Recently I'm learning the use of moving plane method to prove radial symmetry of $C^{2}$ global solution of a PDE in $R^{2}$, and I'm reading a paper where this method is applied: precisely I'm reading "[Monotonicity and symmetry of solutions of fully nonlinear elliptic equations on unbounded domains](https://doi.org/10.1080/03605309108820770)" written by Congming Li.

Since the Laplacian operator is invariant under rotations, we only have to prove the symmetry about a line across origin: thus I try to prove $u(x,y)=u(-x,y)$.

Using maximum principle I have proved that 
$$w \leqslant 0 \ in \ \Sigma(\lambda) \ for \ \lambda \leqslant 0$$
and
$$w>0 \ in \ \Sigma(\lambda) \ for \ \lambda>0.$$
By using Hopf lemma we have
$$w_{1}(0,0)>0 \ for \ y \ \in \mathbb{R},$$
where 
* $\Sigma(\lambda)=\left\{(x, y) \in \mathbb{R}^{2} \mid x<\lambda\right\}$ and 
* $v=u(2 \lambda-x, y)$, $w(x,\lambda)=v-u$.

What I'm confused about is that in his paper Dr.Li says we should use the maximality of 0: trying to follow his work, I set a sequence such that 
$$\lambda^{k} \searrow 0, x^{k} \in \sum\left(\lambda^{k}\right), w\left(x^{k}, \lambda^{k}\right)>0$$
$$w\left(x^{k}, \lambda^{k}\right)=\max _{x \in \Sigma\left(\lambda^{k}\right) \atop 0\leq\lambda \leq \lambda^{k}} w(x, \lambda)>0$$
$$\nabla_{x} w\left(x^{k}, \lambda^{k}\right)=0,\left\{w_{i j}\left(x^{k}, \lambda^{k}\right)\right\} \leq 0$$
The paper said that we can 'assume' $x^{k} \longrightarrow 0 \in \overline{\sum(0)}$, and taking the limit we have
$$\nabla_{x} w(0,0)=0$$
this contradicts to the results I get using Hopf lemma that 
$$w_{1}(0,0)>0 \ \text{ for } y \ \in \mathbb{R}$$
Actually I'm so confused about the construction of this sequence **so my question is**: how can I prove that 
$$
x^{k} \longrightarrow 0 \in \overline{\sum(0)}\;?
$$