Attempt to Hausdorff-ize spaces

Question

Introduction

Let $\mathbf{2cT_x}$ denote the category of second-countable $T_x$ spaces. We all know there is a forgetful functor $\mathsf{F} : \mathbf{2cT_2} → \mathbf{2cT_0}$. I tried to find a functor that goes to the opposite direction; $\mathsf{G} : \mathbf{2cT_0} → \mathbf{2cT_2}$.

Object mapping

For every object $X \in \mathbf{2cT_0}$, let $\mathcal{T}$ be its topology. Let $\mathscr{U}$ be the collection of topologies endowable to the underlying set of $X$, such that:

• Is finer than $\mathcal{T}$, and

• $X$ would be second-countable and Hausdorff.

Proof that $\mathscr{U}$ is nonempty:

Let $\{B_n\}_{n=1}^\infty$ be a countable basis of $X$. Then let $\mathcal{S} = \bigcup\{\{B_n, X \setminus B_n\} : n \in \mathbb{N}\}$. Then $\mathcal{S}$ is a countable subbasis. As such, the basis $\mathcal{R}$ generated by $\mathcal{S}$ is also countable.

The next step is to prove that $\mathcal{R}$ would make $X$ Hausdorff. Since $X$ is $T_0$, for every distinct points $x, y \in X$, there exists an $n \in \mathbb{N}$ where either $x \in B_n \land y \notin B_n$ or $x \notin B_n \land y \in B_n$ holds. Now give $U = B_n$ and $V = X \setminus B_n$. Then $U$ and $V$ are disjoint basis elements of $\mathcal{R}$ that separates $x$ and $y$. As such, the topology generated by $\mathcal{R}$ would be in $\mathscr{U}$.

Now let $\mathcal{U} = \bigcap \mathscr{U}$. Proof that $\mathcal{U} \in \mathscr{U}$:

It's obvious that $\mathcal{U}$ is a topology. Let $X_\mathcal{U}$ denote the space $X$ whose topology is replaced by $\mathcal{U}$. Then $X_\mathcal{U}$ is second-countable. Let $x, y$ be distinct points in $X_\mathcal{U}$. Let $\{B_n\}_{n=1}^\infty$ and $\{C_n\}_{n=1}^\infty$ be countable bases at $x$ and $y$, respectively. Let $\mathcal{V} \in \mathscr{U}$ be a topology containing $\bigcup \{\{B_n, C_n\} : n \in \mathbb{N}\}$. Since $X_\mathcal{V}$ is Hausdorff, there exist $m, n \in \mathbb{N}$ such that $B_m$ and $C_n$ are disjoint open sets in $X_\mathcal{V}$ separating $x$ and $y$. These sets are open also in $X_\mathcal{U}$.

Now give $\mathsf{G}X = X_\mathcal{U}$.

Morphism mapping

For every morphism $f : X → Y$ in $\mathbf{2cT_0}$, let $\mathsf{G}f : \mathsf{G}X → \mathsf{G}Y$ be defined as $\mathsf{G}f(x) = f(x)$. Proof that $\mathsf{G}f$ is continuous:

Since both $\mathsf{G}X$ and $\mathsf{G}Y$ are Hausdorff, continuity is equivalent to sequential continuity. Let $x \in \mathsf{G}X$. Let $\{x_n\}_{n=1}^\infty$ be a sequence in $\mathsf{G}X$ whose limit is $x$. We assert that $\mathsf{G}f(x_n)$ converges to $\mathsf{G}f(x)$.

Since $\mathsf{G}Y$ is Hausdorff, the limit of $\mathsf{G}f(x_n)$ is unique if it exists. Existence is justified by compactness of the sequence. Suppose it converged to a point other than $\mathsf{G}f(x)$, say, $y$. Let $U$ and $V$ be disjoint open sets in $\mathsf{G}Y$ having $\mathsf{G}f(x)$ and $y$, respectively. Now $x_n$ converges within $\mathsf{G}f^{-1}[V]$, yet this set is disjoint from $\mathsf{G}f^{-1}[U]$, where $x$ lives in. This is a contradiction.

Question

Are the proofs correct so far? If so, is $\mathsf{G}$ adjoint to $\mathsf{F}$ in either direction?

Answer 1

Unfortunately, the construction doesn't work. But I think it is good that you play with constructions like this, and I'd like to encourage you to continue. Let me explain the problems. You can ask if something is not clear.

Your argument that $\mathcal{U}$ is second-countable (based on your comment) doesn't work – it is not true that a topology coarser that a second-countable one has to be second-countable. On a countable set, the discrete topology is second-countable, but there are topologies which are not.

Your argument that $\mathcal{U}$ is Hausdorff (based on your comment) doesn't work – it would prove that every $T_1$ first-countable space is Hausdorff, which is not true.

In fact, it is not true that $\mathcal{U}$ is Haudorff. Let $X$ be a countable set with the co-finite topology – that is the coarsest $T_1$ topology. Now take $x_1 ≠ x_2 ∈ X$, and let $X_i$ be $X$ with the topology of a sequence converging to $x_i$. This is a nice compact Hausdorff second-countable topology. But there is no Hausdorff topology coarser than both $X_1$ and $X_2$. You can see this directly: in $X_i$ every neighborhood of $x_i$ contains everything but finitely many points, so no chance to disjointly separate $x_1$ and $x_2$. It also follows from the fact that compact Hausdorff spaces are in a delicate balance – every strictly finer topology is Hausdorff but not compact, and every strictly coarser topology is compact, but not Hausdorff.

There is a Hausdorff modification of a space $X$, but on the other side: it is not a Hausdorff space $Y$ with a continuous map $Y \to X$, but a with a continuous map $X \to Y$, and it will be a certain quotient. More generally and precisely, for every property $\mathcal{P}$ of topological spaces closed under subspaces and products (which includes separation axioms separating points or points and closed sets, so $T_i$ for $i ≤ 3½$ and also zero-dimensionality) and every space $X$, there is a continuous map $f\colon X \to Y$ with $Y ∈ \mathcal{P}$ such that for every other such map $f'\colon X \to Y'$ there is a unique continuous map $g\colon Y \to Y'$ with $f' = g ∘ f$. So this gives the left adjoint to the corresponding inclusion functor.

A sketch how to construct it: every continuous map $g\colon X \to Z$ with $Z ∈ \mathcal{P}$ factorizes as $X \to g[X] ⊆ Z$, and $g[X] ∈ \mathcal{P}$, so we may suppose that $g$ is surjective. Up to isomorphism there is only a set of such maps $\{g_i\colon X \to Z_i\}$. Take their product $g\colon X \to Z := ∏_{i ∈ I} Z_i$. We have $Z ∈ \mathcal{P}$ and every continuous map factorizes (not uniquely) through $g$. We let $f\colon X \to Y := g[X]$ be the factorizing surjection $X \to Y ⊆ Z$. Now every map factorizes through $f$ and also uniquely since $f$ is surjective.

Note that if $\mathcal{P}$ is also closed under finer topologies (includes separating axioms separating points, e.g. $T_i$ for $i ≤ 2$), then $f$ is a quotient map since taking the quotient topology on $Y$ makes is a better candidate, so it is already equal as $Y$ is the best candidate.

Similar construcion would work for a class $\mathcal{P}$ stable only under closed subspaces and products, but stronger than Hausdorff. Then we take $Y$ being the closed image, which is enough for the unique factorization since maps with dense image are epimorphisms for Hausdorff spaces. Note that this variant includes the property of being a Hausdorff compact space, and leads to the Čech–Stone compactification.

Answer 2

Let me add on to user87690's nice answer that in fact no adjoint exists on either side to the forgetful functor $F:\mathbf{2cT_2} \to \mathbf{2cT_0}$. One side is easy: $F$ cannot have a right adjoint because it does not preserve colimits. For instance, in $\mathbf{2cT_2}$, the coequalizer of the inclusion map $\mathbb{R}\setminus\{0\}\to\mathbb{R}$ and a constant map $\mathbb{R}\setminus\{0\}\to\mathbb{R}$ with nonzero value is a singleton, but in $\mathbf{2cT_0}$ it is the two-point quotient space of $\mathbb{R}$ that collapses $\mathbb{R}\setminus\{0\}$ to a single point.

To see that $F$ cannot have a left adjoint, you have to take into account the second-countability restriction. For instance, consider the space $X=(\mathbb{N}\cup\{\infty\})\times\mathbb{N}$ with the following topology. A set $U$ is open if for any $n\in\mathbb{N}$ such that $(\infty,n)\in U$, $(m,n)$ and $(m,n+1)$ are in $U$ for all but finitely many $m\in\mathbb{N}$. It is easy to see this is a second-countable $T_0$ space. If $F$ had a left adjoint $G$, then $G(X)$ would be a second-countable $T_2$ space together with a map $f:X\to G(X)$ which is initial among all maps from $X$ to second-countable $T_2$ spaces. In particular, $f$ would factor through the quotient $q:X\to Y$ that identifies all the points $(\infty,n)$ together, since $(\infty,n)$ and $(\infty,n+1)$ have no disjoint neighborhoods in $X$. This quotient $Y$ is completely regular, and so the induced map $Y\to G(X)$ would have to be an embedding (since every map $Y\to [0,1]$ factors through it and these separate points and closed sets in $Y$). But $Y$ is not second-countable (given any countable collection of neighborhoods of $q(\infty,0)$, you can diagonalize them across the copies of $\mathbb{N}\cup\{\infty\}$ to find another neighborhood not contained in any of them), so this is a contradiction since $G(X)$ is second-countable.