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In Banach-Hilbert Spaces, Vector Measures and Group Representations, P142 7-4.9 Corollary Let $M$ be a closed vector subspace of a normed space $E$. Then for every finite dimensional vector subspace $N$ of $E$, the sum $M+N$ is a closed vector subspace of $E$. This also can be found on Sum of closed subspaces of normed linear space

However, my exercise is not the sum of two subspace, but the sum of subspace of finite dimension with a closed set.

Let $M$ be a vector subspace of a normed space $E$, and $N$ be a finite dimensional vector subspace of $E$. Assume that $M\cap N=\{0\}$, let $K$ be a closed subset in $M$, and $K\cap N=\emptyset$. I would like to know whether $N+K$ is closed.

I feel that it is closed. Is the following proof right?

Let $\left\{ g_{n}\right\} $ be the bases of $N$, and $\left\{ e_{m}\right\} $ be the bases of $M$. For any $a_{i}g_{n}\in N$ and $b_{i}% e_{m}\in K$, then $c_{i}=a_{i}g_{n}+b_{i}e_{m}\in N+K$. If $c_{i}=a_{i}% g_{n}+b_{i}e_{m}\rightarrow c=ag_{n}+be_{m}+dh_{o}\in E$, where $h_{o}$ are the remains bases of $E$. We must have $d=0$, $a_{i}\rightarrow a$ and $b_{i}\rightarrow b$. and thus $c\in N+K$.

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  • $\begingroup$ Consider $\pi : E\to E/N$, the natural quotient map. Since $N$ is a closed subspace, $\pi$ is a continuous and open map. From this, you can conclude that $K+N = \pi^{-1}(\pi(K))$ is a closed set. $\endgroup$ – Prahlad Vaidyanathan Nov 19 '15 at 13:26
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    $\begingroup$ But the image of a closed set under a quotient map need not be closed. E.g., for the first projection $\pi:\mathbb R^2 \to \mathbb R$ and $K=\lbrace (x,y): x>0, y\ge 1/x\rbrace$. $\endgroup$ – Jochen Nov 19 '15 at 14:32
  • $\begingroup$ @Jochen: You are right, and this also answers the OP! $\endgroup$ – Prahlad Vaidyanathan Nov 19 '15 at 16:26
  • $\begingroup$ the examples in math.stackexchange.com/questions/124130/… are from the same bases. But here the two sets come from different set of basis. I still think the sum is closed. $\endgroup$ – Chen Deng-Ta Nov 20 '15 at 8:35
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Here is a counterexample:

  • $E=\ell^2$,
  • $M$ is the space of sequences $(x_1,x_2,\dots)$ with finitely many nonzero elements, and such that $\sum_{k=1}^\infty x_k=0$
  • $N$ is the one-dimensional subspace spanned by the first standard basis vector $e_1$.
  • $K=\{v_n\}$ where $v_n=\sum_{k=1}^n\frac{1}{k}(e_k-e_1)$, $n=1,2,3,\dots$. This is a closed subset of $M$, since the sequence $v_n$ has no convergent subsequence.

On the other hand, the sum $N+K$ contains the sequence $w_n=\sum_{k=1}^n\frac{1}{k}e_k$. This sequence has a limit in $\ell^2$, and this limit, having infinitely many nonzero entries, is not in $M+N$, hence is not in $K+N$.


Note that the space $M$ is not closed in the above example, but you did not require it to be closed in your formulation.

It it was closed, then the sum $M+N$ would be isomorphic to direct sum $M\circ N$, and therefore $K+N$ would be homeomorphic to the product of two closed sets, hence closed.

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