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I have tried to prove that a closure of a connected component of the unity in a topological group is closed, but am not sure of its validity. Since it arose from a sentence in a book on the subject,* while the fact that any open subgroup is also closed does not feature there, and since the following depends upon this fact, I must be cautious.

Statement:
In a (Hausdorff) topological group, denote the connected component of the unity by $\gamma$. Then $\gamma$ is closed.

Try:
Denote the closure of $\gamma$ by $\gamma'$, and let $\Omega$ be an open and closed non-empty subset in it. Let $\gamma^0$ be the interior of $\gamma$, and $\Omega'$= $\Omega \cap \gamma^0$. Then $\Omega'$ is closed and open and contained in $\gamma$, and thus $\Omega' = \gamma$ or $\emptyset$, hence $\Omega$ is a closed subset which contains $\gamma$, hence $\gamma' = \Omega$. Therefore $\gamma$ is closed.
C.Q.F.D.

An argument, which shows the image of the connected component of the unity in the quotient group is the connected component there, follows the statement; it uses the argument of forming the union of two closed sets without common points. But as we cannot assume in advance that $\gamma$ is open, this argument fails to produce the result here. In the end, I would like to express the thank to those who spend their time reading and answering this question.


*The book is L'intégration dans les groupes topologiques et ses applications by A. Weil.

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    I don't think your argument is correct as written. For example, I can take $\Omega=\emptyset$, which is certainly a clopen subset of $\gamma'$, but it need not be true that $\Omega'=\gamma$.2011-11-27
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    @ArturoMagidin: Yes, you are right. In the argument, perhaps it should be said that $\Omega'$ is either $\emptyset$ or $\gamma$. But this might not affect the argument much, can it? Thanks for your pointing this error out.2011-11-27
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    If the intersection is empty, then it cannot be the case that $\Omega$ contains $\gamma$ unless the interior of $\gamma$ is empty; even if the interior is empty, you have no warrant for asserting that $\Omega$ contains $\gamma$ as it currently stands.2011-11-27
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    @ArturoMagidin: I see the point now. Then how can one show this proposition? Or I have per chance to try some degree harder? Could you help me? Thank you very much.2011-11-27
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    @ArturoMagidin: Wait, since $\gamma'$ is the closure of $\gamma$, that intersection cannot be empty, can it? As it must contain some points in the closure, it absolutely has to have a non-empty intersection with $\gamma$ by definition.2011-11-27
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    @ArturoMagidin: Sorry, I know why I was wrong now. It in fact has a non-empty intersection with $\gamma$, but not so with $\gamma^0$. Still thanks for pointing out my error, and, in this case, could you post as an answer? Thanks.2011-11-28

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Maybe the extra hypotheses are clouding things: in any topological space, connected components are closed. This follows from the fact that if $Y$ is a connected subspace of $X$ then $\overline Y$ is also connected.

Here's (most of) a proof of this last fact. If $Y$ is empty then there is nothing to do, so assume otherwise. Now, if $\overline Y$ is the disjoint union of two closed subsets $A$ and $B$ then one of these, let's say $A$, must intersect $Y$. As $Y$ is the disjoint union of relatively closed subsets $A \cap Y$ and $B \cap Y$, we must have $A \cap Y = Y$ because $Y$ is connected. Does this look promising? Let me know if I should say more.

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    Eh, basically I am trying to work out the details here. And the argument above in fact is trying to prove that the closure of the connected component is connected as well. In addition, how to show that $Z$ is closed? Thanks in any case for the answer.2011-11-27
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    @awllower But there's really no need to take a closure -- you already have something closed! I'll include more details.2011-11-27
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    I am looking forward to that. Thanks and regards.2011-11-27
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    Sorry, but I have a question: does it count to be the union of two *relatively closed sets*, rather than two *closed sets*? I encountered the trouble here, as I was trying to use the argument listed here. Still thanks for answering the question.2011-11-27
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    @awllower Yes. For a subset of $X$ to be connected we mean that it is connected in the subspace topology. I was only trying to emphasize that while $A$ is closed in both $X$ and $\overline Y$, $A \cap Y$ might only be closed as a subset of $Y$.2011-11-27
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    Thanks for telling me this general result, along with its proof. For Arturo has already correctly pointed out the error in the post, I think it would be more appropriate to say that he has solved the problem. Thanks again for explaining to me this result, seemingly trivial to you.2011-11-28
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    @awllower: I don't really see the point in posting my remark as an answer, since it leaves the underlying issue unresolved. In that respect, Dylan's response is better.2011-11-28
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    @awllower Whatever you want to do is good with me. I must admit that I couldn't really figure out how to salvage your approach and saw that someone reliable was talking with you about it, so I offered this. [I also hope that the answer wasn't condescending.]2011-11-28
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    I see the point now. Thanks for everyone.2011-11-30