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I'm trying to understand the proof of (ii)\Rightarrow(i) of proposition A.6.2.(1) here. The theorem says that the given definition of "locally compact" is equivalent to a simpler one when the space is Hausdorff. I found the proof quite hard to follow. After a few hours of frustration I'm down to one last detail. Why is F\subset U_1? It seems to me that F could contain limit points of U_1 that aren't in U_1.
Edit: I figured it out. The set V_2 is open in the topology of K_x, so there's an open set V_2' such that V_2=K_x\cap V_2'. This implies that
F=K_x-V_2=K_x-(K_x\cap V_2')=K_x-V_2'.
We also have K_x-U_1\subset V_2\subset V_2', and this implies that
K_x-V_2'\subset K_x-(K_x-U_1)=K_x\cap U_1\subset U_1.
Edit: I figured it out. The set V_2 is open in the topology of K_x, so there's an open set V_2' such that V_2=K_x\cap V_2'. This implies that
F=K_x-V_2=K_x-(K_x\cap V_2')=K_x-V_2'.
We also have K_x-U_1\subset V_2\subset V_2', and this implies that
K_x-V_2'\subset K_x-(K_x-U_1)=K_x\cap U_1\subset U_1.
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