(Reminder: A $T_0$ topological space, also known as a Kolmogorov space, is a space where the topological structure "recognizes" that different points are different: No two points have exactly the same open sets around them.)

For a space that is not $T_0$, we can uniquely form a $T_0$ space from it by taking the Kolmogorov quotient, which just means sending two points to the same point in the quotient iff they are, so to speak, a counterexample to $T_0$-ness. This induces an isomorphism (i.e. order-preserving, union-/finite intersection-compatible bijection) on the topologies of the old space and the new space, and the new space is homeomorphic to any other $T_0$ space with the same topology.

Wikipedia seems to say that we can toggle $T_0$-ness on and off and get analogous theorems in many cases. But it also says, "[...] it may also be easier to allow structures that aren't T0 to get a fuller picture."

In short, I'm wondering:

  • What does point-set topology lose if we require topological spaces to always be $T_0$? (Probably nothing, in pure point-set topology.)
  • What non-$T_0$ spaces in other fields have desirable properties that are lost on Kolmogorov quotient?

(P.S.: I admit that I've been thinking about pointless topology. There's no tag for it?)

  • $\begingroup$ As I understand, the requirement for moving to pointless topology is not $T_0$, but "sober". I'm planning to think about that myself, so, uh, please don't spoil it for me (if it's not necessary for an answer to the public). $\endgroup$
    – leewz
    Oct 23, 2014 at 3:01
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    $\begingroup$ Most of the questions you might ask on pointless topology would fit under either order theory/lattice theory or topos theory. Regarding the main question, I don't know an important use for non-$T_0$ spaces. There are very important non-$T_1$ spaces, namely, spectra, so the distinction is either sharp or wrong. $\endgroup$ Oct 23, 2014 at 3:32
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    $\begingroup$ You are probably right about pure topology. In applied topology, pseudometrics arise naturally, e.g. in function spaces. Of course you can make them metric spaces by taking a quotient, but maybe it's more natural to think of the "points" as functions rather than equivalence classes of functions. $\endgroup$
    – bof
    Oct 23, 2014 at 3:38
  • $\begingroup$ The coarse topology, and some spaces that have some infinite subspaces that have the coarse topology, are useful in constructing examples and counter-examples in point-set topology. $\endgroup$ Jan 20, 2017 at 23:19
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    $\begingroup$ Topological spaces over a given set form a lattice by inclusion order. This is not the case for $T_0$ spaces. I don't know how important that is, but it is certainly something you lose if you add $T_0$ as a requirement. $\endgroup$
    – celtschk
    Feb 20, 2017 at 17:53

1 Answer 1

  1. Losing the indiscrete topology would be unfortunate in several ways. As celtschk points out in the comments, this is the minimal element in the lattice of topologies ordered by inclusion. Moreover, the indiscrete topology gives us a right-adjoint to the forgetful functor from $Top$ to $Set$.

  2. Related to 1.: Given some family of functions $(f_i : X \to Y)_{i \in I}$ and a topology on $Y$, we can construct the initial topology on $X$. If all topologies have to be $T_0$, then we would need to quotient here, which just makes stuff slightly more complicated.

  3. We have separation axioms for points and we have separation axioms for sets. Without the $T_0$ requirement, their interaction becomes more complicated. Some people might like to explore this (I personally don't, but taste differs.)

  4. Related to 2.: If we care about the particular elements of a space, and only consider the topology as a means to understand them better, then taking quotients is annoying. For example, we might be interested in integrable functions, and care about the identity of functions beyond being equal almost everywhere. Still, the topology induced by the pseudodistance is relevant.


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