the VCdim of class of double parameter threshold functions

Let $$H$$ be a family of classifiers such that $$H=\{ h_{a,b} : a,b\in \mathbb{R}\}$$ where $$h_{a,b}(x,y)=1$$ iff $$x\geq a$$ and $$y\geq b$$.

I've proved that for $$C=\{m=(x,y)\}$$, $$H$$ shatters $$C$$.

However, when I try to choose $$C=\{m_1=(x_1,y_1), m_2=(x_2,y_2)\}$$ such that (w.l.o.g) $$x_1 and $$y_1, there is no classifier $$h\in H$$ such that $$h(m_1)=1$$ and $$h(m_2)=0$$.

Does this mean the VCdim of $$H$$ is 1? or is there a way to configure some $$h\in H$$ such that $$h(m_1)=1$$ and $$h(m_2)=0$$?

You cannot assume w.l.o.g that $$x_1 and $$y_1. But you also don't need generality: you only need to find one set of 2 points that $$H$$ shatters to show that the VC dimension is at least 2.
• Are you certain? the definition of shatters states that $H$ shatters $C$ if $|H_C|=2^{|C|}$ (contains all possible combinations). regarding what I'm trying to show, simply put that if we take 2 points in 2D $m_1 , m_2$ where $d(m_1) < d(m_2)$ ($m_1$ is closer to $(0,0)$ than $m_2$), there's no $h\in H$ thats able to categorize $h(m_1)=1$ and $h(m_2)=0$, doesnt this mean that $VCdim(H)=1$? Commented Mar 1 at 22:56
• I see your point, than to show that $VCdim(H)=d$ I must show some group $C$ of points s.t. $|C|=d$ that $H$ shatters, but what about the $|C|=d+1$ part? must I show that for any $C$ of size $d+1$, $H$ doesnt shatter $C$? Commented Mar 2 at 12:05
• I'm quite uncertain on how to prove that the $H$ cannot shatter $|C|=3$, I'm required to choose 3 arbitrary points, Yet i'm unsure on how to position them in order to show that theres not $h\in H$ which gives the correct assignment, care to direct me in this? Commented Mar 2 at 13:42