Suppose that $M, N$ are infinite-dimensional (not necessarily closed) subspaces of a Banach space $X$. If $N$ is contained in the norm closure of $M$, is the subspace $M\cap N$ infinite-dimensional? This question may be stupid or easy.
Thank you!
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asked Oct 30, 2016 at 13:14
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1$\begingroup$ It's enough to consider any non-closed $M$ and $N$ the line generated by a vector in $\bar{M}-M$. $\endgroup$– YCorCommented Oct 30, 2016 at 16:45
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1 Answer
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Not always. Say, both subspaces may be dense, but have trivial intersection. This situation is generic: you may choose two countable dense subsets $A,B$ in a separable Banach space so that their union is linearly independent (choose elements step by step so that both $A,B$ visit each of countably many balls, but elements of $A\cup B$ are linearly independent.) Define $M,N$ as spans of $A,B$.
For concrete example, consider polynomials on $[0,1]$ and polynomials multiplied by, say, $e^x$. In $C[0,1]$ or whatever.
answered Oct 30, 2016 at 13:35