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I don't understand why this is so? Iv'e just seen the proof that a $T_0$ topological group is $T_1$, but don't know how to show that it's $T_{3.5}$.

BTW, the fact that $x\bar{V}=\bar{xV}$, one inclusion is obvious $\bar{xV} \subset x\bar{V}$, cause $xV \subset x\bar{V}$, and $\bar{xV}$ is the least closed set that contains $xV$.

Now the reverse inclusion if I take $y \in x\bar{V}$, then $y=xv$ for some $v \in \bar{V}$, which means there's an open set $O$, s.t $v \in O\cap V \neq \emptyset$, now we have $xv \in x(O\cap V)= (xO \cap xV)$, so $y \in \bar{xV}$, is this about right?

I'm more in need of help with the $T_{3.5}$ property, if you have any reference for this proof, works for me as well.

Thanks.

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Did you try to locate a proof in the literature? –  Rasmus Nov 25 '11 at 9:09
    
Yes, I did, but it doesn't explain why it's $T_3$ or $T_{3.5}$, for example in the next text: carma.newcastle.edu.au/pdf/… in THM 1.0.1, I understand why it proves it to be T1, but not why it's T3, and or T3.5. –  MathematicalPhysicist Nov 25 '11 at 10:19
    
For $x\, \overline{V} = \overline{xV}$, notice that when $f$ is a homeomorphism, then $f(\,\overline{V}\,) = \overline{f(V)}$. –  André Caldas Nov 25 '11 at 10:39
    
Thanks, Andre, didn't think of it this way. –  MathematicalPhysicist Nov 25 '11 at 10:51
    
OK, I see that in the same link I gave, they also prove that it's completely reguler (i.e, T3.5) in Corrolary 3.0.7, I need to be more patient. :-D –  MathematicalPhysicist Nov 26 '11 at 8:47

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