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**HallsofIvy** Well, what definition of "compact" **does** your book have?

The first says that a set, A, is compact if and only if every open cover of A has a finite subcover. That is, given any collection of open sets such that every point of A is in at least one of the open sets, there exist a finite sub-collection of those same open sets such that every point of A is in at least one of the sets in the subcollection.

The second says that a set, A, is compact if and only if every infinite sequence $\displaystyle \{a_n\}$ of points in A converges to a point in A.

To use the first, for example, since f has a finite limit at every point, x, given any $\displaystyle \epsilon> 0$, there exist a number, $\displaystyle \delta_x> 0$ for each x such that if $\displaystyle d(y,x)< \delta_x$ then $\displaystyle |f(x)- f(y)|< \epsilon_x$. let $\displaystyle N_\delta_x(x)$ be the neighborhood about point x with radius $\displaystyle \delta_x$. These form an open cover of K. Since K is compact, there exist a finite sub-cover. Now we have a finite number of points, $\displaystyle x$ such that every point in K is in some $\displaystyle N_x(x)$ which means that for y in $\displaystyle N_x(x)$, $\displaystyle f(x)-\epsilon$< f(y)< f(x)+ \epsilon[/tex]. Since the set of "x"s is finite, there is a largest such "$\displaystyle f(x)+ \epsilon$" and a smallest such "$\displaystyle f(x)- \epsilon$" which are bounds for f(K).