# Quantiles - supremum and infimum

How to prove that

$$\inf\{x \in \mathbb{R}: \mathbb{P}(X \le x) > \alpha \}=\sup \{x \in \mathbb{R}: \mathbb{P}(X

for any random variable $$X$$ and $$\alpha \in (0,1)$$?

If $$P\{X\leq x\}> \alpha$$ and $$P\{X then we must have $$x \geq y$$ because $$x < y$$ would imply $$\alpha , a contardiction. This proves that LHS $$\geq$$RHS. Suppose, if possible, LHS >RHS. Choose $$t$$ such that LHS $$>t>$$ RHS. Conclude that $$P\{X\leq t\} \leq \alpha$$ and $$P\{X\alpha$$ which contradicts the fact that $$P\{X\leq t\}$$ is increasing.