X is a compact subset of $\Re^{n}$. A upper and lower bounded function f is defined that f:X $\rightarrow$ $\Re$. Does a sequence of {$x_n$} always exist so that f($x_n$)$\rightarrow$ $\inf$[f(x)].
I understand that in a bounded subset X of $\Re$, a decreasing sequence {$x_n$} will converge to inf(X). But what I am confused about in my case is whether the behavior of this function f will affect the same results to be applied?
Is this possible to be proved by contradiction? Say, if there no sequence sequence of {$x_n$} so that sequence {f$(x_n)$} converges to inf(f(x)). Then it contradicts the definition of inf(f(x)). Do I miss something?
If, in general (X,d) is a compact metric space, and f is a continuous function from X to R then the infinimum is attained. Else, consider the interval [0,1] with the usual absolute value metric (it's compact) and consider the function f defined on [0,1] that takes the value $-x$ if $ 0\leq x<1$ and takes the value 0 at 1. the infinimum of f is -1 but is never attained!