Mixing
set of all maximum points is either finite or countable
$begingroup$
Try:
Call $M$ to be the set of all maximum points of a given $f: mathbb{R} to mathbb{R}$. We wish to construct $phi: mathbb{N} to M $ so that $phi$ is bijective.
Notice that $x_i$ is maximum of $f$ is we can find $epsilon_i$ so that ....
Maybe we can define $phi(i) = x_i $ and check for bijectivity.
If $phi(i) = phi(k) implies x_i = x_k implies i=k $
Now, any points $x_j$ where $f$ attains its maximum there corresponds an index $j$ so that surjectivity follows and we have bijectivity.
Thus, $M$ is finite our countable.
Is my approach correct?
real-analysis calculus
asked Jan 24 at 2:38
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
add a comment |
$begingroup$
Try:
Call $M$ to be the set of all maximum points of a given $f: mathbb{R} to mathbb{R}$. We wish to construct $phi: mathbb{N} to M $ so that $phi$ is bijective.
Notice that $x_i$ is maximum of $f$ is we can find $epsilon_i$ so that ....
Maybe we can define $phi(i) = x_i $ and check for bijectivity.
If $phi(i) = phi(k) implies x_i = x_k implies i=k $
Now, any points $x_j$ where $f$ attains its maximum there corresponds an index $j$ so that surjectivity follows and we have bijectivity.
Thus, $M$ is finite our countable.
Is my approach correct?
real-analysis calculus
asked Jan 24 at 2:38
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
$begingroup$
Note that the definition allows you to associate to each maximum point an open interval having a certain property such that this property implies that the open intervals are pairwise disjoint. Thus, it is enough to prove that any collection of pairwise disjoint open intervals is at most countable, a result that you might already have available, but if not, think about what you can deduce if you pick a rational number from each of these open intervals (you'll get at most countably many rationals, no two of which belong to the same open interval . . .).
$endgroup$
– Dave L. Renfro
Jan 24 at 9:02
add a comment |
$begingroup$
Try:
Call $M$ to be the set of all maximum points of a given $f: mathbb{R} to mathbb{R}$. We wish to construct $phi: mathbb{N} to M $ so that $phi$ is bijective.
Notice that $x_i$ is maximum of $f$ is we can find $epsilon_i$ so that ....
Maybe we can define $phi(i) = x_i $ and check for bijectivity.
If $phi(i) = phi(k) implies x_i = x_k implies i=k $
Now, any points $x_j$ where $f$ attains its maximum there corresponds an index $j$ so that surjectivity follows and we have bijectivity.
Thus, $M$ is finite our countable.
Is my approach correct?
real-analysis calculus
asked Jan 24 at 2:38
Jimmy SabaterJimmy Sabater
3,023325
$endgroup$
Try:
Call $M$ to be the set of all maximum points of a given $f: mathbb{R} to mathbb{R}$. We wish to construct $phi: mathbb{N} to M $ so that $phi$ is bijective.
Notice that $x_i$ is maximum of $f$ is we can find $epsilon_i$ so that ....
Maybe we can define $phi(i) = x_i $ and check for bijectivity.
If $phi(i) = phi(k) implies x_i = x_k implies i=k $
Now, any points $x_j$ where $f$ attains its maximum there corresponds an index $j$ so that surjectivity follows and we have bijectivity.
Thus, $M$ is finite our countable.
Is my approach correct?
real-analysis calculus
real-analysis calculus
asked Jan 24 at 2:38
Jimmy SabaterJimmy Sabater
3,023325
asked Jan 24 at 2:38
Jimmy SabaterJimmy Sabater
3,023325
asked Jan 24 at 2:38
Jimmy SabaterJimmy Sabater
3,023325
asked Jan 24 at 2:38
Jimmy SabaterJimmy Sabater
3,023325
asked Jan 24 at 2:38
Jimmy SabaterJimmy Sabater
3,023325
3,023325
$begingroup$
Note that the definition allows you to associate to each maximum point an open interval having a certain property such that this property implies that the open intervals are pairwise disjoint. Thus, it is enough to prove that any collection of pairwise disjoint open intervals is at most countable, a result that you might already have available, but if not, think about what you can deduce if you pick a rational number from each of these open intervals (you'll get at most countably many rationals, no two of which belong to the same open interval . . .).
$endgroup$
– Dave L. Renfro
Jan 24 at 9:02
add a comment |
$begingroup$
Note that the definition allows you to associate to each maximum point an open interval having a certain property such that this property implies that the open intervals are pairwise disjoint. Thus, it is enough to prove that any collection of pairwise disjoint open intervals is at most countable, a result that you might already have available, but if not, think about what you can deduce if you pick a rational number from each of these open intervals (you'll get at most countably many rationals, no two of which belong to the same open interval . . .).
$endgroup$
– Dave L. Renfro
Jan 24 at 9:02
$begingroup$
Note that the definition allows you to associate to each maximum point an open interval having a certain property such that this property implies that the open intervals are pairwise disjoint. Thus, it is enough to prove that any collection of pairwise disjoint open intervals is at most countable, a result that you might already have available, but if not, think about what you can deduce if you pick a rational number from each of these open intervals (you'll get at most countably many rationals, no two of which belong to the same open interval . . .).
$endgroup$
– Dave L. Renfro
Jan 24 at 9:02
$begingroup$
Note that the definition allows you to associate to each maximum point an open interval having a certain property such that this property implies that the open intervals are pairwise disjoint. Thus, it is enough to prove that any collection of pairwise disjoint open intervals is at most countable, a result that you might already have available, but if not, think about what you can deduce if you pick a rational number from each of these open intervals (you'll get at most countably many rationals, no two of which belong to the same open interval . . .).
$endgroup$
– Dave L. Renfro
Jan 24 at 9:02
add a comment |
1 Answer
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$begingroup$
No. When you say that $x_i$ is a maximum point, you're assuming that such points are countable.
You can start by proving that the each maximum point is isolated and then use the fact that discrete subsets of $mathbb{R}$ are at most countable.
answered Jan 24 at 3:49
lzralbulzralbu
640512
$endgroup$
add a comment |
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$begingroup$
No. When you say that $x_i$ is a maximum point, you're assuming that such points are countable.
You can start by proving that the each maximum point is isolated and then use the fact that discrete subsets of $mathbb{R}$ are at most countable.
answered Jan 24 at 3:49
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
No. When you say that $x_i$ is a maximum point, you're assuming that such points are countable.
You can start by proving that the each maximum point is isolated and then use the fact that discrete subsets of $mathbb{R}$ are at most countable.
answered Jan 24 at 3:49
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
No. When you say that $x_i$ is a maximum point, you're assuming that such points are countable.
You can start by proving that the each maximum point is isolated and then use the fact that discrete subsets of $mathbb{R}$ are at most countable.
answered Jan 24 at 3:49
lzralbulzralbu
640512
$endgroup$
No. When you say that $x_i$ is a maximum point, you're assuming that such points are countable.
You can start by proving that the each maximum point is isolated and then use the fact that discrete subsets of $mathbb{R}$ are at most countable.
answered Jan 24 at 3:49
lzralbulzralbu
640512
answered Jan 24 at 3:49
lzralbulzralbu
640512
answered Jan 24 at 3:49
lzralbulzralbu
640512
answered Jan 24 at 3:49
lzralbulzralbu
640512
640512
add a comment |
add a comment |
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$begingroup$
Note that the definition allows you to associate to each maximum point an open interval having a certain property such that this property implies that the open intervals are pairwise disjoint. Thus, it is enough to prove that any collection of pairwise disjoint open intervals is at most countable, a result that you might already have available, but if not, think about what you can deduce if you pick a rational number from each of these open intervals (you'll get at most countably many rationals, no two of which belong to the same open interval . . .).
$endgroup$
– Dave L. Renfro
Jan 24 at 9:02