- #1

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[tex]p(x,A)=inf_{y\in A}p(x,y)[/tex]

prove that: f_A(x)=p(x,A) f:X->R is continuous.

basically, if U is open in R, we need to show that f^-1_A(U) is open in X, i.e that

{x in X|p(x,A) in U} is open in X, now because U is open in R, then there's an open interval or half open interval contained in U that it contains p(x,A), i.e p(x,A) is bounded, but from this to show that f^-1(U) is open I'm kind of in a stuck.

any hints?

thanks in advance.