# If $V$ is a vector space, $W$ and $U$ are subspaces of $V$, why is $U^0, W^0 \supseteq ( U \cap W)^0$ true?

If $$\textsf V$$ is a vector space, $$\textsf W$$ and $$\textsf U$$ are subspaces of $$\textsf V$$, why is $$\textsf U^0, \textsf W^0 \supseteq (\textsf U \cap \textsf W)^0$$ true?

My mind tells me that the containment should happen in the other direction : if $$A$$ is a subspace of $$B$$, isn't $$B^{0}\subseteq A^{0}$$?

For clarification $$A^{0} := \{ \phi \in \textsf{V}^* : \, \phi(a) = 0 \textrm{ for all } a \in A \}$$

• You are right, it should be the other direction. – Mark Aug 17 at 15:18

Yes, you are right. If, say, $$V=\mathbb R^2$$, $$U=\mathbb R(1,0)$$, and $$W=\mathbb R(0,1)$$, then $$(U\cap W)^\circ=\{0\}^\circ=(\mathbb R^2)^*$$. And, clearly, $$U^\circ,W^\circ\varsubsetneq(\mathbb R^2)^*$$.