free ultrafilter
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If $omega$ is a free ultrafilter on $mathbb{N}$,$(x_n)$ is a sequence of complex numbers,what is the precise definition of "$lim_{omega}(x_n)$ does not converge to $x$"?
general-topology filters
edited Nov 16 at 23:15
Asaf Karagila♦
300k32421751
asked Nov 16 at 22:12
mathrookie
722512
add a comment |
up vote
0
down vote
favorite
If $omega$ is a free ultrafilter on $mathbb{N}$,$(x_n)$ is a sequence of complex numbers,what is the precise definition of "$lim_{omega}(x_n)$ does not converge to $x$"?
general-topology filters
edited Nov 16 at 23:15
Asaf Karagila♦
300k32421751
asked Nov 16 at 22:12
mathrookie
722512
1
My guess (not confident enough to make it an answer) would be: $x_* omega = { S mid x^{-1}(S) in omega }$ is an ultrafilter on $mathbb{C}$, and we're requiring that ultrafilter not to have limit $x$. So, that would mean that the neighborhood filter $mathcal{N}_x notsubseteq x_* omega$, i.e. there is some $epsilon > 0$ such that ${ n in mathbb{N} mid |x_n - x| < epsilon } notin omega$.
– Daniel Schepler
Nov 16 at 22:26
1
Since $omega$ is often synonymous to $Bbb N$, I would to nominate this to the prize of "worst notation ever". :-)
– Asaf Karagila♦
Nov 16 at 23:16
Yes, write $mathcal{F}$ for the filter, and use $lim_{mathcal{F}} x_n$ instead, as is more usual.
– Henno Brandsma
Nov 17 at 5:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $omega$ is a free ultrafilter on $mathbb{N}$,$(x_n)$ is a sequence of complex numbers,what is the precise definition of "$lim_{omega}(x_n)$ does not converge to $x$"?
general-topology filters
edited Nov 16 at 23:15
Asaf Karagila♦
300k32421751
asked Nov 16 at 22:12
mathrookie
722512
If $omega$ is a free ultrafilter on $mathbb{N}$,$(x_n)$ is a sequence of complex numbers,what is the precise definition of "$lim_{omega}(x_n)$ does not converge to $x$"?
general-topology filters
general-topology filters
edited Nov 16 at 23:15
Asaf Karagila♦
300k32421751
asked Nov 16 at 22:12
mathrookie
722512
edited Nov 16 at 23:15
Asaf Karagila♦
300k32421751
asked Nov 16 at 22:12
mathrookie
722512
edited Nov 16 at 23:15
Asaf Karagila♦
300k32421751
edited Nov 16 at 23:15
Asaf Karagila♦
300k32421751
edited Nov 16 at 23:15
Asaf Karagila♦
300k32421751
300k32421751
asked Nov 16 at 22:12
mathrookie
722512
asked Nov 16 at 22:12
mathrookie
722512
asked Nov 16 at 22:12
mathrookie
722512
722512
1
My guess (not confident enough to make it an answer) would be: $x_* omega = { S mid x^{-1}(S) in omega }$ is an ultrafilter on $mathbb{C}$, and we're requiring that ultrafilter not to have limit $x$. So, that would mean that the neighborhood filter $mathcal{N}_x notsubseteq x_* omega$, i.e. there is some $epsilon > 0$ such that ${ n in mathbb{N} mid |x_n - x| < epsilon } notin omega$.
– Daniel Schepler
Nov 16 at 22:26
1
Since $omega$ is often synonymous to $Bbb N$, I would to nominate this to the prize of "worst notation ever". :-)
– Asaf Karagila♦
Nov 16 at 23:16
Yes, write $mathcal{F}$ for the filter, and use $lim_{mathcal{F}} x_n$ instead, as is more usual.
– Henno Brandsma
Nov 17 at 5:45
add a comment |
1
My guess (not confident enough to make it an answer) would be: $x_* omega = { S mid x^{-1}(S) in omega }$ is an ultrafilter on $mathbb{C}$, and we're requiring that ultrafilter not to have limit $x$. So, that would mean that the neighborhood filter $mathcal{N}_x notsubseteq x_* omega$, i.e. there is some $epsilon > 0$ such that ${ n in mathbb{N} mid |x_n - x| < epsilon } notin omega$.
– Daniel Schepler
Nov 16 at 22:26
1
Since $omega$ is often synonymous to $Bbb N$, I would to nominate this to the prize of "worst notation ever". :-)
– Asaf Karagila♦
Nov 16 at 23:16
Yes, write $mathcal{F}$ for the filter, and use $lim_{mathcal{F}} x_n$ instead, as is more usual.
– Henno Brandsma
Nov 17 at 5:45
1
1
My guess (not confident enough to make it an answer) would be: $x_* omega = { S mid x^{-1}(S) in omega }$ is an ultrafilter on $mathbb{C}$, and we're requiring that ultrafilter not to have limit $x$. So, that would mean that the neighborhood filter $mathcal{N}_x notsubseteq x_* omega$, i.e. there is some $epsilon > 0$ such that ${ n in mathbb{N} mid |x_n - x| < epsilon } notin omega$.
– Daniel Schepler
Nov 16 at 22:26
My guess (not confident enough to make it an answer) would be: $x_* omega = { S mid x^{-1}(S) in omega }$ is an ultrafilter on $mathbb{C}$, and we're requiring that ultrafilter not to have limit $x$. So, that would mean that the neighborhood filter $mathcal{N}_x notsubseteq x_* omega$, i.e. there is some $epsilon > 0$ such that ${ n in mathbb{N} mid |x_n - x| < epsilon } notin omega$.
– Daniel Schepler
Nov 16 at 22:26
1
1
Since $omega$ is often synonymous to $Bbb N$, I would to nominate this to the prize of "worst notation ever". :-)
– Asaf Karagila♦
Nov 16 at 23:16
Since $omega$ is often synonymous to $Bbb N$, I would to nominate this to the prize of "worst notation ever". :-)
– Asaf Karagila♦
Nov 16 at 23:16
Yes, write $mathcal{F}$ for the filter, and use $lim_{mathcal{F}} x_n$ instead, as is more usual.
– Henno Brandsma
Nov 17 at 5:45
Yes, write $mathcal{F}$ for the filter, and use $lim_{mathcal{F}} x_n$ instead, as is more usual.
– Henno Brandsma
Nov 17 at 5:45
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
First know what $x =lim_omega(x_n)$ means, then look at the negation:
$$x = lim_omega(x_n) text{ iff } forall O text{ open with } x in O: {n: x_n in O} in omega$$
So the negation of that is that there exists some open neighbourhood $O_x$ of $x$ such that ${n: x_n in O_x} notin omega$, but as $omega$ is an ultrafilter (so has the property $A notin omega leftrightarrow omegasetminus A in omega$ for all subset $A$ of $omega$) this is equivalent to the fact that ${n: x_n in Xsetminus O_x} in omega$.
So e.g. if $X$ were compact and we'd assume no $x$ is a limit, we'd have an open cover of $X$ of such $O_x$, hence a finite cover of these sets: $O_{x_1},ldots,O_{x_N}$, and then we'd have contradiction as
$$cap_{i=1}^N {n: x_n in Xsetminus O_{x_i}} in omega$$ while the left hand side is empty, as the $O_{x_i}$ form a cover.
answered Nov 16 at 22:35
Henno Brandsma
102k345109
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First know what $x =lim_omega(x_n)$ means, then look at the negation:
$$x = lim_omega(x_n) text{ iff } forall O text{ open with } x in O: {n: x_n in O} in omega$$
So the negation of that is that there exists some open neighbourhood $O_x$ of $x$ such that ${n: x_n in O_x} notin omega$, but as $omega$ is an ultrafilter (so has the property $A notin omega leftrightarrow omegasetminus A in omega$ for all subset $A$ of $omega$) this is equivalent to the fact that ${n: x_n in Xsetminus O_x} in omega$.
So e.g. if $X$ were compact and we'd assume no $x$ is a limit, we'd have an open cover of $X$ of such $O_x$, hence a finite cover of these sets: $O_{x_1},ldots,O_{x_N}$, and then we'd have contradiction as
$$cap_{i=1}^N {n: x_n in Xsetminus O_{x_i}} in omega$$ while the left hand side is empty, as the $O_{x_i}$ form a cover.
answered Nov 16 at 22:35
Henno Brandsma
102k345109
add a comment |
up vote
1
down vote
accepted
First know what $x =lim_omega(x_n)$ means, then look at the negation:
$$x = lim_omega(x_n) text{ iff } forall O text{ open with } x in O: {n: x_n in O} in omega$$
So the negation of that is that there exists some open neighbourhood $O_x$ of $x$ such that ${n: x_n in O_x} notin omega$, but as $omega$ is an ultrafilter (so has the property $A notin omega leftrightarrow omegasetminus A in omega$ for all subset $A$ of $omega$) this is equivalent to the fact that ${n: x_n in Xsetminus O_x} in omega$.
So e.g. if $X$ were compact and we'd assume no $x$ is a limit, we'd have an open cover of $X$ of such $O_x$, hence a finite cover of these sets: $O_{x_1},ldots,O_{x_N}$, and then we'd have contradiction as
$$cap_{i=1}^N {n: x_n in Xsetminus O_{x_i}} in omega$$ while the left hand side is empty, as the $O_{x_i}$ form a cover.
answered Nov 16 at 22:35
Henno Brandsma
102k345109
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First know what $x =lim_omega(x_n)$ means, then look at the negation:
$$x = lim_omega(x_n) text{ iff } forall O text{ open with } x in O: {n: x_n in O} in omega$$
So the negation of that is that there exists some open neighbourhood $O_x$ of $x$ such that ${n: x_n in O_x} notin omega$, but as $omega$ is an ultrafilter (so has the property $A notin omega leftrightarrow omegasetminus A in omega$ for all subset $A$ of $omega$) this is equivalent to the fact that ${n: x_n in Xsetminus O_x} in omega$.
So e.g. if $X$ were compact and we'd assume no $x$ is a limit, we'd have an open cover of $X$ of such $O_x$, hence a finite cover of these sets: $O_{x_1},ldots,O_{x_N}$, and then we'd have contradiction as
$$cap_{i=1}^N {n: x_n in Xsetminus O_{x_i}} in omega$$ while the left hand side is empty, as the $O_{x_i}$ form a cover.
answered Nov 16 at 22:35
Henno Brandsma
102k345109
First know what $x =lim_omega(x_n)$ means, then look at the negation:
$$x = lim_omega(x_n) text{ iff } forall O text{ open with } x in O: {n: x_n in O} in omega$$
So the negation of that is that there exists some open neighbourhood $O_x$ of $x$ such that ${n: x_n in O_x} notin omega$, but as $omega$ is an ultrafilter (so has the property $A notin omega leftrightarrow omegasetminus A in omega$ for all subset $A$ of $omega$) this is equivalent to the fact that ${n: x_n in Xsetminus O_x} in omega$.
So e.g. if $X$ were compact and we'd assume no $x$ is a limit, we'd have an open cover of $X$ of such $O_x$, hence a finite cover of these sets: $O_{x_1},ldots,O_{x_N}$, and then we'd have contradiction as
$$cap_{i=1}^N {n: x_n in Xsetminus O_{x_i}} in omega$$ while the left hand side is empty, as the $O_{x_i}$ form a cover.
answered Nov 16 at 22:35
Henno Brandsma
102k345109
answered Nov 16 at 22:35
Henno Brandsma
102k345109
answered Nov 16 at 22:35
Henno Brandsma
102k345109
answered Nov 16 at 22:35
Henno Brandsma
102k345109
102k345109
add a comment |
add a comment |
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My guess (not confident enough to make it an answer) would be: $x_* omega = { S mid x^{-1}(S) in omega }$ is an ultrafilter on $mathbb{C}$, and we're requiring that ultrafilter not to have limit $x$. So, that would mean that the neighborhood filter $mathcal{N}_x notsubseteq x_* omega$, i.e. there is some $epsilon > 0$ such that ${ n in mathbb{N} mid |x_n - x| < epsilon } notin omega$.
– Daniel Schepler
Nov 16 at 22:26
1
Since $omega$ is often synonymous to $Bbb N$, I would to nominate this to the prize of "worst notation ever". :-)
– Asaf Karagila♦
Nov 16 at 23:16
Yes, write $mathcal{F}$ for the filter, and use $lim_{mathcal{F}} x_n$ instead, as is more usual.
– Henno Brandsma
Nov 17 at 5:45