A query about Atiyah's proof of the convexity of moment map This is about proving connectedness of level sets of moment map of a $\mathbb T^n$ $\implies$ convexity of image of moment map for $\mathbb T^{n+1}$ action. I am following Ana Cannas's wonderful lecture notes but I got stuck at a point in the proof where it says for any two points $p_0,p_1 \in M$ there is a sequence of points $q_n \rightarrow p_0,l_n\rightarrow p_1$ such that $\mu(p_1) - \mu(p_0)$ is in $ker A^t$ for some integer $(n+1) \times (n)$ matrix $A$ with rank n. 
I realised that it'd be enough to get a sequence such that $\mu(p_1) - \mu(p_0)$ has all rational coefficients is enough, but I am confused about how to do this.
I looked at this part in Dusa Mcduff's book and it doesn't elaborate on this topic either, so I have a feeling it might be really trivial but I guess I am missing some trick/ easy observation.
 A: There may be a simpler way of doing this, but here is an approach:
It suffices to consider the case when the moment map is effective (see part 4 HW21). 
To show that any point $q$ of $\mu(M)$ can be approximated by points with rational coordinates, if suffices to prove that $\mu(M)$ has non-empty interior. Indeed, if $q_0$ is an interior point of $\mu(M)$, then $\mu(M)$ contains an open ball $B\subseteq \mu(M)$, and hence (by "rational slope segment convexity" proved before) $\mu(M)$ contains all "rational slope" segments from $q$ to points in $B$, and hence the closure of the set of all such segments. This closure is the solid cone connecting $q$ to the closure of $B$, and it contains points with rational coordinates arbitrarily close to $q$.
To show that $q_0$ exists we just need to find $x\in M$ with $d\mu_i(x)$ linearly independent. Let's call such points $x$ "good", and all others "bad". A "bad" point $y$ is one where exists $X$ such that $<\mu, X>$ is critical at $y$, and so $y$ is a fixed point of some 1 paramter subgroup of the acting torus, so is fixed point of the subtorus that is the closure of this subgroup. This subtorus contains a subcircle generated by rational $X^r$ meaning that $y$ is critical for $<\mu, X^r>$. That means that the set of bad points is the union of critical submanifolds of function $<\mu, X^r>$ as $X^r$ varies over rational vectors. For each individual $X^r$ the critical set is a symplectic submanifold of $M$ (HW21, part 2), and if the action is effective it is proper submanifold, so of codimension at least 2. Since a countable union of codimension 2 submanifolds can not cover $M$ we know that a good point exists and we are done.
