-empty open set with less than $n$ members, contrary to the minimality of $n.$ But now take $p_1,p_2\in Y$ with $p_1\ne p_2.$ We have $p_1\in \overline {\{p_2\}}$ and $p_2\in \overline {\{p_1\}},$ contradicting the $T_0$ property.

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  • $\begingroup$ You've reproved exercise I.1.2(d). $\endgroup$ Dec 25 '17 at 6:40

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