Mixing
Find cardinality of $B = left{ f : mathbb N rightarrow mathbb N mid forall n. f(n)le n) wedge forall m;...
$begingroup$
Find cardinality of $$B = left{ f : mathbb N rightarrow mathbb N mid forall n(f(n)le n) wedge forall m; exists n (f(n) > m) right}$$
My try
I have solved this, but I am not sure if it is correct (I have not a lot of experience in set theory). Can somebody check that or give some tips (or both)?
$|B| < mathfrak{c}$ because $|B| < |mathbb N|^{|mathbb N|}$
from the other hand I can define $G$ injective such as:
$$G: (mathbb N rightarrow left{0,1 right}) rightarrow B $$
$$ G(alpha)(n) = begin{cases}
alpha(n) + G(alpha)(n-1), &text{if }n neq 0 \
0, &text{if }n = 0.
end{cases} $$
The function $G$ increases and is injective, and its power is $mathfrak{c} $ so $|B| = mathfrak{c}$
functions elementary-set-theory
asked Jan 30 at 16:14
VirtualUserVirtualUser
1,237317
$endgroup$
|
show 2 more comments
$begingroup$
Find cardinality of $$B = left{ f : mathbb N rightarrow mathbb N mid forall n(f(n)le n) wedge forall m; exists n (f(n) > m) right}$$
My try
I have solved this, but I am not sure if it is correct (I have not a lot of experience in set theory). Can somebody check that or give some tips (or both)?
$|B| < mathfrak{c}$ because $|B| < |mathbb N|^{|mathbb N|}$
from the other hand I can define $G$ injective such as:
$$G: (mathbb N rightarrow left{0,1 right}) rightarrow B $$
$$ G(alpha)(n) = begin{cases}
alpha(n) + G(alpha)(n-1), &text{if }n neq 0 \
0, &text{if }n = 0.
end{cases} $$
The function $G$ increases and is injective, and its power is $mathfrak{c} $ so $|B| = mathfrak{c}$
functions elementary-set-theory
asked Jan 30 at 16:14
VirtualUserVirtualUser
1,237317
$endgroup$
1
$begingroup$
Note that if $alpha$ takes the value $1$ only finitely many times, then $G(alpha) notin B$.
$endgroup$
– Mees de Vries
Jan 30 at 16:26
1
$begingroup$
Why do you say $|B| < |mathbb{N}|^{|mathbb{N}|}$?
$endgroup$
– Clive Newstead
Jan 30 at 17:06
$begingroup$
It's stil not clear what is $mathfrak{c}$, it should be defined when first used. Which power of $G$ ?
$endgroup$
– Soleil
Jan 30 at 18:21
$begingroup$
continuum - why is it not clear?
$endgroup$
– VirtualUser
Jan 30 at 18:35
$begingroup$
@VirtualUser Because it s not defined in usual set theory / decriptive set theory books, and you did not defined it. Hence $mathfrak{c}:= 2^{mathbb N} = aleph_1$.
$endgroup$
– Soleil
Jan 30 at 18:48
|
show 2 more comments
$begingroup$
Find cardinality of $$B = left{ f : mathbb N rightarrow mathbb N mid forall n(f(n)le n) wedge forall m; exists n (f(n) > m) right}$$
My try
I have solved this, but I am not sure if it is correct (I have not a lot of experience in set theory). Can somebody check that or give some tips (or both)?
$|B| < mathfrak{c}$ because $|B| < |mathbb N|^{|mathbb N|}$
from the other hand I can define $G$ injective such as:
$$G: (mathbb N rightarrow left{0,1 right}) rightarrow B $$
$$ G(alpha)(n) = begin{cases}
alpha(n) + G(alpha)(n-1), &text{if }n neq 0 \
0, &text{if }n = 0.
end{cases} $$
The function $G$ increases and is injective, and its power is $mathfrak{c} $ so $|B| = mathfrak{c}$
functions elementary-set-theory
asked Jan 30 at 16:14
VirtualUserVirtualUser
1,237317
$endgroup$
Find cardinality of $$B = left{ f : mathbb N rightarrow mathbb N mid forall n(f(n)le n) wedge forall m; exists n (f(n) > m) right}$$
My try
I have solved this, but I am not sure if it is correct (I have not a lot of experience in set theory). Can somebody check that or give some tips (or both)?
$|B| < mathfrak{c}$ because $|B| < |mathbb N|^{|mathbb N|}$
from the other hand I can define $G$ injective such as:
$$G: (mathbb N rightarrow left{0,1 right}) rightarrow B $$
$$ G(alpha)(n) = begin{cases}
alpha(n) + G(alpha)(n-1), &text{if }n neq 0 \
0, &text{if }n = 0.
end{cases} $$
The function $G$ increases and is injective, and its power is $mathfrak{c} $ so $|B| = mathfrak{c}$
functions elementary-set-theory
functions elementary-set-theory
asked Jan 30 at 16:14
VirtualUserVirtualUser
1,237317
asked Jan 30 at 16:14
VirtualUserVirtualUser
1,237317
edited Jan 31 at 12:32
VirtualUser
asked Jan 30 at 16:14
VirtualUserVirtualUser
1,237317
asked Jan 30 at 16:14
VirtualUserVirtualUser
1,237317
asked Jan 30 at 16:14
VirtualUserVirtualUser
1,237317
1,237317
1
$begingroup$
Note that if $alpha$ takes the value $1$ only finitely many times, then $G(alpha) notin B$.
$endgroup$
– Mees de Vries
Jan 30 at 16:26
1
$begingroup$
Why do you say $|B| < |mathbb{N}|^{|mathbb{N}|}$?
$endgroup$
– Clive Newstead
Jan 30 at 17:06
$begingroup$
It's stil not clear what is $mathfrak{c}$, it should be defined when first used. Which power of $G$ ?
$endgroup$
– Soleil
Jan 30 at 18:21
$begingroup$
continuum - why is it not clear?
$endgroup$
– VirtualUser
Jan 30 at 18:35
$begingroup$
@VirtualUser Because it s not defined in usual set theory / decriptive set theory books, and you did not defined it. Hence $mathfrak{c}:= 2^{mathbb N} = aleph_1$.
$endgroup$
– Soleil
Jan 30 at 18:48
|
show 2 more comments
1
$begingroup$
Note that if $alpha$ takes the value $1$ only finitely many times, then $G(alpha) notin B$.
$endgroup$
– Mees de Vries
Jan 30 at 16:26
1
$begingroup$
Why do you say $|B| < |mathbb{N}|^{|mathbb{N}|}$?
$endgroup$
– Clive Newstead
Jan 30 at 17:06
$begingroup$
It's stil not clear what is $mathfrak{c}$, it should be defined when first used. Which power of $G$ ?
$endgroup$
– Soleil
Jan 30 at 18:21
$begingroup$
continuum - why is it not clear?
$endgroup$
– VirtualUser
Jan 30 at 18:35
$begingroup$
@VirtualUser Because it s not defined in usual set theory / decriptive set theory books, and you did not defined it. Hence $mathfrak{c}:= 2^{mathbb N} = aleph_1$.
$endgroup$
– Soleil
Jan 30 at 18:48
1
1
$begingroup$
Note that if $alpha$ takes the value $1$ only finitely many times, then $G(alpha) notin B$.
$endgroup$
– Mees de Vries
Jan 30 at 16:26
$begingroup$
Note that if $alpha$ takes the value $1$ only finitely many times, then $G(alpha) notin B$.
$endgroup$
– Mees de Vries
Jan 30 at 16:26
1
1
$begingroup$
Why do you say $|B| < |mathbb{N}|^{|mathbb{N}|}$?
$endgroup$
– Clive Newstead
Jan 30 at 17:06
$begingroup$
Why do you say $|B| < |mathbb{N}|^{|mathbb{N}|}$?
$endgroup$
– Clive Newstead
Jan 30 at 17:06
$begingroup$
It's stil not clear what is $mathfrak{c}$, it should be defined when first used. Which power of $G$ ?
$endgroup$
– Soleil
Jan 30 at 18:21
$begingroup$
It's stil not clear what is $mathfrak{c}$, it should be defined when first used. Which power of $G$ ?
$endgroup$
– Soleil
Jan 30 at 18:21
$begingroup$
continuum - why is it not clear?
$endgroup$
– VirtualUser
Jan 30 at 18:35
$begingroup$
continuum - why is it not clear?
$endgroup$
– VirtualUser
Jan 30 at 18:35
$begingroup$
@VirtualUser Because it s not defined in usual set theory / decriptive set theory books, and you did not defined it. Hence $mathfrak{c}:= 2^{mathbb N} = aleph_1$.
$endgroup$
– Soleil
Jan 30 at 18:48
$begingroup$
@VirtualUser Because it s not defined in usual set theory / decriptive set theory books, and you did not defined it. Hence $mathfrak{c}:= 2^{mathbb N} = aleph_1$.
$endgroup$
– Soleil
Jan 30 at 18:48
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$B$ is the set of functions $mathbb{N} to mathbb{N}$ that are unbounded in absolute terms, but are bounded by the identity function.
Evidently $|B| le mathfrak{c}$ since $B$ is a subset of $mathbb{N}^{mathbb{N}}$.
To see that $mathfrak{c} le |B|$, let $mathcal{P}_{mathsf{inf}}(mathbb{N}) subseteq mathcal{P}(mathbb{N})$ be the set of infinite subsets of $mathbb{N}$. Then $|mathcal{P}_{mathsf{inf}}(mathbb{N})| = mathfrak{c}$, since the set of all finite subsets of $mathbb{mathbb{N}}$ is countable.
For each $U in mathcal{P}_{mathsf{inf}}(mathbb{N})$, define $f_U in B$ inductively by
$$f_U(0) = 0 quad text{and} quad f_U(n+1) = begin{cases} f_U(n) & text{if } n notin U \ f_U(n)+1 & text{if } n in U end{cases}$$
The fact that $f_U$ is unbounded follows from the fact that $U$ is infinite. You can prove by induction that $f_U(n) le n$ for all $n in mathbb{N}$, and that $U mapsto f_U$ defines an injection $mathcal{P}_{mathsf{inf}}(mathbb{N}) to B$.
Hence $mathfrak{c} le |B|$.
answered Jan 30 at 17:12
Clive NewsteadClive Newstead
52k474136
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$B$ is the set of functions $mathbb{N} to mathbb{N}$ that are unbounded in absolute terms, but are bounded by the identity function.
Evidently $|B| le mathfrak{c}$ since $B$ is a subset of $mathbb{N}^{mathbb{N}}$.
To see that $mathfrak{c} le |B|$, let $mathcal{P}_{mathsf{inf}}(mathbb{N}) subseteq mathcal{P}(mathbb{N})$ be the set of infinite subsets of $mathbb{N}$. Then $|mathcal{P}_{mathsf{inf}}(mathbb{N})| = mathfrak{c}$, since the set of all finite subsets of $mathbb{mathbb{N}}$ is countable.
For each $U in mathcal{P}_{mathsf{inf}}(mathbb{N})$, define $f_U in B$ inductively by
$$f_U(0) = 0 quad text{and} quad f_U(n+1) = begin{cases} f_U(n) & text{if } n notin U \ f_U(n)+1 & text{if } n in U end{cases}$$
The fact that $f_U$ is unbounded follows from the fact that $U$ is infinite. You can prove by induction that $f_U(n) le n$ for all $n in mathbb{N}$, and that $U mapsto f_U$ defines an injection $mathcal{P}_{mathsf{inf}}(mathbb{N}) to B$.
Hence $mathfrak{c} le |B|$.
answered Jan 30 at 17:12
Clive NewsteadClive Newstead
52k474136
$endgroup$
add a comment |
$begingroup$
$B$ is the set of functions $mathbb{N} to mathbb{N}$ that are unbounded in absolute terms, but are bounded by the identity function.
Evidently $|B| le mathfrak{c}$ since $B$ is a subset of $mathbb{N}^{mathbb{N}}$.
To see that $mathfrak{c} le |B|$, let $mathcal{P}_{mathsf{inf}}(mathbb{N}) subseteq mathcal{P}(mathbb{N})$ be the set of infinite subsets of $mathbb{N}$. Then $|mathcal{P}_{mathsf{inf}}(mathbb{N})| = mathfrak{c}$, since the set of all finite subsets of $mathbb{mathbb{N}}$ is countable.
For each $U in mathcal{P}_{mathsf{inf}}(mathbb{N})$, define $f_U in B$ inductively by
$$f_U(0) = 0 quad text{and} quad f_U(n+1) = begin{cases} f_U(n) & text{if } n notin U \ f_U(n)+1 & text{if } n in U end{cases}$$
The fact that $f_U$ is unbounded follows from the fact that $U$ is infinite. You can prove by induction that $f_U(n) le n$ for all $n in mathbb{N}$, and that $U mapsto f_U$ defines an injection $mathcal{P}_{mathsf{inf}}(mathbb{N}) to B$.
Hence $mathfrak{c} le |B|$.
answered Jan 30 at 17:12
Clive NewsteadClive Newstead
52k474136
$endgroup$
add a comment |
$begingroup$
$B$ is the set of functions $mathbb{N} to mathbb{N}$ that are unbounded in absolute terms, but are bounded by the identity function.
Evidently $|B| le mathfrak{c}$ since $B$ is a subset of $mathbb{N}^{mathbb{N}}$.
To see that $mathfrak{c} le |B|$, let $mathcal{P}_{mathsf{inf}}(mathbb{N}) subseteq mathcal{P}(mathbb{N})$ be the set of infinite subsets of $mathbb{N}$. Then $|mathcal{P}_{mathsf{inf}}(mathbb{N})| = mathfrak{c}$, since the set of all finite subsets of $mathbb{mathbb{N}}$ is countable.
For each $U in mathcal{P}_{mathsf{inf}}(mathbb{N})$, define $f_U in B$ inductively by
$$f_U(0) = 0 quad text{and} quad f_U(n+1) = begin{cases} f_U(n) & text{if } n notin U \ f_U(n)+1 & text{if } n in U end{cases}$$
The fact that $f_U$ is unbounded follows from the fact that $U$ is infinite. You can prove by induction that $f_U(n) le n$ for all $n in mathbb{N}$, and that $U mapsto f_U$ defines an injection $mathcal{P}_{mathsf{inf}}(mathbb{N}) to B$.
Hence $mathfrak{c} le |B|$.
answered Jan 30 at 17:12
Clive NewsteadClive Newstead
52k474136
$endgroup$
$B$ is the set of functions $mathbb{N} to mathbb{N}$ that are unbounded in absolute terms, but are bounded by the identity function.
Evidently $|B| le mathfrak{c}$ since $B$ is a subset of $mathbb{N}^{mathbb{N}}$.
To see that $mathfrak{c} le |B|$, let $mathcal{P}_{mathsf{inf}}(mathbb{N}) subseteq mathcal{P}(mathbb{N})$ be the set of infinite subsets of $mathbb{N}$. Then $|mathcal{P}_{mathsf{inf}}(mathbb{N})| = mathfrak{c}$, since the set of all finite subsets of $mathbb{mathbb{N}}$ is countable.
For each $U in mathcal{P}_{mathsf{inf}}(mathbb{N})$, define $f_U in B$ inductively by
$$f_U(0) = 0 quad text{and} quad f_U(n+1) = begin{cases} f_U(n) & text{if } n notin U \ f_U(n)+1 & text{if } n in U end{cases}$$
The fact that $f_U$ is unbounded follows from the fact that $U$ is infinite. You can prove by induction that $f_U(n) le n$ for all $n in mathbb{N}$, and that $U mapsto f_U$ defines an injection $mathcal{P}_{mathsf{inf}}(mathbb{N}) to B$.
Hence $mathfrak{c} le |B|$.
answered Jan 30 at 17:12
Clive NewsteadClive Newstead
52k474136
answered Jan 30 at 17:12
Clive NewsteadClive Newstead
52k474136
answered Jan 30 at 17:12
Clive NewsteadClive Newstead
52k474136
answered Jan 30 at 17:12
Clive NewsteadClive Newstead
52k474136
52k474136
add a comment |
add a comment |
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1
$begingroup$
Note that if $alpha$ takes the value $1$ only finitely many times, then $G(alpha) notin B$.
$endgroup$
– Mees de Vries
Jan 30 at 16:26
1
$begingroup$
Why do you say $|B| < |mathbb{N}|^{|mathbb{N}|}$?
$endgroup$
– Clive Newstead
Jan 30 at 17:06
$begingroup$
It's stil not clear what is $mathfrak{c}$, it should be defined when first used. Which power of $G$ ?
$endgroup$
– Soleil
Jan 30 at 18:21
$begingroup$
continuum - why is it not clear?
$endgroup$
– VirtualUser
Jan 30 at 18:35
$begingroup$
@VirtualUser Because it s not defined in usual set theory / decriptive set theory books, and you did not defined it. Hence $mathfrak{c}:= 2^{mathbb N} = aleph_1$.
$endgroup$
– Soleil
Jan 30 at 18:48