I have the following question:
Given a topological space $T$ is possible in general to give a topology to $2^T$ (the power set of $T$) such that this topology in $2^T$ is related to $T$.
If the answer in general is no, are there conditions over $T$ to do this?.
I'm interested to know if given a topological space $T$ and a topology on $2^T$ that is induced by the topology on $T$ one can know some topological properties of $2^T$ knowing that of $T$.
It may be better for you to consider uniform spaces instead of simply topological spaces. If you have a uniform space, then there is a very natural topology that one may put on the power set. Uniform spaces are closely related to topological spaces since one may go back and forth between topological and uniform spaces because uniform spaces are topological spaces with some extra structure. If $(X,\mathcal{U})$ is a uniform space, then $(X,\mathcal{U})$ induces a completely regular topology on $X$ where $U$ is open iff for each $x\in U$ there is an entourage $R\in\mathcal{U}$ with $R[x]\subseteq U$. Furthermore, every completely regular space $X$ can be given a compatible uniformity. For example, if $C$ is a compactification of $X$ such as the Stone-Čech compactification, then $C$ has a unique compatible uniformity. This uniformity on $C$ induces a uniformity on $X$.
Suppose that $(X,\mathcal{U})$ is a uniform space. Let $H(X)$ be the set of closed subsets of $X$. Then we can put a uniformity on $P(X)$ as follows. If $R\in\mathcal{U}$, then let $R^{\sharp}$ be the binary relation on $P(X)$ where $(A,B)\in R$ if and only if $A\subseteq R[B]=\{R[b]|b\in B\}$ and $B\subseteq R[A]$. Let $\widehat{R}$ be the restriction of $R^{\sharp}$ to $H(X)$. Then the system $\{R^{\sharp}|R\in\mathcal{U}\}$ generates a uniformity on $P(X)$, but this uniformity generally does not separate points of $X$. However, if we restrict this uniformity to $H(X)$, we get a separated uniformity on $H(X)$ and this uniformity is generated by the set of entourages $\{\widehat{R}|R\in\mathcal{U}\}$. This uniformity on $H(X)$ inherits some of the properties of your original uniform space $X$. For example, if $(X,d)$ is a metric space, then the hyperspace uniformity on $H(X)$ is induced by a metric $d^{\sharp}$ called the Hausdorff metric. The metric $d^{\sharp}$ is defined by $$d^{\sharp}(C,D)=\max[\sup_{c\in C}d(c,D),\sup_{d\in D}d(d,C)]$$ $$=Max[\sup_{c\in C}\inf_{d\in D}d(c,d),\sup_{d\in D}\sup_{c\in C}d(c,d)].$$ Furthermore, the Hausdorff metric $d^{\sharp}$ is complete whenever the original metric $d$ is complete. Now, if $C$ is a compact space, then $C$ can be given a unique uniform structure. With this uniform structure, the hyperspace $H(C)$ of $C$ remains compact. We say that a uniform space $(X,\mathcal{U})$ is non-Archimedean if it is generated by equivalence relations. The hyperspace of a non-Archimedean uniform space is always non-Archimedean.
The hyperspace uniformity is closely related to the Vietoris topology on a topological space which Steven Landsburg referred to in his answer. The reader is referred to Isbell's book on uniform spaces for more information about hyperspaces of uniform spaces.
It's more customary to consider topologies on the set of closed subsets. There is a well-known construction due to Hausdorff which does this for compact metric spaces, and yields an isometric embedding (given by $x\mapsto \{x\}$ of the original space $X$ into another compact metric space. Vietoris shows that the construction can be expressed in purely topological terms. Fell (Proc AMS 01/1962) modified this construction in a way that continues to yield good results even for nonHausdorff spaces! The construction requires closer attention to fine points of general topology. In particular, since the appropriate ambient space is neither Hausdorff nor regular, the proper definition of local compactness is crucial: here it should be in the sense that every open neighborhood of a point contains a compact subneighborhood. For any topological space $X$, Fell's prescription yields a compact topological space $\mathcal{P}_{closed}(X)$ whose points are the closed subsets of $X$, and which is compact. Moreover, if $X$ is locally compact then:
(1) The space $\mathcal{P}_{closed}(X)$ is Hausdorff.
(2) $A_\eta\rightarrow A$ (convergence of nets of subsets) if and only if $\liminf A_\eta =A=\limsup A_\eta$.
(3) If $X$ is second countable then $\mathcal{P}_{closed}(X)$ is second countable and metrizable.
(4) If $X$ is Hausdorff then $X\ni x\mapsto \{x\}\in \mathcal{P}_{closed}(X)$ is an embedding.
There is an evident retraction $\mathcal{P}(X)\rightarrow\mathcal{P}_{closed}(X)$ given by $A\mapsto \overline{A}$. One might consider endowing $\mathcal{P}(X)$ with the weakest topology such that $\mathcal{P}(X)\ni A \mapsto \overline{A}\in\mathcal{P}_{closed}(X)$ is continuous. Of course this will typically yield a nonHausdorff space.
For a topological space $(T,\tau)$ you might also consider the topology on $\mathcal{P}(T)$ generated by the subbasis
$\mathcal{S} = \{ S \subseteq \mathcal{P}(T): (\bigcup S) \in \tau\}$.
You can embed $T$ in $\mathcal{P}(T)$ by sending $t\in T$ to $\{t\}$.
See my accepted answer to this question and the comments thereon.
A reference to a 1951 survey: Ernest Michael, Topologies on spaces of subsets, TAMS 71(1951), 152-182.
