# "Obviousness" of $A\subset B\Rightarrow A^\circ\supset B^\circ$, where $A^\circ=\{x^*\in X^*:(\forall x\in A)(|\langle x,x^*\rangle|\leq 1)\}$

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?

• In $B$ are more points, thus more conditions Oct 8 '15 at 16:29

As $P(x)$ forall $x \in A$ is automatically true if the property $P(x)$ holds for all $x \in B$.

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