# Basic Ramsey arrow notation property

I’m having difficulties proving the following:

„Let $$a, b, c, d$$ and $$d^{\prime}$$ be cardinals with $$d^{\prime}\le d$$. Then $$a\to(b)_c^d$$ implies $$a\to(b)_c^{d^\prime}$$.“

Here’s what I know: We can assume $$d<\infty$$ and hence $$d\in\mathbb{N}$$ since $$a\to(b)_c^d$$ can’t hold otherwise. Now I’m familiar with the following approach. Take a coloring $$f:[a]^{d^\prime}\to c$$. Then we can construct a coloring $$g:[a]^d\to c$$ by setting $$g(X)$$ to $$f(X^\prime)$$ where $$X^\prime$$ is the subset of the $$d^\prime$$ least elements of the $$d$$-element set $$X$$. Now by assumption there’s a $$g$$-homogeneous subset $$S$$ of $$A$$ which is also supposed to be $$f$$-homogenious. However this last step isn’t clear to me. Let’s take $$d=3$$ and $$d^\prime=2$$. Then for $$s_1 in $$S$$ I get that {$$s_1,s_2$$} and {$$s_2,s_3$$} are colored the same due to

$$f(\{s_1,s_2\})=g(\{s_1,s_2,x_4\})=g(\{s_2,s_3,x_4\})=f(\{s_2,s_3\}).$$

But why must {$$s_3,s_4$$} also have the same color? I think the source I read implicitly assumes $$|S|\ge\infty$$. But then what can we say for the case $$|S|<\infty$$, is there a counterexample? Or maybe I didn’t get the whole idea of this proof.

I hope someone can help me with this.^^

• I found this topic already here. However in the only provided answer explicitly $|S|\ge\infty$ is required. May 27 at 13:40
• The statement is not true in general, despite appearing as an exercise in a prominent textbook. May 30 at 10:45

This is not true in general. For example, trivially $$d\to (d)^d_c$$ for all finite $$c$$ and $$d$$, but when $$d$$ and $$c$$ are greater than $$1$$, $$d\not\to (d)^1_c$$.