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Bounded growth of functions vs bounded growth of functions on countable sets
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I am wondering if the boundedness of growth can be characterized by sequences. I am not sure if I use the term "growth" correctly, or use the correct tags for this question. Here is what I mean.
Let $X$ be an uncountable set and let $F$ be a collection of real-valued functions on $X$ with the following property:
For every countable set $Asubset X$ there is a function $alpha_A:Ato(0,+infty) $ such that for every $fin F$ the set $leftfracf(x)alpha_A(x), xin Aright$ is bounded.
Does there exist a function $alpha:Xto(0,+infty)$ such that for every $fin F$ the set $leftfracf(x)alpha(x), xin Xright$ is bounded?
If the answer is "no", can we remedy the situation by assuming that $X$ is a topological space satisfying some mild conditions and $Fsubset C(X)$?
fa.functional-analysis set-theory gn.general-topology real-analysis
asked 2 hours ago
erz
1,4881714
add a comment |Â
up vote
2
down vote
favorite
I am wondering if the boundedness of growth can be characterized by sequences. I am not sure if I use the term "growth" correctly, or use the correct tags for this question. Here is what I mean.
Let $X$ be an uncountable set and let $F$ be a collection of real-valued functions on $X$ with the following property:
For every countable set $Asubset X$ there is a function $alpha_A:Ato(0,+infty) $ such that for every $fin F$ the set $leftfracf(x)alpha_A(x), xin Aright$ is bounded.
Does there exist a function $alpha:Xto(0,+infty)$ such that for every $fin F$ the set $leftfracf(x)alpha(x), xin Xright$ is bounded?
If the answer is "no", can we remedy the situation by assuming that $X$ is a topological space satisfying some mild conditions and $Fsubset C(X)$?
fa.functional-analysis set-theory gn.general-topology real-analysis
asked 2 hours ago
erz
1,4881714
If I do not misunderstand, then I think the answer is yes for easy reasons: applying your given condition in the special case where each A is a singleton, it follows that for each $xin X$ we have $h(x):=sup f(x) colon fin F<infty$. Then $h:Xto (0,infty)$ is your required upper bound for the collection $F$.
– Yemon Choi
2 hours ago
3
@YemonChoi: I think you do misunderstand --- the sup over $F$ is not supposed to be finite, just the sup over $A$ for each $f$ separately.
– Nik Weaver
2 hours ago
@YemonChoi it is exactly as Nik Weaver says. In fact, $F$ is WLOG a vector space, and the condition should be viewed as existence of a weighted supremum seminorm.
– erz
1 hour ago
1
@NikWeaver Thank you! Apologies to the OP.
– Yemon Choi
23 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am wondering if the boundedness of growth can be characterized by sequences. I am not sure if I use the term "growth" correctly, or use the correct tags for this question. Here is what I mean.
Let $X$ be an uncountable set and let $F$ be a collection of real-valued functions on $X$ with the following property:
For every countable set $Asubset X$ there is a function $alpha_A:Ato(0,+infty) $ such that for every $fin F$ the set $leftfracf(x)alpha_A(x), xin Aright$ is bounded.
Does there exist a function $alpha:Xto(0,+infty)$ such that for every $fin F$ the set $leftfracf(x)alpha(x), xin Xright$ is bounded?
If the answer is "no", can we remedy the situation by assuming that $X$ is a topological space satisfying some mild conditions and $Fsubset C(X)$?
fa.functional-analysis set-theory gn.general-topology real-analysis
asked 2 hours ago
erz
1,4881714
I am wondering if the boundedness of growth can be characterized by sequences. I am not sure if I use the term "growth" correctly, or use the correct tags for this question. Here is what I mean.
Let $X$ be an uncountable set and let $F$ be a collection of real-valued functions on $X$ with the following property:
For every countable set $Asubset X$ there is a function $alpha_A:Ato(0,+infty) $ such that for every $fin F$ the set $leftfracf(x)alpha_A(x), xin Aright$ is bounded.
Does there exist a function $alpha:Xto(0,+infty)$ such that for every $fin F$ the set $leftfracf(x)alpha(x), xin Xright$ is bounded?
If the answer is "no", can we remedy the situation by assuming that $X$ is a topological space satisfying some mild conditions and $Fsubset C(X)$?
fa.functional-analysis set-theory gn.general-topology real-analysis
fa.functional-analysis set-theory gn.general-topology real-analysis
asked 2 hours ago
erz
1,4881714
asked 2 hours ago
erz
1,4881714
asked 2 hours ago
erz
1,4881714
asked 2 hours ago
erz
1,4881714
asked 2 hours ago
erz
1,4881714
1,4881714
If I do not misunderstand, then I think the answer is yes for easy reasons: applying your given condition in the special case where each A is a singleton, it follows that for each $xin X$ we have $h(x):=sup f(x) colon fin F<infty$. Then $h:Xto (0,infty)$ is your required upper bound for the collection $F$.
– Yemon Choi
2 hours ago
3
@YemonChoi: I think you do misunderstand --- the sup over $F$ is not supposed to be finite, just the sup over $A$ for each $f$ separately.
– Nik Weaver
2 hours ago
@YemonChoi it is exactly as Nik Weaver says. In fact, $F$ is WLOG a vector space, and the condition should be viewed as existence of a weighted supremum seminorm.
– erz
1 hour ago
1
@NikWeaver Thank you! Apologies to the OP.
– Yemon Choi
23 mins ago
add a comment |Â
If I do not misunderstand, then I think the answer is yes for easy reasons: applying your given condition in the special case where each A is a singleton, it follows that for each $xin X$ we have $h(x):=sup f(x) colon fin F<infty$. Then $h:Xto (0,infty)$ is your required upper bound for the collection $F$.
– Yemon Choi
2 hours ago
3
@YemonChoi: I think you do misunderstand --- the sup over $F$ is not supposed to be finite, just the sup over $A$ for each $f$ separately.
– Nik Weaver
2 hours ago
@YemonChoi it is exactly as Nik Weaver says. In fact, $F$ is WLOG a vector space, and the condition should be viewed as existence of a weighted supremum seminorm.
– erz
1 hour ago
1
@NikWeaver Thank you! Apologies to the OP.
– Yemon Choi
23 mins ago
If I do not misunderstand, then I think the answer is yes for easy reasons: applying your given condition in the special case where each A is a singleton, it follows that for each $xin X$ we have $h(x):=sup f(x) colon fin F<infty$. Then $h:Xto (0,infty)$ is your required upper bound for the collection $F$.
– Yemon Choi
2 hours ago
If I do not misunderstand, then I think the answer is yes for easy reasons: applying your given condition in the special case where each A is a singleton, it follows that for each $xin X$ we have $h(x):=sup f(x) colon fin F<infty$. Then $h:Xto (0,infty)$ is your required upper bound for the collection $F$.
– Yemon Choi
2 hours ago
3
3
@YemonChoi: I think you do misunderstand --- the sup over $F$ is not supposed to be finite, just the sup over $A$ for each $f$ separately.
– Nik Weaver
2 hours ago
@YemonChoi: I think you do misunderstand --- the sup over $F$ is not supposed to be finite, just the sup over $A$ for each $f$ separately.
– Nik Weaver
2 hours ago
@YemonChoi it is exactly as Nik Weaver says. In fact, $F$ is WLOG a vector space, and the condition should be viewed as existence of a weighted supremum seminorm.
– erz
1 hour ago
@YemonChoi it is exactly as Nik Weaver says. In fact, $F$ is WLOG a vector space, and the condition should be viewed as existence of a weighted supremum seminorm.
– erz
1 hour ago
1
1
@NikWeaver Thank you! Apologies to the OP.
– Yemon Choi
23 mins ago
@NikWeaver Thank you! Apologies to the OP.
– Yemon Choi
23 mins ago
add a comment |Â
1 Answer
1
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This was a fun question! The answer is no.
Let $Omega$ be the set of all countable ordinals. For each limit ordinal $alpha in Omega$ let $f_alpha: Omega to [1,infty)$ be a function which increases to infinity on $[0,alpha)$ and is constantly zero on $[alpha,Omega)$.
For any limit ordinal $alpha in Omega$ the set of $f_beta$ with $beta leq alpha$ a limit ordinal is countable, and therefore there is a function $a_alpha: [0,alpha) to [1,infty)$ such that $f_beta preceq a_alpha$ for all $beta < alpha$ on $[0,alpha)$, where $preceq$ denotes "$leq$ except at finitely many points". Since any $f_beta$ with $beta > alpha$ is bounded on $[0,alpha)$ we get the stated condition.
However, there is no $a: Omega to (0,infty)$ that works. Given any such function $a$, there is an $n in mathbbN$ such that $a(alpha) leq n$ for uncountably many $alpha$. Let $(alpha_k)$ be an increasing sequence of such $alpha$'s and let $alpha$ be their supremum. Then the function $f_alpha$ goes to infinity on $[0,alpha)$ so its quotient with $a$ will still be unbounded.
answered 1 hour ago
Nik Weaver
17.9k143114
Do I understand correctly that $a_alpha$ from your second can be constructed in the following way? Form a sequence $beta_n$ of the limit ordinals smaller than $alpha$ and then say that $a_alpha(beta)=max f_beta_k(beta)$, where $kle m$, where $beta_m$ is the the greatest limit ordinal smaller than $beta$.
– erz
1 hour ago
1
I think so, yes. It's easier to say more generally that any countable family of functions on a countable set can be majorized in this way: wlog the functions are $f_n: mathbbN to (0,infty)$, and then you let $a(n) = rm max(f_1(n), ldots, f_n(n))$.
– Nik Weaver
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This was a fun question! The answer is no.
Let $Omega$ be the set of all countable ordinals. For each limit ordinal $alpha in Omega$ let $f_alpha: Omega to [1,infty)$ be a function which increases to infinity on $[0,alpha)$ and is constantly zero on $[alpha,Omega)$.
For any limit ordinal $alpha in Omega$ the set of $f_beta$ with $beta leq alpha$ a limit ordinal is countable, and therefore there is a function $a_alpha: [0,alpha) to [1,infty)$ such that $f_beta preceq a_alpha$ for all $beta < alpha$ on $[0,alpha)$, where $preceq$ denotes "$leq$ except at finitely many points". Since any $f_beta$ with $beta > alpha$ is bounded on $[0,alpha)$ we get the stated condition.
However, there is no $a: Omega to (0,infty)$ that works. Given any such function $a$, there is an $n in mathbbN$ such that $a(alpha) leq n$ for uncountably many $alpha$. Let $(alpha_k)$ be an increasing sequence of such $alpha$'s and let $alpha$ be their supremum. Then the function $f_alpha$ goes to infinity on $[0,alpha)$ so its quotient with $a$ will still be unbounded.
answered 1 hour ago
Nik Weaver
17.9k143114
Do I understand correctly that $a_alpha$ from your second can be constructed in the following way? Form a sequence $beta_n$ of the limit ordinals smaller than $alpha$ and then say that $a_alpha(beta)=max f_beta_k(beta)$, where $kle m$, where $beta_m$ is the the greatest limit ordinal smaller than $beta$.
– erz
1 hour ago
1
I think so, yes. It's easier to say more generally that any countable family of functions on a countable set can be majorized in this way: wlog the functions are $f_n: mathbbN to (0,infty)$, and then you let $a(n) = rm max(f_1(n), ldots, f_n(n))$.
– Nik Weaver
1 hour ago
add a comment |Â
up vote
3
down vote
accepted
This was a fun question! The answer is no.
Let $Omega$ be the set of all countable ordinals. For each limit ordinal $alpha in Omega$ let $f_alpha: Omega to [1,infty)$ be a function which increases to infinity on $[0,alpha)$ and is constantly zero on $[alpha,Omega)$.
For any limit ordinal $alpha in Omega$ the set of $f_beta$ with $beta leq alpha$ a limit ordinal is countable, and therefore there is a function $a_alpha: [0,alpha) to [1,infty)$ such that $f_beta preceq a_alpha$ for all $beta < alpha$ on $[0,alpha)$, where $preceq$ denotes "$leq$ except at finitely many points". Since any $f_beta$ with $beta > alpha$ is bounded on $[0,alpha)$ we get the stated condition.
However, there is no $a: Omega to (0,infty)$ that works. Given any such function $a$, there is an $n in mathbbN$ such that $a(alpha) leq n$ for uncountably many $alpha$. Let $(alpha_k)$ be an increasing sequence of such $alpha$'s and let $alpha$ be their supremum. Then the function $f_alpha$ goes to infinity on $[0,alpha)$ so its quotient with $a$ will still be unbounded.
answered 1 hour ago
Nik Weaver
17.9k143114
Do I understand correctly that $a_alpha$ from your second can be constructed in the following way? Form a sequence $beta_n$ of the limit ordinals smaller than $alpha$ and then say that $a_alpha(beta)=max f_beta_k(beta)$, where $kle m$, where $beta_m$ is the the greatest limit ordinal smaller than $beta$.
– erz
1 hour ago
1
I think so, yes. It's easier to say more generally that any countable family of functions on a countable set can be majorized in this way: wlog the functions are $f_n: mathbbN to (0,infty)$, and then you let $a(n) = rm max(f_1(n), ldots, f_n(n))$.
– Nik Weaver
1 hour ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This was a fun question! The answer is no.
Let $Omega$ be the set of all countable ordinals. For each limit ordinal $alpha in Omega$ let $f_alpha: Omega to [1,infty)$ be a function which increases to infinity on $[0,alpha)$ and is constantly zero on $[alpha,Omega)$.
For any limit ordinal $alpha in Omega$ the set of $f_beta$ with $beta leq alpha$ a limit ordinal is countable, and therefore there is a function $a_alpha: [0,alpha) to [1,infty)$ such that $f_beta preceq a_alpha$ for all $beta < alpha$ on $[0,alpha)$, where $preceq$ denotes "$leq$ except at finitely many points". Since any $f_beta$ with $beta > alpha$ is bounded on $[0,alpha)$ we get the stated condition.
However, there is no $a: Omega to (0,infty)$ that works. Given any such function $a$, there is an $n in mathbbN$ such that $a(alpha) leq n$ for uncountably many $alpha$. Let $(alpha_k)$ be an increasing sequence of such $alpha$'s and let $alpha$ be their supremum. Then the function $f_alpha$ goes to infinity on $[0,alpha)$ so its quotient with $a$ will still be unbounded.
answered 1 hour ago
Nik Weaver
17.9k143114
This was a fun question! The answer is no.
Let $Omega$ be the set of all countable ordinals. For each limit ordinal $alpha in Omega$ let $f_alpha: Omega to [1,infty)$ be a function which increases to infinity on $[0,alpha)$ and is constantly zero on $[alpha,Omega)$.
For any limit ordinal $alpha in Omega$ the set of $f_beta$ with $beta leq alpha$ a limit ordinal is countable, and therefore there is a function $a_alpha: [0,alpha) to [1,infty)$ such that $f_beta preceq a_alpha$ for all $beta < alpha$ on $[0,alpha)$, where $preceq$ denotes "$leq$ except at finitely many points". Since any $f_beta$ with $beta > alpha$ is bounded on $[0,alpha)$ we get the stated condition.
However, there is no $a: Omega to (0,infty)$ that works. Given any such function $a$, there is an $n in mathbbN$ such that $a(alpha) leq n$ for uncountably many $alpha$. Let $(alpha_k)$ be an increasing sequence of such $alpha$'s and let $alpha$ be their supremum. Then the function $f_alpha$ goes to infinity on $[0,alpha)$ so its quotient with $a$ will still be unbounded.
answered 1 hour ago
Nik Weaver
17.9k143114
answered 1 hour ago
Nik Weaver
17.9k143114
answered 1 hour ago
Nik Weaver
17.9k143114
answered 1 hour ago
Nik Weaver
17.9k143114
17.9k143114
Do I understand correctly that $a_alpha$ from your second can be constructed in the following way? Form a sequence $beta_n$ of the limit ordinals smaller than $alpha$ and then say that $a_alpha(beta)=max f_beta_k(beta)$, where $kle m$, where $beta_m$ is the the greatest limit ordinal smaller than $beta$.
– erz
1 hour ago
1
I think so, yes. It's easier to say more generally that any countable family of functions on a countable set can be majorized in this way: wlog the functions are $f_n: mathbbN to (0,infty)$, and then you let $a(n) = rm max(f_1(n), ldots, f_n(n))$.
– Nik Weaver
1 hour ago
add a comment |Â
Do I understand correctly that $a_alpha$ from your second can be constructed in the following way? Form a sequence $beta_n$ of the limit ordinals smaller than $alpha$ and then say that $a_alpha(beta)=max f_beta_k(beta)$, where $kle m$, where $beta_m$ is the the greatest limit ordinal smaller than $beta$.
– erz
1 hour ago
1
I think so, yes. It's easier to say more generally that any countable family of functions on a countable set can be majorized in this way: wlog the functions are $f_n: mathbbN to (0,infty)$, and then you let $a(n) = rm max(f_1(n), ldots, f_n(n))$.
– Nik Weaver
1 hour ago
Do I understand correctly that $a_alpha$ from your second can be constructed in the following way? Form a sequence $beta_n$ of the limit ordinals smaller than $alpha$ and then say that $a_alpha(beta)=max f_beta_k(beta)$, where $kle m$, where $beta_m$ is the the greatest limit ordinal smaller than $beta$.
– erz
1 hour ago
Do I understand correctly that $a_alpha$ from your second can be constructed in the following way? Form a sequence $beta_n$ of the limit ordinals smaller than $alpha$ and then say that $a_alpha(beta)=max f_beta_k(beta)$, where $kle m$, where $beta_m$ is the the greatest limit ordinal smaller than $beta$.
– erz
1 hour ago
1
1
I think so, yes. It's easier to say more generally that any countable family of functions on a countable set can be majorized in this way: wlog the functions are $f_n: mathbbN to (0,infty)$, and then you let $a(n) = rm max(f_1(n), ldots, f_n(n))$.
– Nik Weaver
1 hour ago
I think so, yes. It's easier to say more generally that any countable family of functions on a countable set can be majorized in this way: wlog the functions are $f_n: mathbbN to (0,infty)$, and then you let $a(n) = rm max(f_1(n), ldots, f_n(n))$.
– Nik Weaver
1 hour ago
add a comment |Â
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If I do not misunderstand, then I think the answer is yes for easy reasons: applying your given condition in the special case where each A is a singleton, it follows that for each $xin X$ we have $h(x):=sup f(x) colon fin F<infty$. Then $h:Xto (0,infty)$ is your required upper bound for the collection $F$.
– Yemon Choi
2 hours ago
3
@YemonChoi: I think you do misunderstand --- the sup over $F$ is not supposed to be finite, just the sup over $A$ for each $f$ separately.
– Nik Weaver
2 hours ago
@YemonChoi it is exactly as Nik Weaver says. In fact, $F$ is WLOG a vector space, and the condition should be viewed as existence of a weighted supremum seminorm.
– erz
1 hour ago
1
@NikWeaver Thank you! Apologies to the OP.
– Yemon Choi
23 mins ago