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This might be a really silly question but i'm curious to know if the following is true:

Let $X$ be an infinite set with topologies $\tau_1$ and $\tau_2$ such that $\tau_1\subset\tau_2$. Is it always possible to find a topology $\tau_3$ such that $\tau_1\subset\tau_3 \subset\tau_2$? If not then can some conditions be imposed on $X$ for which this will always hold true?

If at all we can impose some conditions on $X$, for which the above holds true, then we can obtain an infinite chain of topologies satisfying, $\tau_1\subset\tau_3 \subset\tau_4\subset...\subset\tau_2$

I worked out some examples, in some cases I could come up with such a topology and in some cases I couldn't.

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  • $\begingroup$ No. Some finite spaces are counter examples. $\endgroup$ – William Elliot Jul 12 '17 at 11:19
  • $\begingroup$ @WilliamElliot I would say all finite spaces (with two or more elements) are counterexamples, as any finite space has only finitely many topologies, so there is no way the partial ordering of topologies is dense if there is more than one topology. Also counterexample: The trivial topology on any non-empty set has two elements, and you can easily make topologies with three elements (on any $X$ with more than one point). $\endgroup$ – Arthur Jul 12 '17 at 11:20
  • $\begingroup$ @WilliamElliot Yes I see that, So for a finite point set this is always true. What if $X$ is an infinite set? $\endgroup$ – Naive Jul 12 '17 at 11:24
  • $\begingroup$ @Arthur I missed the case of considering $X$ to be finite and its true for that case. I wanted to know if the above was true for infinite sets in general. $\endgroup$ – Naive Jul 12 '17 at 11:29
  • $\begingroup$ I gave you an example that works for infinite $X$ as well, using the trivial topology. $\endgroup$ – Arthur Jul 12 '17 at 11:30
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For $T_1$ spaces I found this paper.

It refers to a lot of older papers, some of which also study whether some topologies have immediate predecessors (topologies $\tau_1 \subsetneq \tau_2$ with no topology in-between). In seems to be a well-studied subject.

So sometimes it's not possible, sometimes it is, but it's complicated stuff.

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  • $\begingroup$ I'm new to topology, I really don't understand much in that paper. But I get your point that there is no straightforward answer to this question. $\endgroup$ – Naive Jul 13 '17 at 7:28
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Not a thorough answer . Just some examples to consider.

Example 1.

On the set $S=\{0,1\}$ there are just $4$ possible topologies : The coarse topology $T_C,$ the discrete topology $T_D$, and two that are homeomorphic to each other: $T_0=\{\phi,S,\{0\}\}$ and $T_1=\{\phi,S, \{1\}\}$... (Either of these two is called Sierpinski space.)... We have $T_C\subsetneqq T_0\subsetneqq T_D$ and $T_C\subsetneqq T_1\subsetneqq T_D,$ while $T_0,T_1$ are not $\subset$-comparable.

Example 2.

Let $A$ be an infinite set. Let $T$ be a topology on $A$ with no isolated points. That is, $\forall a\in A\;(\{a\}\not \in T).$ For $B\subset A$, let $T_{P(B)}$ be the topology on $A$ generated by the base $T\cup P(B),$ where $P(B)$ is the power-set of $B$ (the set of all subsets of $B$). Let $C=\{x_n:n\in \mathbb N\}$ be an infinite subset of $A$ with $x_i\ne x_j$ when $i\ne j.$ Let $B(n)=\{x_j:j\leq n\}.$ Then $T_{P(B(n))}\subsetneqq T_{P (B(n+1))}\subsetneqq T_{P(C)}.$

Similarly to Example 1, there is no topology strictly between $T_{P(B(1))}$ and $T_{P(B(2))}.$ However there is a topology strictly between $T_{P(B(2))}$ and $T_{P(B(3))}.$

Sub-examples:.... (2i): Let $T$ be the coarse topology on $A$. Then $T_{P(B)}$ is generated by the base $\{A\}\cup P(B).$....(2ii): Let $A=\mathbb R$ and let $T$ be the usual topology on $\mathbb R.$

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Here's an example of a two infinite topologies
with no topology between them.

Let F be a free ultrafilter over an infinite set S.
Let a be a point in S and give S the topology with the base of
B = { {x} | x /= a } union { U in F | a in U }.
This space is called an ultraspace and any topology strictly
finer than an ultraspace is the discrete space.
There is no topology between the two.

Proof. Let A be a set that's not open. If a not in A, then A is
open. So a in A. However, since a is not open, A is not in F.
Since F is an ultrafilter S\A in F. Whence U = {a} union S\A in F.
Thus if A is added to the topology, U cap A = {a} would be open
and the topology would be discrete.

On the other hand, for any multipoint set S, pick a point a from S
and give S the topology { empty set, {a}, S }. S is called an
infraspace and any topology strictly finer than an infraspace is
the indescrete space. There is no topology between the two.


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