Where is the mistake in the induction proof for 'All horses are the same color'? I don't really understand Wikipedia's proof, because 1. if we assign distinct numbering to all the horses, I find  obvious that there can be common elements to two subsets. 


*Also, is the mistake in the initial assumption ($n$ horses always have the same color)? Is this wrong? Because if there are no horses, then there can't be nonzero $n$ horses in the first place. 

 A: If all horses in every set of $25$ horses have the same color,
then all horses in every set of $26$ horses have the same color:
\begin{align}
& \underbrace{ABCDEFGHIJKLMNOPQRSTUVWXY}_{\large\text{All of these have the same color.}} \, Z \\[10pt]
& A\,\overbrace{BCDEFGHIJKLMNOPQRSTUVWXYZ}^{\large\text{All of these have the same color.}}
\end{align}
This gets you from $25$ to $26.$ And similarly from $26$ to $27.$ And so on.
But it does not get you from $1$ to $2:$
$$
\underbrace{\qquad A \qquad} \, \overbrace{\qquad B \qquad}
$$
That is why, if it were said that any two snowflakes are identical, one would conclude that all snowflakes are identical.
The failure to make the step from $1$ to $2$ is why the argument fails.
A: The inductive step falsely assumes that $n \geq 2$.  Note that if you try the inductive step only with 2 horses, it is wrong:  Horse 1 could be red, Horse 2 could be blue, and there is no contradiction.
A: The proof argues that the set $S$ of all horses except the first and the set $T$ of all horses except the last intersect on $M$, the set of all middle horses. Then the first horse has the same color as the middle horses which have the same color as the last horse, so all horses have the same color. The problem is that if there are only $2$ horses, the set of middle horses are empty, so the argument fails for $n=2$. For $n>2$, the proof requires the previous case to hold, so the case for $n=3$ fails because it fails for $n=2$, and so on.
