controlcontrol systemcontrol theory
Let's say we have the basic feedback system of the picture 
We know that the closed loop transfer function is given by $$T(s)=\frac{G(s)}{1+G(s)H(s)}$$ Now, the diagram above is equivalent to an open loop diagram with this transfer function . 
Here's my problem . In the first diagram it's clearly a feedback system while in the second there is no feedback . How is this possible if these are equivalent ? Can a system have feedback or not depending on the perspective? I'm confused .
When evaluating the stability of a control system, it's most insightful to plot the loop gain, either as a Bode plot or as a Nyquist plot. They do not differ in this respect.
The Nyquist plot is useful for employing the Nyquist stability criterion. In summary, loop gain encirclements of the point \$(-1, 0)\$ on the Nyquist plot indicate instability. Unless the system was already unstable (i.e., has RHP poles), in which case a counter-clockwise encirclement must be made for each RHP pole.
On a Bode plot, the usual technique for evaluating stability is to investigate the gain margin and phase margin of the loop gain. If both of these values are greater than zero, then the system is stable (as long as it doesn't have RHP poles). This technique isn't quite as general as the Nyquist criterion, but for the vast majority of control systems it's good enough. It's possible to evaluate the Nyquist criterion by looking at a Bode plot, but it's more difficult.
So, why would you evaluate stability with a Bode plot when the Nyquist criterion is more general? Because the Bode plot gives you a lot of insight that the Nyquist plot doesn't. The Bode plot shows gain and phase versus frequency, helping you identify what frequencies to place compensating poles and zeroes, as well as lending insights into closed-loop response that are impossible to see on a Nyquist plot (such as closed-loop bandwidth).
Finally, once you've determined your system is stable, you can re-use the Bode plot for a meaningful demonstration of closed-loop response as well. Plotting closed-loop gain on a Nyquist plot isn't nearly as meaningful.
Start by putting some equations on it: -
simulate this circuit – Schematic created using CircuitLab
The output of A is called Vout, but clearly ....
\$V_{out} = A(V_{in} - B\cdot V_{out})\$
This boils down to \$\dfrac{V_{out}}{V_{in}} = \dfrac{A}{1+AB} \$
Best Answer
It's eminently possible for an open loop system to have exactly the same TF as a closed loop system: \$ \frac {1}{1+s}\$ could be the TF of a simple series RC circuit with R=1, C=1, or it could be a unity feedback closed loop system with an integrator, \$\frac{1}{s}\$, in the forward path. Or it could be something completely different. It's impossible to tell, from the TF itself, what the primitive structure of the system is.
Equally, putting a Laplace transform into a single box does not mean that the system is open loop. It could be the final abstraction of a primitive closed loop block diagram.
Also, using G for the forward path and H for the feedback path of a CLTF is purely convention. Those letters are not sacrosanct.