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I have read that given any topological space $X$ you can construct a Hausdorff space $h(X)$ as a quotient of $X$ which is universal with respect to maps from $X$ to a Hausdorff space. This means there is a quotient map $q:X\to h(X)$ and if $f:X\to W$ is a map to a Hausdorff space, $W$, then there is a unique map $g:h(X)\to W$ such that $gq=f$.

It seems you can construct the equivalence relation $\sim$ on $X$ so that $X/\sim$ by saying $x\sim y$ $\Leftrightarrow$ $f(x)=f(y)$ for every map $f:X\to W$ to Hausdorff $W$; this is essentially an application of a general adjoint theorem. I would like to know if it is possible to construct $h(X)$ using a transfinite process of taking quotients of $X$.

It seems like the following should be a start:

Let $X_0=X$. Say $x\sim_{0} y$ if $x$ and $y$ can't be separated by disjoint open sets in $X$. Let $\sim_{0}'$ be the transitive closure of this relation so you now have an equivalence relation. Now let $X_1=X_0/\sim_{0}'$.

Since it seems $X_1$ is not necessarily Hausdorff (I am not even sure it is $T_1$) you can construct the equivalence relation $\sim_{1}'$ on $X_1$ and a quotient map $X\to X_2=X_{1}/\sim_{1}'$ by replacing $0,1$ with $1,2$. You should be able to continue this construction by transfinite induction to get a quotient $X_{\alpha}$ of $X$ for each ordinal $\alpha$. For instance, if $\alpha$ is a limit ordinal, define $x\sim_{\alpha}'y$ $\Leftrightarrow$ there is a $\lambda<\alpha$ such that $x\sim_{\lambda}'y$.

Does this transfinite sequence of spaces stabilize to $h(X)$? If so, how does one show this?

If this initial approach is off the mark, is there a more appropriate inductive approach which does stabilize to $h(X)$?

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See also here: – Martin Sleziak Dec 27 '14 at 14:57
up vote 2 down vote accepted

Yes, it will stabilize because there is only about $2^{|X|^2}$ possible equivalence relations on $X$ (and so possible quotients). By going to a large enough ordinal you will exceed this so your sequence must have stabilized somewhere.

A cleaner way of describing your procedure is to realize that you are taking $\sim_{\lambda+1}$ to be the transitive closure of the topologic closure of $\sim_\lambda$, as subsets of $X \times X$.

The relation you need is indeed the smallest closed equivalence relation on $X$. You can either build it from the top as the intersection of all closed equivalence relations (there is always the trivial one, $X \times X$), or from the bottom, start with the diagonal and build a sequence of equivalence relations $(\sim_\alpha)$, which will be stationary as soon as you hit a closed one.

edit : also, defining it as $x \sim y \Leftrightarrow f(x) = f(y)$ for every map to an Hausdorff space is sloppy because you have to quantify over all Hausdorff spaces $W$, so you would need a set of all sets.

The top down construction works best : if there is a map $f$ to Hausdorff $W$, then the diagonal $\Delta \subset W \times W$ is closed since $W$ is Hausdorff, then so is $f^{-1}(\Delta) \subset X \times X$, and it is obviously an equivalence relation, so it contains $\sim$, and thus $f$ factors through $\sim$.

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Thanks, this is helpful. You are right that it is easier to view things in terms of closure in $X\times X$! – J.K.T. Mar 15 '12 at 22:31

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