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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 from a poset. (Although it is not the same.) In this case the poset would be linearly ordered. See also: Specialization preorder. – Martin Sleziak Jun 15 '12 at 11:20
For (2): Isn't such space compact $\Leftrightarrow$ $(\Lambda,\le)$ has a smallest element? – Martin Sleziak Jun 15 '12 at 11:24
@ mean it is not always compact for all linearly ordered $(Λ,≤)$. – K A Khan Jun 15 '12 at 11:37
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. – Asaf Karagila Jun 15 '12 at 11:48
Why is $I$ a topology? Particularly, why is it closed under arbitrary unions? Consider the case $\Lambda = \mathbb{Q}, A_\lambda = (-\infty, \lambda)$. – Niels Diepeveen Jun 16 '12 at 1:57

This was already pointed out in comments, but I'm collecting this into an answer. (In fact, this is closer to a longer comment than to an answer, but the details can be elaborated better here than in a short comment.)

Both examples here deal with your construction in the special case that $X=\Lambda$. And in both cases we work with the sets of the form $(-\infty,a)=\{x\in\Lambda; x<a\}$, which is similar to lower topology.

As noticed by Niels Diepeveen, see his comment, this is not necessarily a topology.

Suppose that we have $X=\Lambda$ and $A_\lambda=(-\infty,\lambda)=\{x\in\Lambda; x<\lambda\}$.

This means that for each bounded subset $D\subseteq\Lambda$ there exists a $\lambda$ such that $\bigcup_{\mu\in D} A_\mu = A_\lambda$. It is relatively easy to show that the last condition is equivalent to $\lambda=\sup_{\mu\in D} \mu$. I.e. the linear order would have to be complete.

A counterexample suggested by Niels is $\Lambda=X=\mathbb Q$. For example if $(q_n)$ is an increasing sequence of rational numbers such that $q_n\to\sqrt{2}$, then $\bigcup (-\infty,q_n)=(-\infty,\sqrt2)$, which is not of the form $A_\lambda$.

There are several possibilities how to circumvent this. E.g. you could take all downward closed sets as open. Or you could take $\{\emptyset,X,A_\lambda; \lambda\in\Lambda\}$ as a base for the topology you want to generate.

Even if this is a topology, it need not necessarily be compact.

Let us take $X=\Lambda=\mathbb Z$.

Again $A_\lambda=\{x\in\mathbb Z; x<\lambda\}=(-\infty,\lambda)\cap\mathbb Z$. I.e. $A_\lambda$'s are down-sets of the linearly ordered set $\mathbb Z$.

Now if $\mathcal C$ is open cover of $X$ then, for each $n\in\mathbb Z$, the cover $\mathcal C$ must contain some $A_\lambda$ with $\lambda>n$. This shows that every open cover is infinite.

And if we take $\mathcal C=\{A_n; n\in\mathbb Z\}=\{(-\infty,n)\cap\mathbb Z; n\in\mathbb Z\}$, we get an open cover which does not have a finite subcover.

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@Martin...Thanx for the answer. – K A Khan Jun 18 '12 at 9:29
@ Martin...can you tell me the areas where such topologies with nested open sets can be found? any article/research paper? – K A Khan Jun 18 '12 at 10:22
Well I have already mention Alexandrov spaces (also called finite generated space). Wikipedia article has some papers among references. However, I don't know whether they were studied for the special case of linear orderings. – Martin Sleziak Jun 18 '12 at 19:17

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