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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 $\tau$.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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