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

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    $\begingroup$ You are right, it should be the other direction. $\endgroup$ – Mark Aug 17 at 15:18
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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, $U^\circ,W^\circ\varsubsetneq(\mathbb R^2)^*$.

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