Two sided normal p-value question From Statistical Inference by Casella and Berger:

Let $X_1 , \dots, X_n$ be a random sample from a $n(\mu, \sigma^2)$ population.  Consider testing $:H_0: \mu = \mu_0$ verses $H_1 : \mu \neq \mu_0$.  $W(X) = |\bar X - \mu_0| / (S / \sqrt n)$ is a test statistic that rejects $H_0$ for large values which has a Student's $t$ distribution with $n-1$ degrees of freedom.
To calculate $p(x) = \sup_{(\mu, \sigma^2) \in \Theta_0} P_{(\mu, \sigma^2)}(W(X) \ge W(x))$ we recognize that the supremum is the same regardless of what value of $\sigma$ is chosen and therefore $p(x) = 2 P (T_{n-1} \ge |\bar x - \mu_0| / (s / \sqrt n))$.

How is $p(x) = 2 P (T_{n-1} \ge |\bar x - \mu_0| / (s / \sqrt n))$?  I'm not following the logic of how the supremum is twice the probability of any $(\mu, \sigma)$ chosen.

The distribution of $T_{n-1}$ is independent of $\sigma$, hence for every value of $\sigma$ the p.value equals the probability of $T_{n-1}$ being larger than $|\bar{x} - \mu_0|/(s/\sqrt{n})$. Due to symmetry of $T_{n-1}$ around $0$, it is suffice to calculate only one sided probaility and multiplying it by $2$.
