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Question about the definition of a topological space
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This might be a stupid question, but since a topology on a set $X$ is a collection of subsets of $X$, $tau$, where for each element, $U in tau$, $U$ is an open set, and since there is also the requirement $Xin tau$, how does this work where $X$ is a closed set like $[0, 1]$?
Also, since the complement of an open set is a closed set, does $X^c neq emptyset$?
general-topology definition
edited Feb 3 at 2:50
J. W. Tanner
4,7721420
asked Feb 2 at 19:01
What whatWhat what
1033
$endgroup$
add a comment |
$begingroup$
This might be a stupid question, but since a topology on a set $X$ is a collection of subsets of $X$, $tau$, where for each element, $U in tau$, $U$ is an open set, and since there is also the requirement $Xin tau$, how does this work where $X$ is a closed set like $[0, 1]$?
Also, since the complement of an open set is a closed set, does $X^c neq emptyset$?
general-topology definition
edited Feb 3 at 2:50
J. W. Tanner
4,7721420
asked Feb 2 at 19:01
What whatWhat what
1033
$endgroup$
$begingroup$
It is common to speak of a space $X$ but a space is actually the pair $(X,tau)$. If $(Y,tau')$ is a space and $Xsubset Y$ then we can have a space $(X,tau)$ where $tau$ may or may not have anything to do with $tau'....$ Also, if $Xsubset Y$ and $tau ={Xcap T : Tin tau'},$ it is common to say "$X$ is a subspace of $Y$", although we really should say $(X,tau)$ is a subspace of $(Y,tau').$
$endgroup$
– DanielWainfleet
Feb 3 at 3:23
add a comment |
$begingroup$
This might be a stupid question, but since a topology on a set $X$ is a collection of subsets of $X$, $tau$, where for each element, $U in tau$, $U$ is an open set, and since there is also the requirement $Xin tau$, how does this work where $X$ is a closed set like $[0, 1]$?
Also, since the complement of an open set is a closed set, does $X^c neq emptyset$?
general-topology definition
edited Feb 3 at 2:50
J. W. Tanner
4,7721420
asked Feb 2 at 19:01
What whatWhat what
1033
$endgroup$
This might be a stupid question, but since a topology on a set $X$ is a collection of subsets of $X$, $tau$, where for each element, $U in tau$, $U$ is an open set, and since there is also the requirement $Xin tau$, how does this work where $X$ is a closed set like $[0, 1]$?
Also, since the complement of an open set is a closed set, does $X^c neq emptyset$?
general-topology definition
general-topology definition
edited Feb 3 at 2:50
J. W. Tanner
4,7721420
asked Feb 2 at 19:01
What whatWhat what
1033
edited Feb 3 at 2:50
J. W. Tanner
4,7721420
asked Feb 2 at 19:01
What whatWhat what
1033
edited Feb 3 at 2:50
J. W. Tanner
4,7721420
edited Feb 3 at 2:50
J. W. Tanner
4,7721420
edited Feb 3 at 2:50
J. W. Tanner
4,7721420
4,7721420
asked Feb 2 at 19:01
What whatWhat what
1033
asked Feb 2 at 19:01
What whatWhat what
1033
asked Feb 2 at 19:01
What whatWhat what
1033
1033
$begingroup$
It is common to speak of a space $X$ but a space is actually the pair $(X,tau)$. If $(Y,tau')$ is a space and $Xsubset Y$ then we can have a space $(X,tau)$ where $tau$ may or may not have anything to do with $tau'....$ Also, if $Xsubset Y$ and $tau ={Xcap T : Tin tau'},$ it is common to say "$X$ is a subspace of $Y$", although we really should say $(X,tau)$ is a subspace of $(Y,tau').$
$endgroup$
– DanielWainfleet
Feb 3 at 3:23
add a comment |
$begingroup$
It is common to speak of a space $X$ but a space is actually the pair $(X,tau)$. If $(Y,tau')$ is a space and $Xsubset Y$ then we can have a space $(X,tau)$ where $tau$ may or may not have anything to do with $tau'....$ Also, if $Xsubset Y$ and $tau ={Xcap T : Tin tau'},$ it is common to say "$X$ is a subspace of $Y$", although we really should say $(X,tau)$ is a subspace of $(Y,tau').$
$endgroup$
– DanielWainfleet
Feb 3 at 3:23
$begingroup$
It is common to speak of a space $X$ but a space is actually the pair $(X,tau)$. If $(Y,tau')$ is a space and $Xsubset Y$ then we can have a space $(X,tau)$ where $tau$ may or may not have anything to do with $tau'....$ Also, if $Xsubset Y$ and $tau ={Xcap T : Tin tau'},$ it is common to say "$X$ is a subspace of $Y$", although we really should say $(X,tau)$ is a subspace of $(Y,tau').$
$endgroup$
– DanielWainfleet
Feb 3 at 3:23
$begingroup$
It is common to speak of a space $X$ but a space is actually the pair $(X,tau)$. If $(Y,tau')$ is a space and $Xsubset Y$ then we can have a space $(X,tau)$ where $tau$ may or may not have anything to do with $tau'....$ Also, if $Xsubset Y$ and $tau ={Xcap T : Tin tau'},$ it is common to say "$X$ is a subspace of $Y$", although we really should say $(X,tau)$ is a subspace of $(Y,tau').$
$endgroup$
– DanielWainfleet
Feb 3 at 3:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are wrong about the definition of topological space. Such a space is a pair $(X,tau)$, where $X$ is a set and $tau$ is a subset of $mathcal{P}(X)$, for which certain conditions must hold ($emptyset,Xintau$ and $X$ is stable with respect to arbitrary unions and to finite intersections). Only after that we define open sets. It's just this: the open sets of a topological space $(X,tau)$ are the elements of $tau$. Besides, the closed sets are those $Ysubset C$ such that $Y^complementintau$.
So, yes, we can perfectly define topologies $tau$ on $mathbb R$ for which $[0,1]$ is an open set. And for such a space, $[0,1]$ may be or not a closed set.
By the way, keep in mind that “open” is not the opposite of “closed”: a subset of a topological space may be open and closed at the same time (the whole space and the empty set, for instance) and may also be neither open nor closed.
answered Feb 2 at 19:07
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Nice suggestion. I've edited my answer.
$endgroup$
– José Carlos Santos
Feb 2 at 19:15
$begingroup$
Just to be clear, we're defining closed sets as sets whose compliments are elements of $tau$?
$endgroup$
– What what
Feb 2 at 19:41
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Feb 2 at 19:42
$begingroup$
Thanks, that helps a lot.
$endgroup$
– What what
Feb 2 at 19:43
add a comment |
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1 Answer
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$begingroup$
You are wrong about the definition of topological space. Such a space is a pair $(X,tau)$, where $X$ is a set and $tau$ is a subset of $mathcal{P}(X)$, for which certain conditions must hold ($emptyset,Xintau$ and $X$ is stable with respect to arbitrary unions and to finite intersections). Only after that we define open sets. It's just this: the open sets of a topological space $(X,tau)$ are the elements of $tau$. Besides, the closed sets are those $Ysubset C$ such that $Y^complementintau$.
So, yes, we can perfectly define topologies $tau$ on $mathbb R$ for which $[0,1]$ is an open set. And for such a space, $[0,1]$ may be or not a closed set.
By the way, keep in mind that “open” is not the opposite of “closed”: a subset of a topological space may be open and closed at the same time (the whole space and the empty set, for instance) and may also be neither open nor closed.
answered Feb 2 at 19:07
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Nice suggestion. I've edited my answer.
$endgroup$
– José Carlos Santos
Feb 2 at 19:15
$begingroup$
Just to be clear, we're defining closed sets as sets whose compliments are elements of $tau$?
$endgroup$
– What what
Feb 2 at 19:41
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Feb 2 at 19:42
$begingroup$
Thanks, that helps a lot.
$endgroup$
– What what
Feb 2 at 19:43
add a comment |
$begingroup$
You are wrong about the definition of topological space. Such a space is a pair $(X,tau)$, where $X$ is a set and $tau$ is a subset of $mathcal{P}(X)$, for which certain conditions must hold ($emptyset,Xintau$ and $X$ is stable with respect to arbitrary unions and to finite intersections). Only after that we define open sets. It's just this: the open sets of a topological space $(X,tau)$ are the elements of $tau$. Besides, the closed sets are those $Ysubset C$ such that $Y^complementintau$.
So, yes, we can perfectly define topologies $tau$ on $mathbb R$ for which $[0,1]$ is an open set. And for such a space, $[0,1]$ may be or not a closed set.
By the way, keep in mind that “open” is not the opposite of “closed”: a subset of a topological space may be open and closed at the same time (the whole space and the empty set, for instance) and may also be neither open nor closed.
answered Feb 2 at 19:07
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
$begingroup$
Nice suggestion. I've edited my answer.
$endgroup$
– José Carlos Santos
Feb 2 at 19:15
$begingroup$
Just to be clear, we're defining closed sets as sets whose compliments are elements of $tau$?
$endgroup$
– What what
Feb 2 at 19:41
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Feb 2 at 19:42
$begingroup$
Thanks, that helps a lot.
$endgroup$
– What what
Feb 2 at 19:43
add a comment |
$begingroup$
You are wrong about the definition of topological space. Such a space is a pair $(X,tau)$, where $X$ is a set and $tau$ is a subset of $mathcal{P}(X)$, for which certain conditions must hold ($emptyset,Xintau$ and $X$ is stable with respect to arbitrary unions and to finite intersections). Only after that we define open sets. It's just this: the open sets of a topological space $(X,tau)$ are the elements of $tau$. Besides, the closed sets are those $Ysubset C$ such that $Y^complementintau$.
So, yes, we can perfectly define topologies $tau$ on $mathbb R$ for which $[0,1]$ is an open set. And for such a space, $[0,1]$ may be or not a closed set.
By the way, keep in mind that “open” is not the opposite of “closed”: a subset of a topological space may be open and closed at the same time (the whole space and the empty set, for instance) and may also be neither open nor closed.
answered Feb 2 at 19:07
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
You are wrong about the definition of topological space. Such a space is a pair $(X,tau)$, where $X$ is a set and $tau$ is a subset of $mathcal{P}(X)$, for which certain conditions must hold ($emptyset,Xintau$ and $X$ is stable with respect to arbitrary unions and to finite intersections). Only after that we define open sets. It's just this: the open sets of a topological space $(X,tau)$ are the elements of $tau$. Besides, the closed sets are those $Ysubset C$ such that $Y^complementintau$.
So, yes, we can perfectly define topologies $tau$ on $mathbb R$ for which $[0,1]$ is an open set. And for such a space, $[0,1]$ may be or not a closed set.
By the way, keep in mind that “open” is not the opposite of “closed”: a subset of a topological space may be open and closed at the same time (the whole space and the empty set, for instance) and may also be neither open nor closed.
answered Feb 2 at 19:07
José Carlos SantosJosé Carlos Santos
174k23134243
edited Feb 2 at 19:15
answered Feb 2 at 19:07
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 19:07
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 19:07
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
$begingroup$
Nice suggestion. I've edited my answer.
$endgroup$
– José Carlos Santos
Feb 2 at 19:15
$begingroup$
Just to be clear, we're defining closed sets as sets whose compliments are elements of $tau$?
$endgroup$
– What what
Feb 2 at 19:41
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Feb 2 at 19:42
$begingroup$
Thanks, that helps a lot.
$endgroup$
– What what
Feb 2 at 19:43
add a comment |
$begingroup$
Nice suggestion. I've edited my answer.
$endgroup$
– José Carlos Santos
Feb 2 at 19:15
$begingroup$
Just to be clear, we're defining closed sets as sets whose compliments are elements of $tau$?
$endgroup$
– What what
Feb 2 at 19:41
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Feb 2 at 19:42
$begingroup$
Thanks, that helps a lot.
$endgroup$
– What what
Feb 2 at 19:43
$begingroup$
Nice suggestion. I've edited my answer.
$endgroup$
– José Carlos Santos
Feb 2 at 19:15
$begingroup$
Nice suggestion. I've edited my answer.
$endgroup$
– José Carlos Santos
Feb 2 at 19:15
$begingroup$
Just to be clear, we're defining closed sets as sets whose compliments are elements of $tau$?
$endgroup$
– What what
Feb 2 at 19:41
$begingroup$
Just to be clear, we're defining closed sets as sets whose compliments are elements of $tau$?
$endgroup$
– What what
Feb 2 at 19:41
1
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Feb 2 at 19:42
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Feb 2 at 19:42
$begingroup$
Thanks, that helps a lot.
$endgroup$
– What what
Feb 2 at 19:43
$begingroup$
Thanks, that helps a lot.
$endgroup$
– What what
Feb 2 at 19:43
add a comment |
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$begingroup$
It is common to speak of a space $X$ but a space is actually the pair $(X,tau)$. If $(Y,tau')$ is a space and $Xsubset Y$ then we can have a space $(X,tau)$ where $tau$ may or may not have anything to do with $tau'....$ Also, if $Xsubset Y$ and $tau ={Xcap T : Tin tau'},$ it is common to say "$X$ is a subspace of $Y$", although we really should say $(X,tau)$ is a subspace of $(Y,tau').$
$endgroup$
– DanielWainfleet
Feb 3 at 3:23