Call a topological space almost-discrete iff it can be obtained from discrete topological spaces by any combination of: (small) products, (small) coproducts, subspaces, quotient spaces.

(Infinite products of discrete topological spaces needn't themselves be discrete, so this isn't trivial.)

Question. Is there a nice "internal" characterization of the almost-discrete topological spaces? I'm looking for a condition on pairs $(X,\tau)$ that makes reference only to the set $X$ and the structure of $\tau$.

  • 1
    $\begingroup$ So the Cantor space is almost discrete? Oy vey. $\endgroup$ – Asaf Karagila Sep 23 '14 at 10:29
  • $\begingroup$ @AsafKaragila, good point. So perhaps every space is almost discrete... (but I hope not.) $\endgroup$ – goblin Sep 23 '14 at 10:33
  • 1
    $\begingroup$ Every compact metric space is the quotient of a Cantor set, so we already have a very large collection of spaces. $\endgroup$ – Dan Rust Sep 23 '14 at 12:05
  • 1
    $\begingroup$ In addition, every completely regular Hausdorff spaces is subspace of $[0,1]^I$ for some $I$, which obtained by infinite product of compact metric space. (I don't certain the meaning of 'small', but I assume the meaning of 'small' as category-theoretical meaning.) $\endgroup$ – Hanul Jeon Sep 23 '14 at 12:08
  • $\begingroup$ @tetori, yes, by "small" I just mean set-sized. But I prefer the terminology "small" to "set-sized" because I find the latter to be philosophically dubious; my position is that every collection is a set, just not necessarily in the universe of interest. But there is always a bigger universe. $\endgroup$ – goblin Sep 23 '14 at 12:44

As the Cantor set $C$ is homeomorphic to $\{0,1\}^\mathbb{N}$, a countable product of two-point discrete spaces, is "almost discrete".

Now, the Sierpinski space $S$ ($\{0,1\}$ with as the only non-trivial open the set $\{0\}$) is a quotient of $C$: take any non-empty open but not closed subset $U$ and send all members of $U$ to $0$, and $C \setminus U$ to $1$; the resulting quotient space (induced by this map) is the Sierpinksi space.

Also, the indiscrete space $I_2 =\{0,1\}$ is also a quotient of $C$: write $C$ as two union of disjoint dense sets $D$ and $X\setminus D$ (both ar enon-open of coursee) and send one to $0$, the other to $1$ and again take the quotient topology induced by this map on $\{0,1\}$, which is indiscrete. So both $S$ and $I_2$ are "almost discrete", and so is any small product of them.

Now, for any space $(X,\tau)$, the family of functions $f_U: X \rightarrow S$, defined by $f_U(x) = 0$ iff $x \in U$, where $U \in \tau$ is non-empty, and not $X$. Also add $g_p : X \rightarrow I_2$, defined by $g(p) = 0$, $g(x) = 1, x \neq p$, for every $x \in X$.

Note that all $f_U$ and all $g_p$ are continuous and this family of functions separates points (using the $g_p$) and separates points from closed sets (using the $f_U$). So their diagonal product into the product of all these copies of $I_2$ and $S$ is an embedding of $X$ into that product. So $X$ is also "almost discrete", as it is a subspace of a (small) product of copies of $S$ and $I_2$.

I.e. all topological spaces are "almost discrete" in this definition. We don't even need sums (coproducts) here. Quotients is too inclusive I think. Closed under products and subspaces already gives us all zero-dimensional Tychonov spaces, when we start with any non-trivial (at least 2 points) discrete space, and exactly those. Quotients adds a large class of spaces, when unrestricted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.