What is the criteria of trapped ideal gas to form Bose-Einstein condensate (BEC)? If we have ideal gas of bosons in a trapped harmonic potential, is the only necessary thing for BEC is a temperature less than the transition temperature? Or is there any other things we should keep in mind. Also, why do we discuss BEC of gas that is trapped? What is the criteria of ideal gas to form Bose-Einstein condensate ? 
 A: If the system is in thermal equilibrium, then the only relevant quantities are the temperature $T$ (or the conjugate variable of entropy $S$), volume $V$ (or pressure $p$), and the chemical potential $\mu$ (or number of particles $N$). From these, one can calculate whether the gas will be condensed or not. For fixed volume and number, this reduces to determining whether the temperature is above or below the critical temperature at that volume and number (this temperature can also depend on e.g. the particle mass, trap geometry, etc..). 
The necessity of discussing trapping is two-fold: first, it is often the case that the experiments are done on trapped gases, so we should try to include this in our calculation. Second, even a free (un-trapped) gas is formally considered as the limit of a trapped gas, with trap size going to infinity (see "particle-in-a-box normalization").
As for the exact criteria, it is effectively to "determine whether all particles are accommodated by the thermal distribution." By this, I mean that one should compute the expected number of particles 
$$ \langle N \rangle = \sum_s \bar{n}_B(E_s),$$
according to the Bose-Einstein distribution $\bar{n}_B(E) = (\exp($\beta$(E-\mu N))+1)^{-1}$ for inverse temperature $\beta$ and chemical potential $\mu$, and compare this to the actual number of particles in the experiment, $N$. If it is possible to choose $\mu<0$ such that all $N$ particles are thermally occupied, then there is no condensation at that temperature, pressure, etc... However, if you cannot satisfy $\langle N\rangle = N$ for any negative $\mu$ at that temperature, then the remaining $N-\langle N\rangle$ particles must be residing in a non-thermal "condensed" state. The critical temperature is the temperature $T_c$ at which $\mu = 0$ is required to maintain $N = \langle N\rangle$. The requirement that $\mu < 0$ stems from the fact that the Bose-Einstein distribution has a singularity at positive energy if $\mu >0$, and that is not possible. 
A: Temperature only?
No. Like also the liquid-gas transition depends on both temperature and pressure.
The BEC happens when the thermal de Broglie wavelength $ \lambda_{\text{th}}\propto n/\sqrt{T}$, $T$ being temperature and $n$ being number density, becomes comparable to the system size - i.e. becomes macroscopic.
The ratio $\propto n/\sqrt{T}$ can also be cast into a more meaningful quantity knows as phase space density $\Lambda \propto n\, \lambda_{\text{th}}^3 \propto n\,T^{-3/2}$ (for a uniform trap, $T^{-3}$ for a harmonic trap). $\Lambda$ basically tells you the combined volume of the cloud in real space ($n$) an in momentum space ($p = h/\lambda$).
Bose Einstein condesation is achieved when $\Lambda \sim 1$.  To achieve this, you need both high $n$ and low $T$.  So no, you cannot just decrease the temperature.  You can actually show that, e.g. in a harmonic trap, decreasing the temperature $T$ leaves the phase space density unchanged - this is analogous to the concept of adiabatic invariants in classical physics.
The usual picture of a BEC is that it is a saturation effect.  The occupacy for bosons $f = \frac{1}{\dots \,-1}$ has is bounded from above for any energy that is not the ground state. So:

*

*you can keep the temperature constant, and increase the number of atoms $N$ until you cannot fit them into the excited states anymore, and they will populate the only state without a cap - the ground state.


*or you can keep the number of atoms $N$ constant and decrease $T$, so as to reduce the number of accessible states ($E \propto k_B \,T$) states, making atoms occupy lower states, and eventually the ground state when they again run in to the cap.
So it's either $N_c$ or $T_c$.  Better summarised as $\Lambda \sim 1$, since experimentally it's hard to control $N$ and $T$ so well that make them constant. Usually, during evaporative cooling, you decrease both $N$ and $T$, but at a different rate so as to have a net increase in $\Lambda$ - a regime called run-away evaporation.
Why trapped?
The BEC is not a (long-term) equilibrium state.  It is metastable.
Why? Do you we see BECs in nature? What do things do when they are cooled down?  They form molecules and solids.
The BEC is dilute, i.e. very low density, for the precise reason that too many particles would interact strongly and solidify.  That's why, experimentally, we choose the "cooling temperature" route rather than the "increase $N$" route.
The rate at which "bonds" create goes (at least) as $n^2$, $n$ being the atom density, as it takes $2$ atoms to form a bond (technically $3$ to conserve energy and momentum).  These losses, along with others (such as the pressure level of the chamber where the BEC is created), are the reason BECs have a finite lifetime.
But as we know, untrapped stuff thends to go on about their own way. And that would decrease $n$ and $\Lambda$. So you need to keep it trapped to keep the number density up (though not too tightly trapped to avoid losing atoms to losses).
Harmonic traps are the most common, but there are also uniform traps (like square wells), hybrid traps etc.  The geometry of the trap affects the energy levels spacing and therefore $N_c$ and $T_c$.
