Annihilator in infinite dimensional vector space

Question

I am working on the following question:

Consider $U$ and $W$ subspaces of a finite-dimensional vector space $V$. Show that $$(U \cap W)^{0} = U^{0} + W^{0}$$ Is it necessary for the $V$ dimension to be finite?

I proved that $(U \cap W)^{0} = U^{0} + W^{0}$ and I know that it is necessary that the dimension of $V$ be finite so that $(U \cap W)^{0} \subset U^{0} + W^{0}$, but I can't find a counterexample show this inclusion fails if the dimension of $V$ is infinite. Can someone help me? Thank you in advance.

Answer 1

Actually this is true in infinite dimensions as well (as long as you accept the Axiom of Choice)!

Here's a quick proof:

Suppose $V$ is a vector space over $k$. Let $A$ be a basis for $U \cap W$, let $B \subseteq U$ be such that $A \cup B$ is a basis for $U$, let $C \subseteq W$ be such that $A \cup C$ is a basis for $W$. One can show that $A \cup B \cup C$ is independent, and spans $U + W$. Then finally let $D \subseteq V$ be such that $I = A \cup B \cup C \cup D$ is a basis for $V$. Then there is an isomorphism $V \cong k^{\oplus I}$ (I think of these as column vectors indexed by $I$ with only finitely many nonzero entries) and $V^* \cong k^I$ (I think of these as row vectors indexed by $I$ with possibly infinitely many nonzero entries). The application of a dual vector to a vector is then just the matrix product of the corresponding row vector and column vector.

Then under the isomorphisms above, $U \cap W$ corresponds to the set of column vectors with zeros outside of the entries indexed by $A$ and $(U \cap W)^0$ corresponds to the set of row vectors with zeros in the entries indexed by $A$. The corresponding result holds for all of the other relevant subspaces.

From this it is clear that every dual vector in $(U \cap W)^0$ can be written as a sum of something in $U^0$ and something in $W^0$. Explicitly, if $(x_i)_{i \in I} \in (U \cap W)^0$, we have $x_i = 0$ for all $i \in A$. Then define

$$y_i = \left\{ \begin{matrix} x_i & \text{ if } & i \notin B \\ 0 & \text{ if } & i \in B \end{matrix} \right.$$

so that $(y_i)_{i \in I} \in U^0$ and

$$z_i = \left\{ \begin{matrix} x_i & \text{ if } & i \notin C \cup D \\ 0 & \text{ if } & i \in C \cup D \end{matrix} \right.$$

so that $(z_i)_{i \in I} \in W^0$. Note then that we have $(y_i)_{i \in I} + (z_i)_{i \in I} = (x_i)_{i \in I}$, as desired.

Answer 2

It is often not that interesting to think of infinite dimensional spaces 'algebraically', so let us consider some norms. Let $$c_0 = \{(a_n)\subseteq\mathbb{R} : a_n\longrightarrow 0\}$$ with the norm $$\lVert (a_n)\rVert_{\infty} = \sup_n\,\, \lvert a_n\rvert.$$ We denote by $c_0^\ast$ the continuous dual of $c_0$ and we consider the annihilators as subspaces of $c_0^\ast$. Now, let us consider the subspaces $U$ and $W$ of $c_0$ given by $$c_{00} = \{(a_n)\subseteq\mathbb{R} : a_n= 0,\forall n\geq N\}$$ and $$W = \text{span}\left(1/n\right)_{n\in\mathbb N}.$$ It is clear that $U\cap W = 0$. Moreover, since $U$ is dense in $c_0$, $U^0 = 0$. Then, $$(U\cap W)^0 = c_0^\ast\quad\text{and}\quad U^0+W^0= W^0.$$ So the question is to determine whether $W^0 = c_0^\ast$, but it is not very hard to see that $W^0$ has codimension 1 in $c_0^\ast = \ell_1$, and therefore $W^0\neq c_0^\ast$.