Weak bisimulation up-to $\approx$ Given this definition of weak bisimilarity:

A configuration relation $\mathcal{R}$ is a weak bisimulation
  provided that whenever $P\ \mathcal{R}\ Q$ and $\alpha$ is $\mu$ or $\tau$ action then:  


$P \to^\alpha P'$ then $Q \Rightarrow^\widehat{\alpha} Q'$ for some $Q'$ s.t. $P'\ \mathcal{R}\ Q'$
$Q \to^\alpha Q'$ then $P \Rightarrow^\widehat{\alpha} P'$ for some $P'$ s.t.
  $P'\ \mathcal{R}\ Q'$


P and Q are weakly bisimilar, written $P \ \approx \ Q$, if $P \ \mathcal{R} \ Q$
  for some configuration relation $\mathcal{R}$.

My questions are:


For the relation $\mathcal{R} = \{(\tau.a, 0)\}$, $\mathcal{R} \nsubseteq \approx$ but $\mathcal{R} \subseteq \, \approx \! \mathcal{R} \approx$. Why is $\mathcal{R} \subseteq \, \approx \! \mathcal{R} \approx$? 
According to Sangiorgi's book, P uses $\tau$ and ends up with $a \approx \tau.a$ and Q has $0$ and ignores the move. I am wondering, how is it that you can ignore the move? I thought that one is forced to do a move...
Could you please provide an detailed explanation of why weak bisimilarity up-to $\approx$ is unsound? (You can use the example above but I would appreciate a detailed explanation


Thanks.
(partial answer)


  
  For the relation $\mathcal{R} = \{(\tau.a, 0)\}$, $\mathcal{R} \nsubseteq \approx$ but $\mathcal{R} \subseteq \, \approx \! \mathcal{R} \approx$. Why is $\mathcal{R} \subseteq \, \approx \! \mathcal{R} \approx$? 
  


Actually, the inclusion $\mathcal{R} \subseteq \, \approx \! \mathcal{R} \approx$ holds for any relation $\mathcal{R}$, and not just the one mentioned. Indeed, if we have $P \mathcal{R} Q$, then we also have $P \approx P \mathcal{R} Q \approx Q$, by reflexivity of weak bisimulation.


  
  According to Sangiorgi's book, P uses $\tau$ and ends up with $a \approx \tau.a$ and Q has $0$ and ignores the move. I am wondering, how is it that you can ignore the move? I thought that one is forced to do a move...
  


You have to check that $0\xrightarrow{{\hat\tau}} 0$, which (if I remember correctly) is defined as $0\xrightarrow{{\tau}}^* 0$ which holds since the Kleene's star includes "zero steps".

I seem to remember that the "$\hat\alpha$" move definition made a special case for $\hat\tau$, specifically to allow for zero steps.