Q: (Baby Rudin Chap $2$ Prob $18$) Is there a nonempty perfect set in $\mathbb R$ which contains no rational number?
Proof (which is obviously wrong since there are counterexamples):
Consider that such a set exists. It follows that it must have a irrational point. By the definition of perfect set, this point must be a limit point. By definition of limit point, we get that a rational number must be part of this set (since there is a rational between any two numbers). This contradicts the property of the set. Hence, no such set can exist.
What is wrong with this "proof"? I must be missing some basic definition.
The problem is here:
If $x$ is a limit point of $A\subseteq\mathbb{R}$ then there must be a sequence $(y_i)_{i\in\mathbb{N}}$ of points in $A$ with limit $x$ ... but all those points could be irrational! That is, we could "skip over" each rational.
It might help to consider the picture with "rational" and "irrational" switched: there are irrationals between any two numbers, but $$0,{1\over 2}, {3\over 4}, {5\over 6}, ...$$ is a sequence containing no irrationals which converges to the rational $1$. It's an easy exercise to "flip this around" and get a sequence of irrationals converging to an irrational.