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I'm trying to understand the proof of (ii)[itex]\Rightarrow[/itex](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 [itex]F\subset U_1[/itex]? It seems to me that F could contain limit points of [itex]U_1[/itex] that aren't in [itex]U_1[/itex].
Edit: I figured it out. The set [itex]V_2[/itex] is open in the topology of [itex]K_x[/itex], so there's an open set [itex]V_2'[/itex] such that [itex]V_2=K_x\cap V_2'[/itex]. This implies that
[tex]F=K_x-V_2=K_x-(K_x\cap V_2')=K_x-V_2'.[/tex]
We also have [itex]K_x-U_1\subset V_2\subset V_2'[/itex], and this implies that
[tex]K_x-V_2'\subset K_x-(K_x-U_1)=K_x\cap U_1\subset U_1.[/tex]
Edit: I figured it out. The set [itex]V_2[/itex] is open in the topology of [itex]K_x[/itex], so there's an open set [itex]V_2'[/itex] such that [itex]V_2=K_x\cap V_2'[/itex]. This implies that
[tex]F=K_x-V_2=K_x-(K_x\cap V_2')=K_x-V_2'.[/tex]
We also have [itex]K_x-U_1\subset V_2\subset V_2'[/itex], and this implies that
[tex]K_x-V_2'\subset K_x-(K_x-U_1)=K_x\cap U_1\subset U_1.[/tex]
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