Mixing
Bounding constants in terms of functions
Clash Royale CLAN TAG#URR8PPP
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Problem statement:
Assume there are two constants $alpha, beta in mathbbR$ and two functions $$f: Atimes B rightarrow mathbbR text and g: Atimes B rightarrow mathbbR$$ non-negative, continuous, and bounded, $A$ and $B$ non-empty, such that $forall b in B, exists ain A$ with $alpha leq f(a,b)$ and $forall a in A, exists b in B$ with $beta leq g(a,b)$, then there exists a pair $(a,b) in A times B$ such that $$alpha leq f(a,b) text and beta leq g(a,b).$$
My attempt:
The statement seems like it should be trivial, but I find myself making what seems like a circular argument.
Proof:
Let $bin B$, then by assumption there is an $a_b in A$ such that $alpha leq f(a_b, b)$. Now, with $a_b$, there exists a $b_a_b in B$ such that $beta leq g(a_b, b_a_b)$. If $b=b_a_b$, then $g(a_b, b) = g(a_b, b_a_b)$ and
$$alpha leq f(a_b, b) text and beta leq g(a_b, b)$$
and so $(a_b, b)$ is a pair that satisfies the statement.
Comment:
My issue is, can I just write "if $b = b_a_b$"? Nothing prevents this from happening and since I just need existence this argument seems valid...
logic
asked Jul 18 at 19:04
Stephen Diadamo
112
 |Â
show 1 more comment
up vote
2
down vote
favorite
Problem statement:
Assume there are two constants $alpha, beta in mathbbR$ and two functions $$f: Atimes B rightarrow mathbbR text and g: Atimes B rightarrow mathbbR$$ non-negative, continuous, and bounded, $A$ and $B$ non-empty, such that $forall b in B, exists ain A$ with $alpha leq f(a,b)$ and $forall a in A, exists b in B$ with $beta leq g(a,b)$, then there exists a pair $(a,b) in A times B$ such that $$alpha leq f(a,b) text and beta leq g(a,b).$$
My attempt:
The statement seems like it should be trivial, but I find myself making what seems like a circular argument.
Proof:
Let $bin B$, then by assumption there is an $a_b in A$ such that $alpha leq f(a_b, b)$. Now, with $a_b$, there exists a $b_a_b in B$ such that $beta leq g(a_b, b_a_b)$. If $b=b_a_b$, then $g(a_b, b) = g(a_b, b_a_b)$ and
$$alpha leq f(a_b, b) text and beta leq g(a_b, b)$$
and so $(a_b, b)$ is a pair that satisfies the statement.
Comment:
My issue is, can I just write "if $b = b_a_b$"? Nothing prevents this from happening and since I just need existence this argument seems valid...
logic
asked Jul 18 at 19:04
Stephen Diadamo
112
If $A$ and $B$ are empty, then the preconditions hold vacuously but the consequence fails as there simply isn't any pair in $Atimes B$.
– Derek Elkins
Jul 18 at 19:16
For your comment, you can certainly say "if $b=b_a_b$ then" but you have to also handle the $bneq b_a_b$ case, or you have to give an argument for what that case can't happen. You can't just say "if it is the case that..." because the obvious retort is then, "well, what if it isn't the case?"
– Derek Elkins
Jul 18 at 19:25
Let $alpha=beta=1$ and $A=B=0,1$. Define $f(a,b)=a+bpmod 2$ and $g(a,b)=1- f(a,b)pmod 2$. $f$ and $g$ satisfy the preconditions ($f(a,1-a)=1$ and $g(a,a)=1$), but $f+g$ is the constantly $1$ function and thus less than $alpha+beta=2$. Perhaps you also have an unstated continuity assumption.
– Derek Elkins
Jul 18 at 20:11
I modified the statement slightly, sorry for the unnecessary confusion.
– Stephen Diadamo
Jul 18 at 20:35
Well, my counter-example was non-negative and bounded. So you will need to make use of the continuity assumption. The fact that your proof attempt would, if it worked, work just as well for non-continuous functions shows that something is missing/wrong and suggests a direction to look to find out what may need to be added.
– Derek Elkins
Jul 18 at 20:38
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem statement:
Assume there are two constants $alpha, beta in mathbbR$ and two functions $$f: Atimes B rightarrow mathbbR text and g: Atimes B rightarrow mathbbR$$ non-negative, continuous, and bounded, $A$ and $B$ non-empty, such that $forall b in B, exists ain A$ with $alpha leq f(a,b)$ and $forall a in A, exists b in B$ with $beta leq g(a,b)$, then there exists a pair $(a,b) in A times B$ such that $$alpha leq f(a,b) text and beta leq g(a,b).$$
My attempt:
The statement seems like it should be trivial, but I find myself making what seems like a circular argument.
Proof:
Let $bin B$, then by assumption there is an $a_b in A$ such that $alpha leq f(a_b, b)$. Now, with $a_b$, there exists a $b_a_b in B$ such that $beta leq g(a_b, b_a_b)$. If $b=b_a_b$, then $g(a_b, b) = g(a_b, b_a_b)$ and
$$alpha leq f(a_b, b) text and beta leq g(a_b, b)$$
and so $(a_b, b)$ is a pair that satisfies the statement.
Comment:
My issue is, can I just write "if $b = b_a_b$"? Nothing prevents this from happening and since I just need existence this argument seems valid...
logic
asked Jul 18 at 19:04
Stephen Diadamo
112
Problem statement:
Assume there are two constants $alpha, beta in mathbbR$ and two functions $$f: Atimes B rightarrow mathbbR text and g: Atimes B rightarrow mathbbR$$ non-negative, continuous, and bounded, $A$ and $B$ non-empty, such that $forall b in B, exists ain A$ with $alpha leq f(a,b)$ and $forall a in A, exists b in B$ with $beta leq g(a,b)$, then there exists a pair $(a,b) in A times B$ such that $$alpha leq f(a,b) text and beta leq g(a,b).$$
My attempt:
The statement seems like it should be trivial, but I find myself making what seems like a circular argument.
Proof:
Let $bin B$, then by assumption there is an $a_b in A$ such that $alpha leq f(a_b, b)$. Now, with $a_b$, there exists a $b_a_b in B$ such that $beta leq g(a_b, b_a_b)$. If $b=b_a_b$, then $g(a_b, b) = g(a_b, b_a_b)$ and
$$alpha leq f(a_b, b) text and beta leq g(a_b, b)$$
and so $(a_b, b)$ is a pair that satisfies the statement.
Comment:
My issue is, can I just write "if $b = b_a_b$"? Nothing prevents this from happening and since I just need existence this argument seems valid...
logic
asked Jul 18 at 19:04
Stephen Diadamo
112
edited Jul 18 at 20:33
asked Jul 18 at 19:04
Stephen Diadamo
112
asked Jul 18 at 19:04
Stephen Diadamo
112
asked Jul 18 at 19:04
Stephen Diadamo
112
112
If $A$ and $B$ are empty, then the preconditions hold vacuously but the consequence fails as there simply isn't any pair in $Atimes B$.
– Derek Elkins
Jul 18 at 19:16
For your comment, you can certainly say "if $b=b_a_b$ then" but you have to also handle the $bneq b_a_b$ case, or you have to give an argument for what that case can't happen. You can't just say "if it is the case that..." because the obvious retort is then, "well, what if it isn't the case?"
– Derek Elkins
Jul 18 at 19:25
Let $alpha=beta=1$ and $A=B=0,1$. Define $f(a,b)=a+bpmod 2$ and $g(a,b)=1- f(a,b)pmod 2$. $f$ and $g$ satisfy the preconditions ($f(a,1-a)=1$ and $g(a,a)=1$), but $f+g$ is the constantly $1$ function and thus less than $alpha+beta=2$. Perhaps you also have an unstated continuity assumption.
– Derek Elkins
Jul 18 at 20:11
I modified the statement slightly, sorry for the unnecessary confusion.
– Stephen Diadamo
Jul 18 at 20:35
Well, my counter-example was non-negative and bounded. So you will need to make use of the continuity assumption. The fact that your proof attempt would, if it worked, work just as well for non-continuous functions shows that something is missing/wrong and suggests a direction to look to find out what may need to be added.
– Derek Elkins
Jul 18 at 20:38
 |Â
show 1 more comment
If $A$ and $B$ are empty, then the preconditions hold vacuously but the consequence fails as there simply isn't any pair in $Atimes B$.
– Derek Elkins
Jul 18 at 19:16
For your comment, you can certainly say "if $b=b_a_b$ then" but you have to also handle the $bneq b_a_b$ case, or you have to give an argument for what that case can't happen. You can't just say "if it is the case that..." because the obvious retort is then, "well, what if it isn't the case?"
– Derek Elkins
Jul 18 at 19:25
Let $alpha=beta=1$ and $A=B=0,1$. Define $f(a,b)=a+bpmod 2$ and $g(a,b)=1- f(a,b)pmod 2$. $f$ and $g$ satisfy the preconditions ($f(a,1-a)=1$ and $g(a,a)=1$), but $f+g$ is the constantly $1$ function and thus less than $alpha+beta=2$. Perhaps you also have an unstated continuity assumption.
– Derek Elkins
Jul 18 at 20:11
I modified the statement slightly, sorry for the unnecessary confusion.
– Stephen Diadamo
Jul 18 at 20:35
Well, my counter-example was non-negative and bounded. So you will need to make use of the continuity assumption. The fact that your proof attempt would, if it worked, work just as well for non-continuous functions shows that something is missing/wrong and suggests a direction to look to find out what may need to be added.
– Derek Elkins
Jul 18 at 20:38
If $A$ and $B$ are empty, then the preconditions hold vacuously but the consequence fails as there simply isn't any pair in $Atimes B$.
– Derek Elkins
Jul 18 at 19:16
If $A$ and $B$ are empty, then the preconditions hold vacuously but the consequence fails as there simply isn't any pair in $Atimes B$.
– Derek Elkins
Jul 18 at 19:16
For your comment, you can certainly say "if $b=b_a_b$ then" but you have to also handle the $bneq b_a_b$ case, or you have to give an argument for what that case can't happen. You can't just say "if it is the case that..." because the obvious retort is then, "well, what if it isn't the case?"
– Derek Elkins
Jul 18 at 19:25
For your comment, you can certainly say "if $b=b_a_b$ then" but you have to also handle the $bneq b_a_b$ case, or you have to give an argument for what that case can't happen. You can't just say "if it is the case that..." because the obvious retort is then, "well, what if it isn't the case?"
– Derek Elkins
Jul 18 at 19:25
Let $alpha=beta=1$ and $A=B=0,1$. Define $f(a,b)=a+bpmod 2$ and $g(a,b)=1- f(a,b)pmod 2$. $f$ and $g$ satisfy the preconditions ($f(a,1-a)=1$ and $g(a,a)=1$), but $f+g$ is the constantly $1$ function and thus less than $alpha+beta=2$. Perhaps you also have an unstated continuity assumption.
– Derek Elkins
Jul 18 at 20:11
Let $alpha=beta=1$ and $A=B=0,1$. Define $f(a,b)=a+bpmod 2$ and $g(a,b)=1- f(a,b)pmod 2$. $f$ and $g$ satisfy the preconditions ($f(a,1-a)=1$ and $g(a,a)=1$), but $f+g$ is the constantly $1$ function and thus less than $alpha+beta=2$. Perhaps you also have an unstated continuity assumption.
– Derek Elkins
Jul 18 at 20:11
I modified the statement slightly, sorry for the unnecessary confusion.
– Stephen Diadamo
Jul 18 at 20:35
I modified the statement slightly, sorry for the unnecessary confusion.
– Stephen Diadamo
Jul 18 at 20:35
Well, my counter-example was non-negative and bounded. So you will need to make use of the continuity assumption. The fact that your proof attempt would, if it worked, work just as well for non-continuous functions shows that something is missing/wrong and suggests a direction to look to find out what may need to be added.
– Derek Elkins
Jul 18 at 20:38
Well, my counter-example was non-negative and bounded. So you will need to make use of the continuity assumption. The fact that your proof attempt would, if it worked, work just as well for non-continuous functions shows that something is missing/wrong and suggests a direction to look to find out what may need to be added.
– Derek Elkins
Jul 18 at 20:38
 |Â
show 1 more comment
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If $A$ and $B$ are empty, then the preconditions hold vacuously but the consequence fails as there simply isn't any pair in $Atimes B$.
– Derek Elkins
Jul 18 at 19:16
For your comment, you can certainly say "if $b=b_a_b$ then" but you have to also handle the $bneq b_a_b$ case, or you have to give an argument for what that case can't happen. You can't just say "if it is the case that..." because the obvious retort is then, "well, what if it isn't the case?"
– Derek Elkins
Jul 18 at 19:25
Let $alpha=beta=1$ and $A=B=0,1$. Define $f(a,b)=a+bpmod 2$ and $g(a,b)=1- f(a,b)pmod 2$. $f$ and $g$ satisfy the preconditions ($f(a,1-a)=1$ and $g(a,a)=1$), but $f+g$ is the constantly $1$ function and thus less than $alpha+beta=2$. Perhaps you also have an unstated continuity assumption.
– Derek Elkins
Jul 18 at 20:11
I modified the statement slightly, sorry for the unnecessary confusion.
– Stephen Diadamo
Jul 18 at 20:35
Well, my counter-example was non-negative and bounded. So you will need to make use of the continuity assumption. The fact that your proof attempt would, if it worked, work just as well for non-continuous functions shows that something is missing/wrong and suggests a direction to look to find out what may need to be added.
– Derek Elkins
Jul 18 at 20:38