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In Sierpiński topology the open sets are linearly ordered by set inclusion, i.e. If $S=\{0,1\}$, then the Sierpiński topology on $S$ is the collection $\{ϕ,\{1\},\{0,1\} \}$ such that $$\phi\subset\{0\}\subset\{0,1\}$$ we can generalize it by defining a topology analogous to Sierpiński topology with nested open sets on any arbitrary non-empty set as follows: Let $X$ be a non-empty set and $I$ a collection of some nested subsets of $X$ indexed by a linearly ordered set $(\Lambda,\le)$ such that $I$ always contains the void set $ϕ$ and the whole set $X$, i.e. $$I=\{\emptyset,A_\lambda,X:A_\lambda\subset X ,\lambda\in\Lambda\}$$ such that $A_\mu⊂A_\nu$ whenever $\mu\le\nu$.

Then it is easy to show that $I$ qualifies as a topology on $X$.

My questions are -

(1) Is there a name for such a topology in general topology literature?

(2) Is there any research paper studying such type of compact, non-Hausdorff and connected chain topologies?

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    This seems to be similar to obtaining [Alexandroff topology](http://en.wikipedia.org/wiki/Alexandroff_topology) from a poset. (Although it is not the same.) In this case the poset would be linearly ordered. See also: [Specialization preorder](http://en.wikipedia.org/wiki/Specialization_preorder).2012-06-15
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    For (2): Isn't such space compact $\Leftrightarrow$ $(\Lambda,\le)$ has a smallest element?2012-06-15
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    @ Martin...you mean it is not always compact for all linearly ordered $(Λ,≤)$.2012-06-15
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    If you put $(\Lambda,\le)=(\mathbb Z,\le)$ with the usual order and $A_\lambda=\{x\in\mathbb Z; x\ge\lambda\}$ then it is not compact. Maybe I misunderstood your question (2). I thought that you are asking *when* such spaces are compact. (And the other properties.)2012-06-15
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    The topology is $S=\{\varnothing,\{1\},\{0,1\}\}$. The empty set is an element, not a singleton. Also $\phi$ is not the empty set. You can have `\varnothing` or `\emptyset` for the symbol of the empty set instead.2012-06-15
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    @Martin $A_λ\subset X$ and $Λ$ is the index set.Please read the whole formulation again.2012-06-15
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    @ Asaf thanx. I edited.2012-06-15
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    I my example $X=\Lambda=\mathbb Z$. I still think it's an example of a non-compact space obtained by your construction. If needed, we can continue the discussion about this example [here](http://chat.stackexchange.com/rooms/3784/generalized-sierpinski-space), in order to avoid putting too many comments at your question.2012-06-15
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    what is then your definition of $A_\lambda$? since $A_\lambda=\{x\in\mathbb Z; x\ge\lambda\}$ does not satisfy $A_μ⊂A_ν$ whenever $μ≤ν$.2012-06-15
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    Ok, I should have taken $\mathbb Z$ with the opposite order, not the usual order. The other possibility would be putting $A_\lambda=\{x\in\mathbb Z; x\le\lambda\}$ and leave the usual order.2012-06-15
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    Why is $I$ a topology? Particularly, why is it closed under arbitrary unions? Consider the case $\Lambda = \mathbb{Q}, A_\lambda = (-\infty, \lambda)$.2012-06-16
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    @NielsDiepeveen what is $X$? since $A_\lambda\subset X$2012-06-16
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    @MartinSleziak in this type of topology every open cover must contain $X$ itself. This is also applicable to your example as well.2012-06-16
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    BTW the same question was posted at MO, but it was closed there: http://mathoverflow.net/questions/99783/on-the-compactness-of-a-certain-chain-topology-closed2012-06-20

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