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Math Help - Closed and compact sets

  1. #1
    mms
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    Closed and compact sets

    let <br />
(E,\left\| . \right\|)<br />
be a normed vector space, and <br />
A,B \subset E<br /> <br />

    Show that if A is compact and B is closed then A+B is closed
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mms View Post
    let <br />
(E,\left\| . \right\|)<br />
be a normed vector space, and <br />
A,B \subset E<br /> <br />

    Show that if A is compact and B is closed then A+B is closed
    This generalized substantially, but regardless. This is not a particularly easy problem as I recall. Have any leads? I remember having to prove a lemma about disjoint sets around the additive identity elements and compact sets or something.
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    Let {c}_{n}={a}_{n}+{b}_{n} be a sequence in A+B converging to a point c\in E, where {a}_{n}\in A, {b}_{n}\in B and {c}_{n}\in A+B. Since A is compact, there is a subsequence {a}_{n(k)} of {a}_{n} that converges to some a\in A. Since {c}_{n(k)} also converges (its a subsequence of a convergent sequence), it follows that {b}_{n(k)}={c}_{n(k)}-{a}_{n(k)} also converges, say to b\in B ( B is closed). Hence {c}_{n(k)} converges to a+b, which implies that c=a+b (every subsequence of a convergent sequence converges to the same limit). Since a\in A,b\in B, it follows that c\in A+B, which means that A+B is closed.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by becko View Post
    Let {c}_{n}={a}_{n}+{b}_{n} be a sequence in A+B converging to a point c\in E, where {a}_{n}\in A, {b}_{n}\in B and {c}_{n}\in A+B. Since A is compact, there is a subsequence {a}_{n(k)} of {a}_{n} that converges to some a\in A. Since {c}_{n(k)} also converges (its a subsequence of a convergent sequence), it follows that {b}_{n(k)}={c}_{n(k)}-{a}_{n(k)} also converges, say to b\in B ( B is closed). Hence {c}_{n(k)} converges to a+b, which implies that c=a+b (every subsequence of a convergent sequence converges to the same limit). Since a\in A,b\in B, it follows that c\in A+B, which means that A+B is closed.
    Wow, that's a very nice proof! I was thinking way to generally for my own good. If you're interested the same idea holds for any arbitrary topological group.
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    Not sure what a topological group is ... will study it in the near future.

    Quote Originally Posted by Drexel28 View Post
    If you're interested the same idea holds for any arbitrary topological group.
    You mean the proof is also valid generally? Or just the result?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by becko View Post
    Not sure what a topological group is ... will study it in the near future.
    Formally it's 3-tuple \left(G,\mathfrak{J},\star\right) such that \left(G,\mathfrak{J}\right) is a topological space and \left(G,\star\right) a group, with the condition that the map g,h)\mapsto gh^{-1}" alt="\alpha:G\times G\to Gg,h)\mapsto gh^{-1}" /> is continuous.



    You mean the proof is also valid generally? Or just the result?
    The result. That's why I was talking nonsense about hard lemmas and such. You need that for the general theorem.
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  7. #7
    mms
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    Quote Originally Posted by becko View Post
    Let {c}_{n}={a}_{n}+{b}_{n} be a sequence in A+B converging to a point c\in E, where {a}_{n}\in A, {b}_{n}\in B and {c}_{n}\in A+B. Since A is compact, there is a subsequence {a}_{n(k)} of {a}_{n} that converges to some a\in A. Since {c}_{n(k)} also converges (its a subsequence of a convergent sequence), it follows that {b}_{n(k)}={c}_{n(k)}-{a}_{n(k)} also converges, say to b\in B ( B is closed). Hence {c}_{n(k)} converges to a+b, which implies that c=a+b (every subsequence of a convergent sequence converges to the same limit). Since a\in A,b\in B, it follows that c\in A+B, which means that A+B is closed.

    hmm i have a doubt, does the subsequence converge to the same limit of the convergent sequence (considering that we are in a normed vector space of any dimension)
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  8. #8
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    All subsequences of a convergenct sequence converge to the same limit. This holds in an arbitrary metric space. Any normed vector space is also a metric space.
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