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Let $ \mathcal{H} $ be a Hilbert space and $ M $ be a closed subspace, prove that $ M + x_0\mathbb{R} $ is a closed subspace.

It was easy enough to prove that $ M + x_0\mathbb{R} $ is a vector subspace, but I got into trouble trying to prove it's closed.

I tried defining the sequence $ x_n + a_n x_0 $ (where $ x_n \in M $ and $ a_n \in \mathbb{R} $) assuming it converges to $ v $, in order to show that $ v\in M + x_0\mathbb{R} $. My approach is to show that both $ x_n $ and $ a_n $ have to be Cauchy sequences in their respective subspace. Doing so, it's easy to imply that $ v = \lim x_n + x_0 \lim a_n$. However, I wasn't able to do so (and to be completely honest, I'm not even convinced that has to hold, but I could not come up with any other approach).

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Hint: Assume $x_0 \notin M$. Can $a_n$ be unbounded? If not, you can choose a convergent subsequence $a_{n_k}$. Then $a_{n_k}$ converges, hence $x_{n_k}$ converges, too. –  t.b. Nov 12 '11 at 14:50
    
Just emphasizing that you should follow @t.b.'s advice. You may be getting stuck because if you do not explicitly assume that $x_0$ is not in $M$, the decomposition of any $v \in M + x_0 \mathbb{R}$ into a sum $x + a x_0$ (with $x \in M$ and $a \in \mathbb{R}$) is non-unique, and if you have a convergent sequence $(v_n)$ of vectors in $M + x_0 \mathbb{R}$, the sequences $(x_n)$ in $M$ and $(a_n)$ in $\mathbb{R}$ coming from chosen representations $v_n = x_n + a_n x_0$ need not converge. The point: it is not possible to prove that they converge in complete generality. –  leslie townes Feb 11 '12 at 1:14

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