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I'm working through Folland's Abstract Harmonic Analysis again and find myself at a bit stuck on a certain part of a proof. The result is

If $H$ is a closed subgroup of a topological group $G$ (not necessarily Hausdorff), then $G/H$ is Hausdorff.

As per usual, for $xH$, $yH$ distinct in $G/H$, we want to find closed neighborhoods of both that do not intersect. $xHy^{-1}$ does not include $e$ since $xH$ and $yH$ are distinct and is also closed since $H$ is closed (and $xHy^{-1}$ is just a translation of $H$ which preserves closedness).

Since $xHy^{-1}$ is a closed set not containing $e$, we can find an open neighborhood $U$ of $e$ that does not intersect it. We can then find a symmetric neighborhood inside $U$ containing $e$, call it $V$.

$VV$ and $xHy^{-1}$ do not intersect which follows by virtue of $VV\subseteq U$. Since $V = V^{-1}$ and $HH = H,$ $e \not\in VxH(Vy)^{-1} = (VxH)(VyH)^{-1}$ which makes sense. Here's where I get tripped up. Folland then somehow makes the assertion that $VxH$ and $VyH$ do not intersect from the fact that $e$ is not in the product of these two sets.

What's the logic here?

Cameron L. Williams
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    Maybe I'm missing something, but if there were some $g\in VxH\cap VyH$, wouldn't it then follow that $e=gg^{-1}\in (VxH)(VyH)^{-1}$? – carmichael561 May 19 '18 at 02:06
  • Ah yep that's the logic I was missing! Thanks so much. – Cameron L. Williams May 19 '18 at 15:22
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    It took me a while to figure out why $VV\subseteq U$. So here is an explanation for people like me. Since the multiplication is continuous at $e$, there exists a neighborhood $V'$ of $e$ such that $V'V'\subseteq U$. Define $V=V'\cap V'^{-1}$. – zxcv Jul 05 '24 at 02:14

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Turning carmichael561's comment into an answer:

The key piece is that if there is some $g\in VxH\cap VyH$, then $g = v_1 xh_1 = v_2 y h_2$ and thus $g^{-1} = (v_1 xh_1)^{-1} = (v_2 y h_2)^{-1}$. Multiplying these together gives

$$ e = gg^{-1} = (v_1 xh_1)(v_2 y h_2)^{-1} \in (VxH)(VyH)^{-1}$$

which is a contradiction since $e$ is not in the set. This more or less completes the proof as the two cosets are separated, proving Hausdorffness.

Cameron L. Williams
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