modification of the Ramsey number This was a question on my exam which I was struggling with:

Show that there is a number $T(a,b)$ for given positive integers $a,b$, such that for any set $N$ with $|N|\ge T(a,b)$ and any subset $E\subseteq \{h\subseteq N: |h|=3\}$, at least one of the following is true:
  -There is a subset $X\subseteq N,|X|\ge a$ such that $\{h\subseteq X: |h|=3\}\subseteq E$
  -There is a subset $Y\subseteq N,|Y|\ge b$ such that $\{h\subseteq Y: |h|=3\}\cap E=\emptyset$.

My idea was to color all the 3-subsets of $E$ red. Then the Ramsey number $R(3,a,b)$ says there is a subset that contains $a$ red 3-subsets, or $b$ blue ones. From this the claim follow. But this seems way to easy and I'm pretty sure my reasoning is wrong, I don't see where though.
 A: This is just Ramsey's theorem for $2$ colors and hypergraphs of dimension $3$ the existence of Ramsey's theorem for $2$ colors and dimension $n\geq3$ can be proven using induction. I will just prove the step from $n=2$ to $n=3$ but it can be easily modified. $R(a,b)$ is just the usual Ramsey number on two colors and dimension $2$.The $3-$subsets of color 1 are those pertaining to $E$ and the $3-$subsets of color $2$ are those not pertaining to $E$
The claim is that $R_3(a,b)\leq R(R_3(a-1,b),R_3(a,b-1))+1$
The proof is by dividing in two cases, each of which has two subcases. notice that a vertex $v$ has $R(R_3(a-1,b),R_3(a,b-1))$ neighbors. We color the edge $uw$ blue if the triangle $uvw$ is blue and red if triangle $uvw$ is red. Since $v$ has $R(R_3(a-1,b),R_3(a,b-1))$ neigbors, its neighbors must make a red complete graph of size $R_3(a-1,b)$ or a blue complete graph of size $R_3(a,b-1)$.
Case 1: the neighbors of $v$ make a red complete graph of size $R_3(a-1,b)$. Then there are two cases: that complete graph contains a subsetset of size $a-1$ with only red triangles inside it. If this happens, since all of the edges are red, we can add vertex $v$ and all triangles of the new set will also be red, this set has size $a$. The other case is that the complete graph has a subset of size $b$ with only blue triangles, but this also satisfies the problem.
Case 2: the neighbors of $v$ make a complete blue graph of size $R_3(a,b-1)$. Then there are two cases: that complete graph contains a subsetset of size $a$ with only red triangles inside it. or it contains a subset of size $b-1$ with only blue triangles, but in this case we can add $v$ and since all the edges of that complete graph are blue, so are all the triangles with $v$. So we now have a subset of size $b$ containing only blue triangles.
This completes the proof.
