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Let $f : X \rightarrow [0,\infty)$ and $A$ , $B\,$ two subsets of $X$ such that $A \cap B \neq \emptyset$.
If I have that $ \inf \limits_{X \setminus A} f \,>\, \inf \limits_B f $, does it imply that $\inf \limits_B f = \inf \limits_{A \cap B} f $ ??
Intuitevely it makes sense to me but I can't prove it. Any thoughts ? Or counterexapmple ?
If $\inf_B f \neq \inf_{A\cap B} f$ then $\inf_B f < \inf_{A\cap B} f$.
On the other hand $$ \inf_B f = \min \{ \inf_{A\cap B} f , \, \inf_{B\setminus A} f\}. $$ Hence $\inf_B f = \inf_{B\setminus A} f$.
However $\inf_{B\setminus A} f>\inf_{X\setminus A} f >\inf_B f$, contradiction.
