Spring 2008 #1

Problem Statement: Let K be a compact subset of \mathbb{R} and f:K\rightarrow (0,\infty) a continuous function. Show there exists a \delta>0 such that f(x)\geq \delta for every x\in K.

Proof: Since K is compact and f:K\rightarrow (0,\infty) is continuous it follows that f(K)=\{f(k)|k\in K\} is compact. Bolzano-Weierstrass gives us that f(K) is closed and bounded. This means there is both an upper and a lower bound for f(K). Since f:K\rightarrow (0,\infty) is continuous it follows that there is a x_o\in K such that f(x_o) is a minimum since f is continuous on a compact set. Let \delta=f(x_0). Then \delta is a lower bound for f(K). We know that \delta\neq 0 since each f(x)>0 which means that the minimum value of f must be strictly greater than 0.

Thus, there exists a \delta>0 such that f(x)\geq \delta for every x\in K.

Reflection: The key for this problem is that continuous functions map compact sets to compact sets. Then we know that the function attains it’s minimum value since f is a continuous function on a compact set. Since we were given that f(x)>0 for every x\in K we know that this minimum value is greater than 0.

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This entry was posted in Analysis, Compact, Continuity, Math. Bookmark the permalink.

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