Shows by an example that a $\gamma$-hyperconnected space in topology may not be hyperconnected. In a topological space $(X,T)$ a subset A of X is said to be preopen if $A \subset int(cl(A))$ and $\gamma$-open if $A \cap B$ is preopen for every preopen set B in $X.$ $(X, T)$ is said to be hyperconnected if closure of any open set gives $X$ and $\gamma$-hyperconnected if $\gamma$-closure of each $\gamma$-open set gives $X$.
 A: Let me summarize the picture. We start with some topology $T$ on a set $X$. The topology induces a finer family of all preopen sets, say $pT$. This in general does not have to be a topology. So we may fix this by introducing the collection of all $γ$-open sets – $γT$. It seems to me that $γT$ is a topology.
In general, we may say that a set $X$ is $(T_1, T_2)$-hyperconnected (where $T_1$ and $T_2$ are topologies on $X$) if for every nonempty sets $U ∈ T_1$, $V ∈ T_2$ we have $U ∩ V ≠ ∅$. This way, 


*

*$(X, T)$ is hyperconnected if it is $(T, T)$-hyperconnected,

*$(X, T)$ is $γ$-hyperconnected if it is $(γT, γT)$-hyperconnected,

*and in the comments we have also considered the condition of being $(T, γT)$-hyperconnected.


Since we have $Τ ⊆ γT ⊆ pT$, we have also the implications $γ$-hyperconnected $\implies$ $(T, γT)$-hyperconnected $\implies$ hyperconnected.
If $T$ is indiscrete, then both $pT$ and $γT$ are discrete. This shows $(T, γT)$-hyperconnected $\not\Rightarrow$ $γ$-hyperconnected.
It turns out that $X$ is hyperconnected if and only if it is $(T, γT)$-hyperconnected. For every nonempty preopen set $A$ there is an open set $U$ such that $A ⊆ U$ and $A$ is dense in $U$. So if $X$ is hyperconnected, for every nonempty open set $V$ we have $U ∩ V ≠ ∅$, and so $A ∩ V ≠ ∅$ since $A$ is dense in $U$. That means $X$ is even $(T, pT)$-hyperconnected.
So it is really like this: $γ$-hyperconnected $\implies$ hyperconnected, and the indiscrete space is a counterexample for the other implication.
