# Theorem 3.27 of Rudin's Functional Analysis

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, $$\Lambda y=\lim_j\Lambda y_j=\int_Q(\Lambda f)\,d\mu.$$