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In the chapter on completions in Atiyah Macdonald, they define a topological abelian group. Let $G$ be such a group. Denote $H$ to be the set of intersection of neighbourhoods of $0$. Then it is shown that $H$ is a subgroup. Then they say $G/H$ is Hausdorff because points are closed. I don't quite understand this.

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What is the question? Why points are closed in $G/H$, or why this implies that $G/H$ is Hausdorff (or both)? –  Jonas Meyer Feb 1 '11 at 23:31
    
Why points are closed implies $G/H$ is $T_2$? –  Dev Bappa Feb 2 '11 at 0:13
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3 Answers 3

up vote 5 down vote accepted

Suppose $G$ is a topological group such that $\{1\}$ is a closed set. Let $x$, $y\in G$ be two distinct elements. Since $x^{-1}y\neq1$ and $\{1\}$ is closed, there exists an open neighborhood $U$ of $1$ such that $x^{-1}y\not\in U$, so that $y\not\in xU$.

Now, it is easy to prove using the axioms of topological groups that there exists an open neighborhood $V$ of $1$ such that $VV^{-1}\subseteq U$. And, finally, one can see easily that $xV\cap yV=\emptyset$, so that $xV$ and $yV$ are disjoint open neighborhoods of $x$ and $y$, respectively. $G$ is therefore Hausdorff.

In fact, on a topological group even the fact that the topology be $T_0$ (a weaker condition that $T_1$,which is the one we used above) is enough to prove that it is Hausdorff. You can do that as an exercise :)

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Thanks for the elaborate answer. I am still unable to see why $xV\cap yV=\emptyset$. –  Dev Bappa Feb 2 '11 at 0:13
    
OK. I guess if you choose a $V$ as above, then consider the neighbourhood of $1$ given by $V\cap (G-{{xy^{-1}}})$, then the intersection in my comment above is nonempty. –  Dev Bappa Feb 2 '11 at 0:27
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G/H will be Hausdorff, essentially because quotienting by H identifies all the points that can't be discerned from 0.

Lemma : If $U \subset G$ is open, $U = U+H$ :

Let $x \in U$. $U-x$ is a neighboorhood of 0, so $H \subset U-x$. Thus $H+x \subset U-x+x = U$. Since we have this for all $x$, $U = U+H$.

Let $x$ be an element of G/H. $x$ is the projection of some $y$ from G, with $y \not\in H$. Since $y \not\in H$, it means that there exists a neighboorhood $U$ of 0 that doesn't contain $y$. Then, as Mariano said, since the application $x \rightarrow x-x$ is continuous, there exists an open $V$ containing 0 such that $V-V \subset U$. Finally, since we have $y \not\in V-V$, $y+V \cap V = \emptyset$, and $y+V+H \cap V+H = \emptyset$.

This shows that, in the quotient G/H, 0 and x are separated.

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Points are closed in $\mathbb{R}$ with the usual topology. Points are always closed in Hausdorff spaces. In topological groups the converse holds. –  Jonas Meyer Feb 2 '11 at 0:50
    
Uh I wonder how I wrote that o_o. Yes they are. Let that be a lesson to not do math in the middle of the night. –  mercio Feb 2 '11 at 0:53
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You should edit your answer to mark the erroneous claims. –  Mariano Suárez-Alvarez Feb 2 '11 at 3:04
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Other ways to go about it:

Topological groups $G$ are homogeneous (for every $x$ and $y$ in $G$ we have a homeomorphism of $G$ that maps $x$ to $y$) and homogeneous $T_0$ spaces are $T_1$ (easy exercise). Suppose there is a net $(x_i), i \in I$ that converges to 2 points $x, y$. Then the net $((x_i)^{-1} * x_i), i \in I$ is constantly $e$ but also converges (using continuity of the group operations) to $x * y^{-1}$ which must equal $e$ as $\{e\}$ is closed in a $T_1$ space. Hence $x = y$ and $G$ is Hausdorff (unique net limits).

In fact we can get Tychonoff (completely regular, $T_{3\frac{1}{2}}$) and the most general way to go about that is to use that the topology is induced by a uniformity. All uniform spaces that are $T_0$ are Tychonoff.

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The same argument without mentioning nets: A topological space is Hausdorff if and only if its diagonal is closed. The function $f: G \times G \to G$ given by $(x,y) = x^{-1}y$ is continuous. If $\{1\}$ is closed then so is the diagonal $f^{-1}(1)$. –  t.b. Feb 2 '11 at 9:17
    
@Theo, thanks, I forgot about that one. –  Henno Brandsma Feb 2 '11 at 12:21
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