# Partition of unity from RCA Rudin

Let me ask the following question: How Rudin applies Theorem 2.7 in the begining? He take some $x\in K$ then $x\in V_i$ where $i=i(x)$. What's next?

• He is taking $\{x\}$ to be the compact set of theorem $2.7$. Then $x \in V_i$ for some $i$. Therefore the $W_x$ exists. Jun 19, 2016 at 13:03
• @MaikPickl, Precisely $\{x\}\subset V_i$ where $\{x\}$ is compact since it's a one-point set. Using theorem 2.7 we get that exists neighborhood $W_x$ with compact closure such that $\{x\}\subset W_x \subset \overline W_x\subset V_i$. Right?
– ZFR
Jun 19, 2016 at 13:06
• That's the idea. You could argue that the notation is misleading since the $K$ in 2.13 is not the $K$ he is reffering to in 2.7., but after all it's just notation. :) Jun 19, 2016 at 13:08
• @MaikPickl, Yes you're right! But you can write you comment as answer! I'd accepted it as the best answer
– ZFR
Jun 19, 2016 at 13:11

Allright, just so the question doesn't appear as unanswered: He is taking $\{x\}$ as the compact set of theorem 2.7., since $x \in V_i$ for some $i$ you have $\{x\}\subset V_i$. Since $\{x\}$ is compact theorem 2.7. applies and you find an open $W_x$ with $\{x\} \subset W_x \subset \overline{W_x} \subset V_i$ where $\overline{W_x}$ is compact.