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Extreme value theorem: help with contradiction
$begingroup$
I have a problem understanding the last part of the usual proof of the extreme value theorem (found for example here: Extreme Value Theorem proof help)
It is this part that I have trouble understanding:
Since $g(x)=dfrac1{M−f(x)}leq K$ is equivalent to $f(x)leq M−dfrac1K$, we have contradicted the fact that $M$ was assumed to be the least upper bound of $f$ on $[a,b]$. Hence, there must be a balue $cin[a,b]$ such that $f(c)=M$.
Why is it that we are sure that there exist a c on the interval where the maximum is attained? I mean, don't we just know that the new sup is $M-dfrac1K$, but that it necessarily will not attain a max on the interval? Or do we have to assume that the function $g$ that we create attains a maximum for $f$ to attain a maximum?
real-analysis proof-explanation extreme-value-theorem
edited Jan 11 at 11:23
José Carlos Santos
160k22126232
asked Jan 11 at 11:14
user5744148user5744148
112
$endgroup$
add a comment |
$begingroup$
I have a problem understanding the last part of the usual proof of the extreme value theorem (found for example here: Extreme Value Theorem proof help)
It is this part that I have trouble understanding:
Since $g(x)=dfrac1{M−f(x)}leq K$ is equivalent to $f(x)leq M−dfrac1K$, we have contradicted the fact that $M$ was assumed to be the least upper bound of $f$ on $[a,b]$. Hence, there must be a balue $cin[a,b]$ such that $f(c)=M$.
Why is it that we are sure that there exist a c on the interval where the maximum is attained? I mean, don't we just know that the new sup is $M-dfrac1K$, but that it necessarily will not attain a max on the interval? Or do we have to assume that the function $g$ that we create attains a maximum for $f$ to attain a maximum?
real-analysis proof-explanation extreme-value-theorem
edited Jan 11 at 11:23
José Carlos Santos
160k22126232
asked Jan 11 at 11:14
user5744148user5744148
112
$endgroup$
add a comment |
$begingroup$
I have a problem understanding the last part of the usual proof of the extreme value theorem (found for example here: Extreme Value Theorem proof help)
It is this part that I have trouble understanding:
Since $g(x)=dfrac1{M−f(x)}leq K$ is equivalent to $f(x)leq M−dfrac1K$, we have contradicted the fact that $M$ was assumed to be the least upper bound of $f$ on $[a,b]$. Hence, there must be a balue $cin[a,b]$ such that $f(c)=M$.
Why is it that we are sure that there exist a c on the interval where the maximum is attained? I mean, don't we just know that the new sup is $M-dfrac1K$, but that it necessarily will not attain a max on the interval? Or do we have to assume that the function $g$ that we create attains a maximum for $f$ to attain a maximum?
real-analysis proof-explanation extreme-value-theorem
edited Jan 11 at 11:23
José Carlos Santos
160k22126232
asked Jan 11 at 11:14
user5744148user5744148
112
$endgroup$
I have a problem understanding the last part of the usual proof of the extreme value theorem (found for example here: Extreme Value Theorem proof help)
It is this part that I have trouble understanding:
Since $g(x)=dfrac1{M−f(x)}leq K$ is equivalent to $f(x)leq M−dfrac1K$, we have contradicted the fact that $M$ was assumed to be the least upper bound of $f$ on $[a,b]$. Hence, there must be a balue $cin[a,b]$ such that $f(c)=M$.
Why is it that we are sure that there exist a c on the interval where the maximum is attained? I mean, don't we just know that the new sup is $M-dfrac1K$, but that it necessarily will not attain a max on the interval? Or do we have to assume that the function $g$ that we create attains a maximum for $f$ to attain a maximum?
real-analysis proof-explanation extreme-value-theorem
real-analysis proof-explanation extreme-value-theorem
edited Jan 11 at 11:23
José Carlos Santos
160k22126232
asked Jan 11 at 11:14
user5744148user5744148
112
edited Jan 11 at 11:23
José Carlos Santos
160k22126232
asked Jan 11 at 11:14
user5744148user5744148
112
edited Jan 11 at 11:23
José Carlos Santos
160k22126232
edited Jan 11 at 11:23
José Carlos Santos
160k22126232
edited Jan 11 at 11:23
José Carlos Santos
160k22126232
160k22126232
asked Jan 11 at 11:14
user5744148user5744148
112
asked Jan 11 at 11:14
user5744148user5744148
112
asked Jan 11 at 11:14
user5744148user5744148
112
112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because we don't have a new $sup f$. By definition, $M=sup f$ and that proof proves that if there was no such $c$, then we would have $bigl(forall xin[a,b]bigr):f(x)leqslant M-frac1K$, which is impossible, because it would follow from that that $sup fleqslant M-frac1K<M=sup f.$$$
answered Jan 11 at 11:21
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Okey, so the existence of the contradiction of that M is not the sup, also contradicts our assumption that f does not attain a maximum on the interval?
$endgroup$
– user5744148
Jan 11 at 11:50
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Jan 11 at 11:53
$begingroup$
So, is that because of the nature of the contradiction proof. That if one our assumptions is proved wrong, then all of our assumptions are wrong?
$endgroup$
– user5744148
Jan 11 at 12:12
$begingroup$
No. We make an assumption. Then, if we deduce from it a statement that we know that it is false, the assumption is false.
$endgroup$
– José Carlos Santos
Jan 11 at 12:14
add a comment |
$begingroup$
We have $M= sup f([a,b])$. If we assume that there is no $c in [a,b]$ with $f(c)=M$, then we have $f(x) <M$ for all $x in [a,b]$ . In the above proof this leads to $f(x) le M- frac{1}{K}$ for all $x in [a,b]$. Hence
$M = sup f([a,b]) le M- frac{1}{K}$, a contradiction.
answered Jan 11 at 11:22
FredFred
46k1848
$endgroup$
$begingroup$
Yes, it is that contradiction that I have a problem with. How come that we can say that just because we have found a contradiction regarding the sup, we can also say that it is a contradiction of our assumption that f does not attain a maximum?
$endgroup$
– user5744148
Jan 11 at 11:53
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
Because we don't have a new $sup f$. By definition, $M=sup f$ and that proof proves that if there was no such $c$, then we would have $bigl(forall xin[a,b]bigr):f(x)leqslant M-frac1K$, which is impossible, because it would follow from that that $sup fleqslant M-frac1K<M=sup f.$$$
answered Jan 11 at 11:21
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Okey, so the existence of the contradiction of that M is not the sup, also contradicts our assumption that f does not attain a maximum on the interval?
$endgroup$
– user5744148
Jan 11 at 11:50
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Jan 11 at 11:53
$begingroup$
So, is that because of the nature of the contradiction proof. That if one our assumptions is proved wrong, then all of our assumptions are wrong?
$endgroup$
– user5744148
Jan 11 at 12:12
$begingroup$
No. We make an assumption. Then, if we deduce from it a statement that we know that it is false, the assumption is false.
$endgroup$
– José Carlos Santos
Jan 11 at 12:14
add a comment |
$begingroup$
Because we don't have a new $sup f$. By definition, $M=sup f$ and that proof proves that if there was no such $c$, then we would have $bigl(forall xin[a,b]bigr):f(x)leqslant M-frac1K$, which is impossible, because it would follow from that that $sup fleqslant M-frac1K<M=sup f.$$$
answered Jan 11 at 11:21
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Okey, so the existence of the contradiction of that M is not the sup, also contradicts our assumption that f does not attain a maximum on the interval?
$endgroup$
– user5744148
Jan 11 at 11:50
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Jan 11 at 11:53
$begingroup$
So, is that because of the nature of the contradiction proof. That if one our assumptions is proved wrong, then all of our assumptions are wrong?
$endgroup$
– user5744148
Jan 11 at 12:12
$begingroup$
No. We make an assumption. Then, if we deduce from it a statement that we know that it is false, the assumption is false.
$endgroup$
– José Carlos Santos
Jan 11 at 12:14
add a comment |
$begingroup$
Because we don't have a new $sup f$. By definition, $M=sup f$ and that proof proves that if there was no such $c$, then we would have $bigl(forall xin[a,b]bigr):f(x)leqslant M-frac1K$, which is impossible, because it would follow from that that $sup fleqslant M-frac1K<M=sup f.$$$
answered Jan 11 at 11:21
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
Because we don't have a new $sup f$. By definition, $M=sup f$ and that proof proves that if there was no such $c$, then we would have $bigl(forall xin[a,b]bigr):f(x)leqslant M-frac1K$, which is impossible, because it would follow from that that $sup fleqslant M-frac1K<M=sup f.$$$
answered Jan 11 at 11:21
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 11 at 11:21
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 11 at 11:21
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 11 at 11:21
José Carlos SantosJosé Carlos Santos
160k22126232
160k22126232
$begingroup$
Okey, so the existence of the contradiction of that M is not the sup, also contradicts our assumption that f does not attain a maximum on the interval?
$endgroup$
– user5744148
Jan 11 at 11:50
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Jan 11 at 11:53
$begingroup$
So, is that because of the nature of the contradiction proof. That if one our assumptions is proved wrong, then all of our assumptions are wrong?
$endgroup$
– user5744148
Jan 11 at 12:12
$begingroup$
No. We make an assumption. Then, if we deduce from it a statement that we know that it is false, the assumption is false.
$endgroup$
– José Carlos Santos
Jan 11 at 12:14
add a comment |
$begingroup$
Okey, so the existence of the contradiction of that M is not the sup, also contradicts our assumption that f does not attain a maximum on the interval?
$endgroup$
– user5744148
Jan 11 at 11:50
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Jan 11 at 11:53
$begingroup$
So, is that because of the nature of the contradiction proof. That if one our assumptions is proved wrong, then all of our assumptions are wrong?
$endgroup$
– user5744148
Jan 11 at 12:12
$begingroup$
No. We make an assumption. Then, if we deduce from it a statement that we know that it is false, the assumption is false.
$endgroup$
– José Carlos Santos
Jan 11 at 12:14
$begingroup$
Okey, so the existence of the contradiction of that M is not the sup, also contradicts our assumption that f does not attain a maximum on the interval?
$endgroup$
– user5744148
Jan 11 at 11:50
$begingroup$
Okey, so the existence of the contradiction of that M is not the sup, also contradicts our assumption that f does not attain a maximum on the interval?
$endgroup$
– user5744148
Jan 11 at 11:50
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Jan 11 at 11:53
$begingroup$
Yes, that's it.
$endgroup$
– José Carlos Santos
Jan 11 at 11:53
$begingroup$
So, is that because of the nature of the contradiction proof. That if one our assumptions is proved wrong, then all of our assumptions are wrong?
$endgroup$
– user5744148
Jan 11 at 12:12
$begingroup$
So, is that because of the nature of the contradiction proof. That if one our assumptions is proved wrong, then all of our assumptions are wrong?
$endgroup$
– user5744148
Jan 11 at 12:12
$begingroup$
No. We make an assumption. Then, if we deduce from it a statement that we know that it is false, the assumption is false.
$endgroup$
– José Carlos Santos
Jan 11 at 12:14
$begingroup$
No. We make an assumption. Then, if we deduce from it a statement that we know that it is false, the assumption is false.
$endgroup$
– José Carlos Santos
Jan 11 at 12:14
add a comment |
$begingroup$
We have $M= sup f([a,b])$. If we assume that there is no $c in [a,b]$ with $f(c)=M$, then we have $f(x) <M$ for all $x in [a,b]$ . In the above proof this leads to $f(x) le M- frac{1}{K}$ for all $x in [a,b]$. Hence
$M = sup f([a,b]) le M- frac{1}{K}$, a contradiction.
answered Jan 11 at 11:22
FredFred
46k1848
$endgroup$
$begingroup$
Yes, it is that contradiction that I have a problem with. How come that we can say that just because we have found a contradiction regarding the sup, we can also say that it is a contradiction of our assumption that f does not attain a maximum?
$endgroup$
– user5744148
Jan 11 at 11:53
add a comment |
$begingroup$
We have $M= sup f([a,b])$. If we assume that there is no $c in [a,b]$ with $f(c)=M$, then we have $f(x) <M$ for all $x in [a,b]$ . In the above proof this leads to $f(x) le M- frac{1}{K}$ for all $x in [a,b]$. Hence
$M = sup f([a,b]) le M- frac{1}{K}$, a contradiction.
answered Jan 11 at 11:22
FredFred
46k1848
$endgroup$
$begingroup$
Yes, it is that contradiction that I have a problem with. How come that we can say that just because we have found a contradiction regarding the sup, we can also say that it is a contradiction of our assumption that f does not attain a maximum?
$endgroup$
– user5744148
Jan 11 at 11:53
add a comment |
$begingroup$
We have $M= sup f([a,b])$. If we assume that there is no $c in [a,b]$ with $f(c)=M$, then we have $f(x) <M$ for all $x in [a,b]$ . In the above proof this leads to $f(x) le M- frac{1}{K}$ for all $x in [a,b]$. Hence
$M = sup f([a,b]) le M- frac{1}{K}$, a contradiction.
answered Jan 11 at 11:22
FredFred
46k1848
$endgroup$
We have $M= sup f([a,b])$. If we assume that there is no $c in [a,b]$ with $f(c)=M$, then we have $f(x) <M$ for all $x in [a,b]$ . In the above proof this leads to $f(x) le M- frac{1}{K}$ for all $x in [a,b]$. Hence
$M = sup f([a,b]) le M- frac{1}{K}$, a contradiction.
answered Jan 11 at 11:22
FredFred
46k1848
answered Jan 11 at 11:22
FredFred
46k1848
answered Jan 11 at 11:22
FredFred
46k1848
answered Jan 11 at 11:22
FredFred
46k1848
46k1848
$begingroup$
Yes, it is that contradiction that I have a problem with. How come that we can say that just because we have found a contradiction regarding the sup, we can also say that it is a contradiction of our assumption that f does not attain a maximum?
$endgroup$
– user5744148
Jan 11 at 11:53
add a comment |
$begingroup$
Yes, it is that contradiction that I have a problem with. How come that we can say that just because we have found a contradiction regarding the sup, we can also say that it is a contradiction of our assumption that f does not attain a maximum?
$endgroup$
– user5744148
Jan 11 at 11:53
$begingroup$
Yes, it is that contradiction that I have a problem with. How come that we can say that just because we have found a contradiction regarding the sup, we can also say that it is a contradiction of our assumption that f does not attain a maximum?
$endgroup$
– user5744148
Jan 11 at 11:53
$begingroup$
Yes, it is that contradiction that I have a problem with. How come that we can say that just because we have found a contradiction regarding the sup, we can also say that it is a contradiction of our assumption that f does not attain a maximum?
$endgroup$
– user5744148
Jan 11 at 11:53
add a comment |
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