Countability, separability and Lindelöf of $[-1,1]$ with another topology.
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I am trying to solve the following exercise.
Study the countability axioms, separability axioms and the Lindelöf property in $(X,tau)$, where $X=[-1,1]$ and $Uintau$ if and only if $0notin U$ or $(-1,1)subset U$.
Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.
If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).
Is this right or I am leaving something?
general-topology proof-verification
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up vote
0
down vote
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I am trying to solve the following exercise.
Study the countability axioms, separability axioms and the Lindelöf property in $(X,tau)$, where $X=[-1,1]$ and $Uintau$ if and only if $0notin U$ or $(-1,1)subset U$.
Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.
If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).
Is this right or I am leaving something?
general-topology proof-verification
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve the following exercise.
Study the countability axioms, separability axioms and the Lindelöf property in $(X,tau)$, where $X=[-1,1]$ and $Uintau$ if and only if $0notin U$ or $(-1,1)subset U$.
Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.
If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).
Is this right or I am leaving something?
general-topology proof-verification
I am trying to solve the following exercise.
Study the countability axioms, separability axioms and the Lindelöf property in $(X,tau)$, where $X=[-1,1]$ and $Uintau$ if and only if $0notin U$ or $(-1,1)subset U$.
Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.
If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).
Is this right or I am leaving something?
general-topology proof-verification
general-topology proof-verification
asked Sep 2 at 10:37
mkspk
546617
546617
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1 Answer
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A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.
So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.
If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.
So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.
If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).
add a comment |Â
up vote
2
down vote
accepted
A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.
So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.
If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.
So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.
If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).
A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.
So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.
If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).
edited Sep 2 at 13:23
answered Sep 2 at 12:44
Henno Brandsma
93.4k342101
93.4k342101
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