Question
We observe $x$, the maximum of $n$ values in a random sample from the uniform distribution between $0$ and $c$, where $c > 0$. Find an exact lower range $100(1 - \alpha)\%$ confidence interval for $c$.
The Answer
The maximum $X$ of $Y_1, \dots, Y_n \overset{iid}{\sim} U(0, c)$ has cumulative density function $$\begin{aligned} F(x) & = \mathbb{P}(X \leq x)\\[2 mm] & = \mathbb{P}(Y_1 \leq x)^n\\[2 mm] & = \left(\frac x c\right)^n. \end{aligned}$$ Setting this to $\alpha$ yields $$x = c\alpha^{\frac 1 n}.$$ Thus, $$\alpha = \mathbb{P}\left(X \leq c\alpha^{\frac 1 n}\right)$$ and noting that $X < c$, we have $$\begin{aligned} 1 - \alpha & = 1 - \mathbb{P}\left(X \leq c\alpha^{\frac 1 n}\right)\\[2 mm] & = \mathbb{P}\left(X > c\alpha^{\frac 1 n}\right)\\[2 mm] & = \mathbb{P}\left(X\alpha^{-\frac 1 n} > c\right)\\[2 mm] & = \mathbb{P}\left(X < c < X\alpha^{-\frac 1 n}\right), \end{aligned}$$ so the $(1 - \alpha)$ confidence interval for $c$ is $$\left(x, x\alpha^{-\frac 1 n}\right).$$
However, my professor also covered the lower range confidence interval as follows.
Let $$\begin{aligned} I & = [L, U]\\[2 mm] & = [g(Y_1, \dots, Y_n), h(Y_1, \dots, Y_n)] \end{aligned}$$ be a $100(1 - \alpha)\%$ confidence interval for $\theta$.
We say that $I$ is a lower range confidence interval for $\theta$ if $$\mathbb{P}(\theta \leq U) = 1 - \alpha.$$
Then, $$\mathbb{P}(\theta > U) = \alpha$$ and $$L = -\infty$$ and we call $U$ the $(1 - \alpha)$ upper confidence limit for $\theta$.
Using the above definition of a lower range confidence interval, shouldn't the step where the solution gives $$\alpha = \mathbb{P}\left(X \leq c\alpha^{\frac 1 n}\right)$$ be $$1 - \alpha = \mathbb{P}\left(X \leq c\alpha^{\frac 1 n}\right)$$ instead, since we want $$\mathbb{P}(\theta \leq U) = 1 - \alpha?$$
Until now, I find upper and lower range confidence intervals to be very confusing, so any intuitive explanations (regarding the solution and the definition in general) will be greatly appreciated!
