Are rvs $X$ and $2X$ independent? Suppose you have a random variable $X$. Intuitively one thinks that the random variables $X$ and $2X$ will not be independent.
But I think it is not so if $X$ follows a point mass distribution, that is $\exists x_0 \in \mathbb{R}$ such that $P(X = x_0) = 1$
Claim: Let $X$ be a discrete RV that follows a point mass ($x_0$) distribution, $X$ and $2X$ are independent.
Proof: Need to show that $P(X=a, 2X=b) = P(X=a)P(2X=b)$, for all values $a,b \in \mathbb{R}$
Case 1: $a=x_0, b=2x_0$ then $P(X=a, 2X=b) = P(X=x_0, 2X=2x_0) = P(X=x_0) = 1$ and $P(X=a)P(2X=b) = P(X=x_0)P(2X=2x_0) = 1$
Case 2: $a=x_0, b \neq 2x_0$ then $P(X=a, 2X = b) = P(X=x_0, 2X = b (\neq 2x_0)) = 0$ and $P(X=a)P(2X=b) = P(X=x_0)P(2X = b) = 0$
Similarly, $P(X=a, 2X=b) = P(X=a)P(2X=b) = 0$ for the other two cases also.
Are there any other corner cases where such counter-intuitive thing happens?

Any two constant rvs are always independent. This can be seen easily following the comments above from @BGM.
Formally, let $X$ and $Y$ be two rvs (defined on the same probability space). Say $X \equiv x_0$ and $Y \equiv y_0$
Denote by $E_x$ the event $X \leq x$. Then $P(E_x) = 0$ if $x < x_0$ and $P(E_x) = 1$ otherwise. Similarly we denote by $E'_y$ the corresponding events for $Y$.
Then notice that $P(E_x \cap E'_y) = P(E_x)P(E'_y)$. Can be easily proved as follows:
Case 1: $x < x_0$. $P(E_x) = 0 \implies P(E_x \cap E'_y) = 0$
Case 2: $y < y_0$. $P(E'_y) = 0 \implies P(E_x \cap E'_y) = 0$
Case 3: $x \geq x_0$ and $y \geq y_0$. Only need to show that intersection of probability $1$ sets is also of probability $1$, which is left as an exercise for the reader :)
