Proposition: Any topological group $(G, \tau)$ which is a $T_1$-space is also a Hausdorff space.

Part of Proof:

Let $x$ and $y$ be distinct points of $G$. Then $x^{-1}y \neq e$ (identity element).

The set $G$ \ $\{x^{-1}y\}$ is an open neighborhood of $e$ and so there exists an open symmetric neighborhood $V$ of $e$ such that

$V^2 \subseteq$ $G$ \ $\{x^{-1}y\}$

Thus $x^{-1}y \not\in V^2$

Now $xV$ and $yV$ are open neighborhoods of $x$ and $y$, respectively.

Can anyone explain the last row in this proof?

Why they are open neighborhoods.

My knowledge of group theory is not what it should be.

I should add that

$V^2 = \{v_1v_2 : v_1 \in V, v_2 \in V \}$

  • 1
    $\begingroup$ In a topological group, the multiplication map $g\mapsto xg$ is a homeomorphism for any fixed $x\in G$. So the open set $V$ maps to open sets $xV$ or $yV$ under these homeomorphisms. $\endgroup$ – Ben West Feb 5 '16 at 0:22
  • $\begingroup$ Thx, now it is obvious to me! $\endgroup$ – JKnecht Feb 5 '16 at 0:34

Since $G$ is a topological group, the map $f:z \to x^{-1}z$ is continuous. Thus $f^{-1}(V)$ is open. Now $f^{-1}(V) = \{z \mid f(z) \in V \}= \{z \mid x^{-1}z \in V \} = xV$. Thus $xV$ is open.


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