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Let $\langle X,X^{\ast}\rangle$ be a dual pair. For a subset $A$ of $X$ define $$A^\circ:=\{x^{\ast}\in X^{\ast} :|\langle x,x^{\ast}\rangle|\le 1\text{, for all $x\in A$}\}.$$ If $A\subset B \subset X$ are non-empty, then $A^\circ \supset B^\circ$.

The proof of this property seems to always be described as "obvious". So I ask, why is it obvious?

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    $\begingroup$ In $B$ are more points, thus more conditions $\endgroup$ – user251257 Oct 8 '15 at 16:29
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As $P(x)$ forall $x \in A$ is automatically true if the property $P(x)$ holds for all $x \in B$.

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$A\subseteq B$. So something that is true of every member of $B$ is true of every member of $A$.

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