Let $ V $ be an inner product space. Assume $ dim(V)=\infty $
prove/disprove $ \left(U\cap W\right)^{\perp}=U^{\perp}+W^{\perp} $
Well, I proved that this statement is correct if $ dim\left(v\right)<\infty $.
And I have seen a post here with the same question, but I could not understand the counterexample that was provided there, since it used the term "Hilbert space" and we have'nt learned about it yet. Is there a simple counter example for this statement? Thanks.
Best Answer
Let $V$ be the vector space of sequences $x = (x_1, x_2, \ldots)$ of real numbers such that $\sum_ix_i^2$ converges, the usual dot product gives an inner product. Let $U$ be the set of sequences for which only finitely many of the $x_i$ are nonzero and let $W$ be the set of sequences of the form $\left(\frac{a}{1}, \frac{a}{2}, \frac{a}{3}, \ldots\right)$ for some $a$.
Now $U^\perp = 0$ because anything in $U^\perp$ must be orthogonal to all the standard basis vectors $(1, 0, 0, \ldots)$, $(0, 1, 0, \ldots)$, $\ldots$, hence it must be zero in every coordinate. Also $U \cap W = 0$ because every nonzero vector in $W$ is nonzero in infinitely many places. So $(U \cap W)^\perp = U^\perp + W^\perp$ reduces to $0^\perp = 0 + W^\perp$ and then $V = W^\perp$. But this last equation is not true, for example all those standard basis vectors are not in $W^\perp$.