Hello,
First off, the theorem is "if f is continuous on [a,b] then f is bounded above on [a,b]. What is the name of this theorem? He doesn't say but I'm sure it has a popular name.
Anyway, it's on page 115. Let me type what he wrote, then I'll say which part I don't understand.
Proof: Let A= {x: a≤ x ≤ b and f is bounded above on [a,x]}
Before I continue (because this is one of two things I don't understand here), why do we assume f is bounded above on [a,x]? What if f is not bounded above on [a,x]? Also, how do we KNOW f is bounded above on [a,x]? We are trying to prove a function is bounded above by assuming it is bounded above? Doesn't make sense to me, I must be missing something.
Anyway, continuing on:
Clearly, A≠ null set, because a is in A, and A is bounded above (by b), so A has a least upper bound, Ω (in the book it is alpha but it looks too similiar to the letter a so I'm gonna roll with Ω)
Our first step is to prove that we actually have Ω=b. Suppose instead Ω < b. By theorem 1 (which I already proved on the previous page) there is δ>0 such that f is bounded on (Ω-σ, Ω+σ). Since Ω is sup A there is some Xo in A satisfying Ω-σ<Xo<Ω. This means x is bounded on [a,Xo]. But if X1 is any number with Ω<X1<Ω+δ, then f is also bounded on [X0,X1]. Therefore x is bounded on [a,X1] so X1 is in A, contradicting the fact that Ω is an upper bound of A.
Ok, so I understand why what he just showed contradicts that Ω is an upper bound of A. I just don't understand what any of that has to do with Ω<b. What about when Ω=b, how does all of that fail to show a contradiction?
Thanks in advance
First off, the theorem is "if f is continuous on [a,b] then f is bounded above on [a,b]. What is the name of this theorem? He doesn't say but I'm sure it has a popular name.
Anyway, it's on page 115. Let me type what he wrote, then I'll say which part I don't understand.
Proof: Let A= {x: a≤ x ≤ b and f is bounded above on [a,x]}
Before I continue (because this is one of two things I don't understand here), why do we assume f is bounded above on [a,x]? What if f is not bounded above on [a,x]? Also, how do we KNOW f is bounded above on [a,x]? We are trying to prove a function is bounded above by assuming it is bounded above? Doesn't make sense to me, I must be missing something.
Anyway, continuing on:
Clearly, A≠ null set, because a is in A, and A is bounded above (by b), so A has a least upper bound, Ω (in the book it is alpha but it looks too similiar to the letter a so I'm gonna roll with Ω)
Our first step is to prove that we actually have Ω=b. Suppose instead Ω < b. By theorem 1 (which I already proved on the previous page) there is δ>0 such that f is bounded on (Ω-σ, Ω+σ). Since Ω is sup A there is some Xo in A satisfying Ω-σ<Xo<Ω. This means x is bounded on [a,Xo]. But if X1 is any number with Ω<X1<Ω+δ, then f is also bounded on [X0,X1]. Therefore x is bounded on [a,X1] so X1 is in A, contradicting the fact that Ω is an upper bound of A.
Ok, so I understand why what he just showed contradicts that Ω is an upper bound of A. I just don't understand what any of that has to do with Ω<b. What about when Ω=b, how does all of that fail to show a contradiction?
Thanks in advance
via Physics Forums RSS Feed http://www.physicsforums.com/showthread.php?t=720509&goto=newpost
0 commentaires:
Enregistrer un commentaire