# Thread: supremum on closed section proof

1. ## supremum on closed section proof

f is continues on [a,b]
A={x in [a,b]|f(x)=f(a)}
prove that f(x) has a maximum
f(supA)=f(a)
??

first of all there are two kinds of maximums here
1st a supremum which is inside [a,b] is called maximum

and maximum of the function values
correct??

2. ## Re: supremum on closed section proof

Originally Posted by transgalactic
f is continues on [a,b]
A={x in [a,b]|f(x)=f(a)}
prove that f(x) has a maximum
f(supA)=f(a)
??

first of all there are two kinds of maximums here
1st a supremum which is inside [a,b] is called maximum

and maximum of the function values
correct??

Extreme value theorem - Wikipedia, the free encyclopedia

3. ## Re: supremum on closed section proof

Originally Posted by transgalactic
f is continues on [a,b]
A={x in [a,b]|f(x)=f(a)}
prove that f(x) has a maximum
f(supA)=f(a)?
Because $\left( {\forall x \in A} \right)\left[ {x \leqslant b} \right]$, you know that $\beta=\sup\{A\}$ exists and $\beta\in[a,b]$.
If it is true $\beta\in A$ you are done.
Else If you assume $\beta\notin A$ that leads to a contradiction to the continuity of $f$.

BTW: This not the Extreme value theorem.