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







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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















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







share|cite|improve this question


















  • 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













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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









Jack J.

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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














  • 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















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