Mixing
Countable basis of topology
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I know that the family $$tilde{I}=bigg{(a,b);|;-infty<a<b<+inftybigg}$$ is a basis of a single topology on $mathbb{R}$, that is $mathcal{T}(mathbb{R})$ and the family
$$overline{I}:={(a,b);|;-infty<a<b<+infty}cup{[-infty,b);|;-infty<b<+infty}cup{(a,+infty];|;-infty<a<+infty}$$
is a basis of a single topology on $mathbb{overline{R}=mathbb{R}cup{{pminfty}}}$ that is $mathcal{T}(mathbb{overline{R}})$.
Question
How can I claim that these bases are countable?
If I consider the family $$tilde{I_1}=bigg{(a,b);|;a,binmathbb{Q};,-infty<a<b<+inftybigg},$$
then $tilde{I}supseteq tilde{I_1}$, does the vice versa also holds?
In the same way if I consider the family $$overline{I_1}:={(a,b);|;a,binmathbb{Q};,-infty<a<b<+infty}cup{[-infty,b);|;-infty<b<+infty}cup{(a,+infty];|;-infty<a<+infty}$$ then,
$overline{I}supseteq overline{I_1}$, does the vice versa also holds?
Thanks!
general-topology
asked yesterday
Jack J.
4301317
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0
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I know that the family $$tilde{I}=bigg{(a,b);|;-infty<a<b<+inftybigg}$$ is a basis of a single topology on $mathbb{R}$, that is $mathcal{T}(mathbb{R})$ and the family
$$overline{I}:={(a,b);|;-infty<a<b<+infty}cup{[-infty,b);|;-infty<b<+infty}cup{(a,+infty];|;-infty<a<+infty}$$
is a basis of a single topology on $mathbb{overline{R}=mathbb{R}cup{{pminfty}}}$ that is $mathcal{T}(mathbb{overline{R}})$.
Question
How can I claim that these bases are countable?
If I consider the family $$tilde{I_1}=bigg{(a,b);|;a,binmathbb{Q};,-infty<a<b<+inftybigg},$$
then $tilde{I}supseteq tilde{I_1}$, does the vice versa also holds?
In the same way if I consider the family $$overline{I_1}:={(a,b);|;a,binmathbb{Q};,-infty<a<b<+infty}cup{[-infty,b);|;-infty<b<+infty}cup{(a,+infty];|;-infty<a<+infty}$$ then,
$overline{I}supseteq overline{I_1}$, does the vice versa also holds?
Thanks!
general-topology
asked yesterday
Jack J.
4301317
1
Your first two basis are not countable. You don't want $tilde I=tilde I_1$, you want that every interval in $tilde I$ is included in an interval in $tilde I_1$, and vice versa.
– Nicolas FRANCOIS
yesterday
@NicolasFRANCOISI read on my book that $mathcal{T}(mathbb{R})$ and $mathcal{T}(mathbb{overline{R}})$ are a a countable basis, why?
– Jack J.
yesterday
If $a$ and $b$ in your initial definitions can take any value in $mathbb R$, then your sets are uncountable. That doesn't mean that the topology of $mathbb R$ is uncountable, as you show by restricting yourself to bounds in $mathbb Q$. The basis are uncountable, not the topology.
– Nicolas FRANCOIS
yesterday
2
The correct statement is that there exists a countable basis for the standard topology on $mathbb R$. That topology has many different bases, some of which are countable, and some of which are uncountable.
– Lee Mosher
yesterday
2
@JackJ. Because we can approximate all reals by rationals.
– Henno Brandsma
yesterday
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that the family $$tilde{I}=bigg{(a,b);|;-infty<a<b<+inftybigg}$$ is a basis of a single topology on $mathbb{R}$, that is $mathcal{T}(mathbb{R})$ and the family
$$overline{I}:={(a,b);|;-infty<a<b<+infty}cup{[-infty,b);|;-infty<b<+infty}cup{(a,+infty];|;-infty<a<+infty}$$
is a basis of a single topology on $mathbb{overline{R}=mathbb{R}cup{{pminfty}}}$ that is $mathcal{T}(mathbb{overline{R}})$.
Question
How can I claim that these bases are countable?
If I consider the family $$tilde{I_1}=bigg{(a,b);|;a,binmathbb{Q};,-infty<a<b<+inftybigg},$$
then $tilde{I}supseteq tilde{I_1}$, does the vice versa also holds?
In the same way if I consider the family $$overline{I_1}:={(a,b);|;a,binmathbb{Q};,-infty<a<b<+infty}cup{[-infty,b);|;-infty<b<+infty}cup{(a,+infty];|;-infty<a<+infty}$$ then,
$overline{I}supseteq overline{I_1}$, does the vice versa also holds?
Thanks!
general-topology
asked yesterday
Jack J.
4301317
I know that the family $$tilde{I}=bigg{(a,b);|;-infty<a<b<+inftybigg}$$ is a basis of a single topology on $mathbb{R}$, that is $mathcal{T}(mathbb{R})$ and the family
$$overline{I}:={(a,b);|;-infty<a<b<+infty}cup{[-infty,b);|;-infty<b<+infty}cup{(a,+infty];|;-infty<a<+infty}$$
is a basis of a single topology on $mathbb{overline{R}=mathbb{R}cup{{pminfty}}}$ that is $mathcal{T}(mathbb{overline{R}})$.
Question
How can I claim that these bases are countable?
If I consider the family $$tilde{I_1}=bigg{(a,b);|;a,binmathbb{Q};,-infty<a<b<+inftybigg},$$
then $tilde{I}supseteq tilde{I_1}$, does the vice versa also holds?
In the same way if I consider the family $$overline{I_1}:={(a,b);|;a,binmathbb{Q};,-infty<a<b<+infty}cup{[-infty,b);|;-infty<b<+infty}cup{(a,+infty];|;-infty<a<+infty}$$ then,
$overline{I}supseteq overline{I_1}$, does the vice versa also holds?
Thanks!
general-topology
general-topology
asked yesterday
Jack J.
4301317
asked yesterday
Jack J.
4301317
asked yesterday
Jack J.
4301317
asked yesterday
Jack J.
4301317
asked yesterday
Jack J.
4301317
4301317
1
Your first two basis are not countable. You don't want $tilde I=tilde I_1$, you want that every interval in $tilde I$ is included in an interval in $tilde I_1$, and vice versa.
– Nicolas FRANCOIS
yesterday
@NicolasFRANCOISI read on my book that $mathcal{T}(mathbb{R})$ and $mathcal{T}(mathbb{overline{R}})$ are a a countable basis, why?
– Jack J.
yesterday
If $a$ and $b$ in your initial definitions can take any value in $mathbb R$, then your sets are uncountable. That doesn't mean that the topology of $mathbb R$ is uncountable, as you show by restricting yourself to bounds in $mathbb Q$. The basis are uncountable, not the topology.
– Nicolas FRANCOIS
yesterday
2
The correct statement is that there exists a countable basis for the standard topology on $mathbb R$. That topology has many different bases, some of which are countable, and some of which are uncountable.
– Lee Mosher
yesterday
2
@JackJ. Because we can approximate all reals by rationals.
– Henno Brandsma
yesterday
|
show 1 more comment
1
Your first two basis are not countable. You don't want $tilde I=tilde I_1$, you want that every interval in $tilde I$ is included in an interval in $tilde I_1$, and vice versa.
– Nicolas FRANCOIS
yesterday
@NicolasFRANCOISI read on my book that $mathcal{T}(mathbb{R})$ and $mathcal{T}(mathbb{overline{R}})$ are a a countable basis, why?
– Jack J.
yesterday
If $a$ and $b$ in your initial definitions can take any value in $mathbb R$, then your sets are uncountable. That doesn't mean that the topology of $mathbb R$ is uncountable, as you show by restricting yourself to bounds in $mathbb Q$. The basis are uncountable, not the topology.
– Nicolas FRANCOIS
yesterday
2
The correct statement is that there exists a countable basis for the standard topology on $mathbb R$. That topology has many different bases, some of which are countable, and some of which are uncountable.
– Lee Mosher
yesterday
2
@JackJ. Because we can approximate all reals by rationals.
– Henno Brandsma
yesterday
1
1
Your first two basis are not countable. You don't want $tilde I=tilde I_1$, you want that every interval in $tilde I$ is included in an interval in $tilde I_1$, and vice versa.
– Nicolas FRANCOIS
yesterday
Your first two basis are not countable. You don't want $tilde I=tilde I_1$, you want that every interval in $tilde I$ is included in an interval in $tilde I_1$, and vice versa.
– Nicolas FRANCOIS
yesterday
@NicolasFRANCOISI read on my book that $mathcal{T}(mathbb{R})$ and $mathcal{T}(mathbb{overline{R}})$ are a a countable basis, why?
– Jack J.
yesterday
@NicolasFRANCOISI read on my book that $mathcal{T}(mathbb{R})$ and $mathcal{T}(mathbb{overline{R}})$ are a a countable basis, why?
– Jack J.
yesterday
If $a$ and $b$ in your initial definitions can take any value in $mathbb R$, then your sets are uncountable. That doesn't mean that the topology of $mathbb R$ is uncountable, as you show by restricting yourself to bounds in $mathbb Q$. The basis are uncountable, not the topology.
– Nicolas FRANCOIS
yesterday
If $a$ and $b$ in your initial definitions can take any value in $mathbb R$, then your sets are uncountable. That doesn't mean that the topology of $mathbb R$ is uncountable, as you show by restricting yourself to bounds in $mathbb Q$. The basis are uncountable, not the topology.
– Nicolas FRANCOIS
yesterday
2
2
The correct statement is that there exists a countable basis for the standard topology on $mathbb R$. That topology has many different bases, some of which are countable, and some of which are uncountable.
– Lee Mosher
yesterday
The correct statement is that there exists a countable basis for the standard topology on $mathbb R$. That topology has many different bases, some of which are countable, and some of which are uncountable.
– Lee Mosher
yesterday
2
2
@JackJ. Because we can approximate all reals by rationals.
– Henno Brandsma
yesterday
@JackJ. Because we can approximate all reals by rationals.
– Henno Brandsma
yesterday
|
show 1 more comment
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1
Your first two basis are not countable. You don't want $tilde I=tilde I_1$, you want that every interval in $tilde I$ is included in an interval in $tilde I_1$, and vice versa.
– Nicolas FRANCOIS
yesterday
@NicolasFRANCOISI read on my book that $mathcal{T}(mathbb{R})$ and $mathcal{T}(mathbb{overline{R}})$ are a a countable basis, why?
– Jack J.
yesterday
If $a$ and $b$ in your initial definitions can take any value in $mathbb R$, then your sets are uncountable. That doesn't mean that the topology of $mathbb R$ is uncountable, as you show by restricting yourself to bounds in $mathbb Q$. The basis are uncountable, not the topology.
– Nicolas FRANCOIS
yesterday
2
The correct statement is that there exists a countable basis for the standard topology on $mathbb R$. That topology has many different bases, some of which are countable, and some of which are uncountable.
– Lee Mosher
yesterday
2
@JackJ. Because we can approximate all reals by rationals.
– Henno Brandsma
yesterday