I know that for example the point $\frac{1}{4}$ is in the Cantor set. But how do I show that is not an endpoint of any interval in the Cantor set?
My attempt: At step $n$, we remove the open interval $$ (\frac{1}{3^n},\frac{2}{3^n})\cup (\frac{4}{3^n},\frac{5}{3^n}) \cup \dots \cup (\frac{3^n -2}{3^n},\frac{3^n -1 }{3^n}). $$ Therefore, for $\frac{1}{4}$ to be an endpoint of an interval in the Cantor set, it must exists an $n\in \mathbb{N}$ such that either $\frac{3^n-2}{3^n}=\frac{1}{4}$ or $\frac{3^n-1}{3^n}=\frac{1}{4}$. But if we do the calculations for each of these two cases. There exists no such $n \in \mathbb{N}$. Therefore, $\frac{1}{4}$ cannot be an endpoint.
Is this a valid reasoning, or am I missing something?
You are looking at only the last interval in your collection. The correct argument is to say that $\frac 1 4$ is not of the form $\frac k {3^{n}}$ for any positive integers $k$ and $n$: $4k=3^{n}$ would imply that $2$ divides $3^{n}$ which is not true. Since all end points of the intervals removed in the construction of the Cantor set are of the form $\frac k {3^{n}}$ we have proved that $\frac 1 4$ is not an end point of any of those intervals.