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So, the question goes like this "Prove that the intersection of an arbitrary nonempty collection of subgroups of $G$ is again a subgroup of $G$ (do not assume the collection is countable)."

At first, I thought of approaching this question with a normal mathematical induction. ( Assume true for base case, prove that if it's true for $n$ number of set, then it must be true for $n+1$ number of set and so on so forth) However, the bold part of the question confused me. Isn't it the case that if we use MI to prove this, the set is countable? ( although, it could be countably infinite ). Or, does the question actually ask us not to assume the number of sets is finite instead?

Thanks in advance for the pointers.

  • 3
    I think the authors **are** "Dummit and **Foote**"; **not** "Dummit and **Forte**"2011-06-14
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    by definition is OK.2011-06-14
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    Mathematical induction would not work even for a countably infinite collection. The collection is supposed to be "arbitrary", not assumed finite or countable. The argument is more direct than induction.2011-06-14
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    Thanks a lot for the help guys.2011-06-14
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    Well, you could prove it by transfinite induction, I suppose. (Not that you should or that it is at all necessary or helpful to do it that way...)2011-06-14
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    @biralol You have received some answers, and I think they answer your question. Please accept the answers given.2011-08-02

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