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$M = \{(x_n) \in \ell_2 :\sum_{n=1}^\infty x_n = 0\}$

It is obvious that it is a linear set. But I don't know how to prove it is closed. I try to prove the complement is open, but it doesn't work. Can anyone help me?

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    What is your definition of subspace? It's not usually required that they're closed. For example that would make it hard to talk about dense linear subspaces (e.g. $C_c(\mathbb R) \subset L_1(\mathbb R)$).2012-10-14
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    @kahen In the book, subspace should be linear and closed.2012-10-14
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    Are you sure this is true?2012-10-14
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    @abatkai I don't if it is closed. If it isn't,can you give me a counterexample?2012-10-14
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    You are speaking of the kernel of the functional $(x_n)\mapsto \sum_{n=1}^{\infty}x_n$, which is linear but obviously not continuous. Does this help?2012-10-14
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    @abatkai I know what you mean.But why it is not continuous?2012-10-14
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    Take a sequance like $x_n=1/n$. It must be clear that you can approximate it by finite sequences. Can you approximate it by sequences belonging to your set?2012-10-14
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    @abatkai thanks a lot.2012-10-14
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    ... and furthermore the domain of that functional is not the whole space ...2012-10-14

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