Sorry if this problem is repeated.

Is there a nonempty perfect set in $\mathbb{R}^1$ which contains no rational number?

**Proof sketch:**
This set must be uncountable because *any nonempty perfect set in $\mathbb{R}^k$ is uncountable.* 

We know that Cantor set (C) is perfect set on a line. If we take $E=C+\sqrt{2}=\{x+\sqrt{2}: x\in C\}$. Will this set be an answer to above problem?