I have a small question about the proof of Theorem 3.27 of Rudin's Functional Analysis. I understood everything else in the proof, but there's one small piece I'm not able to figure out.
So what we are trying to prove is that there exist $y \in H$, where $H = co(f(Q))$ such that $\Lambda y = \int_{Q}(\Lambda f)d\mu \; \forall \Lambda \in X^*$. Rudin then defined $E_L$ to be the set of $y \in \bar H$ that satisfies the previous integral $\forall \Lambda \in L$ where $L$ is finite and $L = \{\Lambda_1, ... \Lambda_n\}$.
He then claims that each $E_L$ is closed by continuity of $\Lambda$. This is the part I don't understand - why is $E_L$ closed? Does $\cup\Lambda {y}$ for all y satisfying the integral need to be a closed set?
Thanks in advance to everyone for helping out!
Here is the whole theorem and proof by Rudin:
First, you have $$ E_L=\bigcap_{j=1}^nE_{\Lambda_j}. $$ So it is enough to show that a single $E_{\Lambda}$ is closed.
A set is closed if and only if it contains all its limit points. Suppose that $\{y_j\}\subset E_L$ is a net that converges to $y\in\bar H$. Then, because $\Lambda$ is continuous, $$ \Lambda y=\lim_j\Lambda y_j=\int_Q(\Lambda f)\,d\mu. $$