Can we treat ETS(ANA) as a subset of ETS(AAA) I experimented with two series, one without trend and the other with trend. I found ets(ANA) gave better result in former (with trend) and ets (AAA) gave better results in later. But what I expected is, ets(AAA) should be providing as good a result as ets(ANA) in former series too. I thought this as I believed ets(AAA) to provide a better result in both cases because it is a superset model for all ets additive models. 
I used R forecast package to derive results.
(sorry for asking such a naive question)
 A: It depends what you mean by "better". The equations for ETS(A,A,A) are (https://otexts.org/fpp2/ets.html):
\begin{align*}
y_t&=\ell_{t-1}+b_{t-1}+s_{t-m} + \varepsilon_t\\
\ell_t&=\ell_{t-1}+b_{t-1}+\alpha \varepsilon_t\\
b_t&=b_{t-1}+\beta \varepsilon_t,\\
s_t &= s_{t-m} +\gamma\varepsilon_t.
\end{align*}
So setting $b_0=0$ and $\beta=0$, we obtain the ETS(A,N,A) model:
\begin{align*}
y_t&=\ell_{t-1}+s_{t-m} + \varepsilon_t\\
\ell_t&=\ell_{t-1}+\alpha \varepsilon_t\\
s_t &= s_{t-m} +\gamma\varepsilon_t.
\end{align*}
Therefore, the fitted model should always have smaller residual variance for the ETS(A,A,A) model. However, in practice, the optimization of the parameters sometimes means a local optimum is found and the ETS(A,N,A) model fits better.
But looking at residual variance is not a good way to select a model. You need to think about predictive accuracy, and a more complicated model may not give better predictions even if it fits better. 
That is why we use the AIC, which penalizes for model complexity. The model which minimizes the AIC is optimal (at least asymptotically) for one-step forecasts.
