Existence of a topology satisfying a certain condition 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. 
 A: 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.
A: 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 $\mathbb R.$
A: 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. 

