Does a stretch of a set $E\subset\mathbb{R}$ about a condensation point of $E$ intersect $E?$

A point $p$ in a metric space $X$ is said to be a condensation point of a set $E\subset X$ if every neighbourhood of $p$ contains uncountably many points of $E.$ [Note that $p$ need not be in $E.$]

Suppose $0$ is a condensation point of a set $E\subset\mathbb{R}.$ Let$\alpha>0.$ Then does $\{\alpha x: x\in E\}\cap E \neq \emptyset ?$

Something I know about condensation points is that the set of condensation points of an uncountable set is perfect.

Maybe the Cantor set with a well-chosen $\alpha$ is a counter-example?

Or maybe a counter-example can be constructed using the Axiom of choice?