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If $X$ is a complete topological vector space, Y is a dense subspace (so $\overline{Y}=X$), Z is a closed subspace, it is possible that $Y\cap Z=\{0\}$? This is definitely possible for subsets in topological spaces (with intersection being empty), but not sure about complete TVS, or even more particular in complete metric spaces, Banach or Hilbert spaces.

Edit: Thanks for the answer Nate.

Updated questions:

1) Is it possible to find $Y$ dense that non-trivially intersects any $Z$ of dimension at least 2?

2) Does any infinite dimensional closed $Z$ intersect non-trivially any dense $Y$?

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Part 2) needs elaboration –  copper.hat May 8 '12 at 3:42
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2 Answers

up vote 6 down vote accepted

Yes; if $Y$ is any proper dense subspace of $X$, then there exists a vector $x \notin Y$. Set $Z$ to be the one-dimensional subspace spanned by $x$.

This can also happen with $Z$ infinite dimensional. Try $X = C([0,1])$, $Y$ the set of all polynomials, and $Z$ the set of functions which vanish on all of $[0,1/2]$. Then $Y$ is dense (by the Weierstrass approximation theorem), $Z$ is closed, and $Y \cap Z = \{0\}$ (since any nonzero polynomial can only vanish at finitely many points). This answers your updated question 2.

Edit: For your updated question 1, assuming the axiom of choice, there is a dense subspace whose intersection with any subspace of dimension at least two is nontrivial. Let $f : X \to \mathbb{R}$ be an unbounded linear functional (which one can construct given a Hamel basis) and let $Y$ be its kernel. $Y$ has codimension 1 so any two-dimensional subspace intersects it nontrivially. (More explicitly, if $x,y$ are linearly independent, then $f(x) y - f(y) x \in Y$ so the span of $x,y$ intersects $Y$.)

I claim $Y$ must also be dense. First, $Y$ is not closed; if it were, then $f$ would be continuous (standard fact). So $Y \subsetneq \bar{Y}$. But since $Y$ has codimension 1 the only subspace which properly contains it is $X$ itself. (To put it another way, let $y \in \bar{Y} \setminus Y$, so $f(y) \ne 0$. Given any $x \in X$, we write $x = (x - \frac{f(x)}{f(y)} y) + \frac{f(x)}{f(y)}y$ and we have shown that $x$ is in the span of $Y$ and $y$, hence is in $\bar{Y}$.)

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Of course! Thank you. I edited the question and ask something which is perhaps harder :) –  Markus May 8 '12 at 3:34
    
Excellent, now my question is completely answered. Thanks! –  Markus May 8 '12 at 4:33
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It can only happen in infinite dimensions. Finite dimensional subspaces of TVSs are closed.

Another simple example in $l_1$ is the subspace of sequences with only a finite number of non-zero terms. This is easily seen to be a subspace, dense in $l_1$, but the point $x=(\frac{1}{2^0},\frac{1}{2^1}, \frac{1}{2^2},...)$ is not in the subspace. Then $\mathbb{sp} \{ x \}$ is closed but doesn't intersect.

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