$\tilde{P}$ is a refinement of $P$. $m_j\le \tilde{m_p}$?

Question

$\tilde{P}$ is a refinement of $P$. $P=\{x_0,....,x_n\}$ and $\tilde{P}=\{x_{k_0},...,x_{k_n}\}$, and $x_{k_j}=x_j.$ $m_j = \inf\{f(x):x_{j-1}\le x \le x_j\}$, and $\tilde{m_p}=\inf\{f(x_p):x_{k_{j-1}}\le x_p \le x_{k_j}\}$. My textbook says $m_j\le \tilde{m_p}$, but shouldn't it be the opposite? because $P \subset \tilde {P}$.

Edit: I add the photo in case the above explanation is not enough.

Answer 1

Let's start with the observation that if $I$ is an interval, and $I' \subset I$ is a subinterval, then $\inf \{f(x) : x \in I\} \leq \inf \{f(x) : x \in I'\}$, since the $\inf$ on the whole interval is clearly smaller than the $\inf$ on the contained interval.

Now let

$$ m_j = \inf\{f(x):x_{j-1}\le x \le x_j\}$$.

If we refine this interval, we get a bunch of points

$$\tilde x_{0} \ \tilde x_{1} \ ... \ \tilde x_{m}$$

with $\tilde x_0 = x_{j-1}$ and $\tilde x_{m} = x_j$

Now, for any $p < m$, we have

$$(\tilde x_p, \tilde x_{p+1}) \subset (x_{j-1}, x_j) $$

And so

$$\inf \{f(x) : x \in (x_{j-1}, x_j) \} \leq \inf \{f(x) : x \in x_{p}, x_{p+1} \} $$