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I am not that familiar with arguments using ordinals so this may be pretty simple. Let $T$ be a topology on a set $X$. For each ordinal $\alpha$ let $T_{\alpha}$ be a topology on $X$ satisfying $T\subseteq T_{\alpha+1}\subsetneq T_{\alpha}$ (notice the proper inclusion) and $T_{\alpha}=\bigcap_{\beta<\alpha}T_{\beta}$ when $\alpha$ is a limit ordinal. Apparently, this should imply that there is some ordinal $\gamma$ such that $T=T_{\gamma}$. Why is this?

Edit: I see there is a problem with the above question. I will try to right this here. Henno is right that I am referring to a certain $f$ that makes a specific topology containing $T$ coarser but so that the new topology still contains $T$.

New question:

We start with a topology $T_0$ on $X$ containing $T$ and have $T\subseteq T_{\alpha+1}=f(T_{\alpha})$ for successor ordinals. We have $T_{\alpha}=\bigcap_{\beta<\alpha}T_{\beta}$ for limit ordinals. We also have that $T_{\alpha+1}=T$ whenever $T_{\alpha+1}=T_{\alpha}$. Apparently there should be a $\gamma$ such that $T=T_{\gamma}$. Why is this?

2 Answers 2

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The reason is that as you proceed with your recursion, you are gradually throwing out objects from your set. Since you only have a set of objects to begin with, you must eventually run out of things to throw out. The only way this can happen, under the assumptions you have set up, is if you have hit the bottom topology $T$.

One can turn this idea into a proof by noting that if the recursion never hit $T$, then you would have an injection from the class of ordinals into the power set of $T_0$, by the map $\alpha\mapsto (T_\alpha-T_{\alpha+1})$. (Or you could also select an element from this set and get a map directly into $T_0$.) This would contradict Hartog's theorem among others, that for every set there is an ordinal that does not inject into it.

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    Thanks. This is very helpful.2011-03-07
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For every $\alpha$ we have $T_\alpha\subseteq P(P(X))$, if you have a decreasing chain of anything longer than $2^{2^X}$ then it has to stabilize somewhere by a simple argument of cardinality.

Since all the topologies include $T$ it has to be that at some point we reached it.

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    Good point. I have edited the question.2011-03-07