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Given $X$ be a topological space (or maybe a topological group), I am curious about is there a coarser or finer topology of $X$ or some quotient space such that it become first (or second) countable.

Is there something like this? "first (or second) countablization"?

For example, let $G$ be a non-compact locally compact group, and its "first (or second) countablization" is also non-compact locally compact group?

If this concept doesn't exist, can somebody give some intuition to explain why? Some thing like "If you try to verify the existence of "first (or second) countablization" by zorn's lemma you will encounter some ... problem."?

More precisely, I characterize this concept by universal property: Let $X$ be a topological space, and $\widetilde{X}$ be first (or second) countablization of $X$, then given any $Y$ be first (or second) countable with a continuous map $\phi:X\to Y$, then $\phi$ factor through $\widetilde{X}$.

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    $\begingroup$ Well, the discrete topology is first-countable and the indiscrete topology is second-countable (and therefore also first-countable). So you probably want to put some more requirements into the question. $\endgroup$ – Andreas Blass Nov 29 '16 at 3:59
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I assume by "first (or second) countablization" you mean an adjunction.

You could show that products in the category of first (or second) countable spaces are the usual product. Thus, since products of first (or second) countable spaces are in general not first (or second) countable there should be any reasonable right adjoint functor. (Right adjoints preserve products)

In the case of left adjoints you can easily rule out second-countable since there is no chance for disjoint unions to be second-countable. For first countable spaces look at $\Bbb R/ \Bbb N$ (collapsing $\Bbb N$) which is a coequalizer, thus the "first countablization" $\Bbb R$ can not be $\Bbb R$ since $\Bbb R/ \Bbb N$ is not first countable. Since you will have functions to first countable spaces all nice properties separation of $\Bbb R$ would get lost. Such a functor if it exists wouldn't be nice.

Andreas comment also rules out the use of Zorns lemma in most cases since you easily can construct chains with no maximal elements.

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  • $\begingroup$ I think that your argument is perfect, but I still don't understand what are you trying to say in the example $\Bbb R/ \Bbb N$. Can you explain to me in another way again in comment?(or can you explain $\Bbb R/ \Bbb N$ look like? Is it identifies all $\Bbb N$ in to one point?) On the other hand, do you means the "first countablization" of $\Bbb R/ \Bbb N$ is $\Bbb R$? $\endgroup$ – yoyo Nov 29 '16 at 12:50
  • $\begingroup$ If " the first countablization of $\Bbb R/ \Bbb N$ is $\Bbb R$" is true, I need to give you an apologize for my question didn't explain clear enough, I will edit again. $\endgroup$ – yoyo Nov 29 '16 at 12:56
  • $\begingroup$ The coeqalizer of the map sending $\Bbb N$ to $0$ and the inclusion is $\Bbb R / \Bbb N$. So if the countablization is left adjoint $\Bbb N$ has to be mapped to $\Bbb N$ since it is discrete, but the quotient has to be first countable, so the "first countablization" of $\Bbb R$ has to be ugly.(I am not talking about the "first countablization" of $\Bbb R / \Bbb N$) What I wanted to say is: I don't know if it exsits, but if it won't be nice. $\endgroup$ – user60589 Nov 29 '16 at 13:01

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