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Definition: Let $(X, \mathcal{T})$ be a topological space, where the set $X$ has more than one element. Suppose that for every pair of distinct elements $a, b \in X$, there exists a separation $(A,B)$ of $X$ such that $a \in A$ and $b \in B$. Then we say $(X, \mathcal{T})$ is very disconnected.

Is this condition (being "very disconnected") equivalent to another, well-known one?

The definition above is my own, but I suspect it is equivalent to some pre-existing notion (e.g., a $T_{n \frac{1}{2}}$ space for some $n$). Here are a few propositions that I have proved about v.d. spaces:

  1. Any very disconnected space is disconnected.

  2. Any discrete space is very disconnected.

  3. There are very disconnected spaces that are not discrete.

  4. If a space is very disconnected, then all singletons are closed.

  5. All singletons closed does not imply the space is very disconnected.

Benjamin Dickman
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  • Do you mean "totally disconnected" perhaps? – Asaf Karagila Jul 03 '13 at 17:31
  • A separation is a decomposition into two disjoint sets that are simultaneously open and closed? – Daniel Fischer Jul 03 '13 at 17:41
  • @DanielFischer Right: $(A,B)$ is a separation of $(X, \mathcal{T})$ provided $A \cap B = \emptyset$; both $A,B$ are nonempty, clopen sets; and $A \cup B = X$. – Benjamin Dickman Jul 03 '13 at 17:44
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    @AsafKaragila My recollection is that "very disconnected" is different from "totally disconnected." If I have misremembered and you can see (show) them as the same, then I would gladly accept that as an answer. – Benjamin Dickman Jul 03 '13 at 17:46
  • No, I was just asking whether you may have meant "totally disconnected". Both terms would translate almost the same in Hebrew, so I wanted to be sure that a cross-lingual barrier didn't cause any issues here. – Asaf Karagila Jul 03 '13 at 17:48
  • Okay, then a very disconnected space is certainly totally disconnected. I'm not sure yet whether it's in fact equivalent, or stronger. I suspect it's stronger, but can't come up with an example quickly. – Daniel Fischer Jul 03 '13 at 17:48
  • @AsafKaragila No, the definition here ("very disconnected") is one that I came up with as a part of a top. course that used the closure axioms. Probably v.d. implies t.d. or vice-versa (I see Daniel Fischer has now made a remark to this effect.) – Benjamin Dickman Jul 03 '13 at 17:49
  • @DanielFischer Consider $(\mathbb{Z}^+, \mathcal{T})$ where all finite sets are closed, and all infinite sets containing $1$ are closed. I believe this gives a very disconnected space. Furthermore, the nontrivial set ${2, 3, 4, \ldots}$ is a connected subset: it cannot be written as a disjoint union of two closed sets, since at least one of them must be infinite but not contain $1$. Thus, I do not think v.d. implies t.d. – Benjamin Dickman Jul 03 '13 at 17:59
  • ${2,, 3,, \ldots}$ is not connected. Disconnected means a set can be written as a disjoint union of two relatively closed/open subsets. Any partition of that set into one finite set and its complement is such. – Daniel Fischer Jul 03 '13 at 18:04
  • Oh, right, the subset would be given the subspace topology; thanks. – Benjamin Dickman Jul 03 '13 at 18:08
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    I think this property is normally called being totally separated. It is stronger than being totally disconnected. – Chris Eagle Jul 03 '13 at 18:08
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    @ChrisEagle A ha, this looks like what I was looking for. (The one difference being that my definition has $|X| > 1$.) Thanks. (If you include this as an answer I will accept it.) – Benjamin Dickman Jul 03 '13 at 18:10

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You can define a relation where $x\sim y$ if there is no separation of the space $X$ between $x$ and $y$, or equivalently, each clopen set containing $x$ also contains $y$. It is very easy to check that this is an equivalence relation and that the equivalence class of $x$ is the intersection of all clopen neighborhoods. This class is called quasi-component of $x$.

This relation is coarser than the relation defining connectedness, so a quasi-component is a disjoint union of components. There are conditions under which the components and the quasi-components coincide, for example when $X$ is compact Hausdorff, or when there are only finitely many quasi-components. It also holds if the components are open (which is the case for locally connected spaces or spaces with only a finite number of components.)

An example of a totally disconnected space, i.e. all components are singletons, where the quasi-components are not the singletons is the sequence $$\left\{\frac1n\mid n\in\Bbb N\right\}\cup\{0,0'\}$$ converging to two distinct zeros, where the neighborhood base of $0$ is given by the intervals $[0,\epsilon),\ \epsilon>0$ and similarly for $0'$. Then each $0$ is a component but the quasi-component of $0$ is $\{0,0'\}$

Edit: @Chris Eagle points out that a space where the quasi-components are the singletons is called totally separated.

Stefan Hamcke
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