# Proof of the Extreme Value Theorem

1. Dec 16, 2008

### jgens

1. The problem statement, all variables and given/known data

Essentially, prove the Extreme Value Theorem.

2. Relevant equations

n/a

3. The attempt at a solution

Proof: Let a function f(x) be continuous on the closed interval [a,b]. Moreover, define a set A such that A={x ϵ [a,b]}. Since f(x) satisfies the condition for the Boundedness Theorem, sup{f(A)}= M. If f(x) does not attain its supremum in [a,b] then it must asymptotically converge to some valve infinitesimally near M in the interval [a,b]. We may define this value such that f(c_n ) > M-1/n, where as n→∞, f(c_n )→f(c). We are then left with the resultant inequality: M-1/n < f(c_n ) <M, and by the Squeeze Theorem, lim(x→c)f(x) = M. Therefore, under the initial premise of continuity, f(x) attains its supremum M at some value x=c within the closed interval [a,b].

Given that f(x) satisfies the Boundedness Theorem, we may also claim inf{f(A)} = m. If f(x) does not attain its infimum in [a,b] then it must asymptotically converge to some value infinitesimally near m in the interval [a,b]. We may define this value such that f(d_n) < m+1/n where as n→∞, f(d_n )→f(d). We are then left with the resultant inequality: m+1/n > f(d_n ) > m, and by the Squeeze Theorem, lim(x→d)f(x) = m. Therefore, under the initial premise of continuity, f(x) attains its infimum m at some value x=d within the closed interval [a,b].

Therefore, f(d)≤f(x)≤f(c) for all x ϵ [a,b]. Q.E.D.

Any suggestions would be appreciated since I'm fairly certain the above proof is incorrect - I'm not certain how I would fix any errors though.

Thanks.

2. Dec 16, 2008

### HallsofIvy

Staff Emeritus
Your proof looks pretty good to me. Essentially, because f is continuous on the closed and bounded interval [a,b], f([a,b]) is also bounded and so has sup and inf (the "boundedness" theorem). f([a,b]) is also closed, which is what you show with your limit argument.