I'm having trouble understanding a variation of the famous "all horses are the same color" problem, and would like some help. While I am familiar with the original problem, and know what's wrong with it, this one is formulated a bit differently, and I'm a bit confused about how to approach it.
Hyp: When one horse in a group is red, so are all the others.
The proof that I'm supposed to debunk, then goes as follows:
For N = 1, the claim is true.
We assume that the claim goes for groups of the size N.
Now, let one in a group of N+1 horses be red. We consider this horse together with the N-1 others. We can conculde that these are also red, because of the assumption. Because of that, there is at least one red horse in the group of these N-1 horses, together with the one that we haven't considered, which means that the last horse is also red.
Can I apply the same principle where N=2 and the two subsets have no common elements? Because the way I look at it, when I try to do that and have $H_0$ and $H_1$, and argue that I'm saying that if I consider $H_0$ and all the other horses except $H_1$, if one is red, so are the others, and I have my red horse. Then I take all those nonexistent red horses together with $H_1$, the assumption tells us that $H_1$ is red too, which is wrong, since I can't compare a horse to nonexistent horses.
Now, I could just turn that in and be done with it, but I'd really like to understand what's going on? Namely, why can't someone argue that they can simply look at $H_0$ and $H_1$ together, and if $H_0$ is red, then the assumption tells us that $H_1$ too must be red. I feel like I'm close to understanding this, but still have a tiny bit missing.
Thank you for your help!