Mixing
Why do we need singletons to be closed in the definition of Normal/Regular spaces?
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In the book of Topology by Munkres, at page 193, it is given that
Definition. Suppose that one-point sets are closed in X. Then X is said to be regular if for each pair consisting of a point x and a
closed set B disjoint from x, there exist disjoint open sets
containing x and B, respectively. The space X is said to be normal if
for each pair A, B of disjoint closed sets of X, there exist disjoint
open sets containing A and B, respectively.
It is clear that a regular space is Hausdorff, and that a normal space
is regular. (We need to include the condition that one-point sets be
closed as part of the definition of regularity and normality in order
for this to be the case. A two-point space in the indiscrete topology
satisfies the other part of the definitions of regularity and
normality, even though it is not Hausdorff.)
But, how does assuming one-point sets are closed allow regular (normal) space to be Hausdorff(regular) ?
general-topology
asked Jan 3 at 6:48
onurcanbektasonurcanbektas
3,36611036
$endgroup$
add a comment |
$begingroup$
In the book of Topology by Munkres, at page 193, it is given that
Definition. Suppose that one-point sets are closed in X. Then X is said to be regular if for each pair consisting of a point x and a
closed set B disjoint from x, there exist disjoint open sets
containing x and B, respectively. The space X is said to be normal if
for each pair A, B of disjoint closed sets of X, there exist disjoint
open sets containing A and B, respectively.
It is clear that a regular space is Hausdorff, and that a normal space
is regular. (We need to include the condition that one-point sets be
closed as part of the definition of regularity and normality in order
for this to be the case. A two-point space in the indiscrete topology
satisfies the other part of the definitions of regularity and
normality, even though it is not Hausdorff.)
But, how does assuming one-point sets are closed allow regular (normal) space to be Hausdorff(regular) ?
general-topology
asked Jan 3 at 6:48
onurcanbektasonurcanbektas
3,36611036
$endgroup$
$begingroup$
Because you can take B={y}, a different point, in your definition of regular.
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– Gaffney
Jan 3 at 6:51
1
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@Gaffney In case when $B= {y}$ is open ? I didn't full get it .
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– onurcanbektas
Jan 3 at 6:53
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If it's open then you can't conclude that the space is Hausdorff. You don't conclude anything.
$endgroup$
– Gaffney
Jan 3 at 7:04
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@Gaffney Yes, I know that, but I still don't understand what are you trying to say in your first comment.
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– onurcanbektas
Jan 3 at 7:05
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@Gaffney Why did you deleted your answer ?
$endgroup$
– onurcanbektas
Jan 3 at 7:20
add a comment |
$begingroup$
In the book of Topology by Munkres, at page 193, it is given that
Definition. Suppose that one-point sets are closed in X. Then X is said to be regular if for each pair consisting of a point x and a
closed set B disjoint from x, there exist disjoint open sets
containing x and B, respectively. The space X is said to be normal if
for each pair A, B of disjoint closed sets of X, there exist disjoint
open sets containing A and B, respectively.
It is clear that a regular space is Hausdorff, and that a normal space
is regular. (We need to include the condition that one-point sets be
closed as part of the definition of regularity and normality in order
for this to be the case. A two-point space in the indiscrete topology
satisfies the other part of the definitions of regularity and
normality, even though it is not Hausdorff.)
But, how does assuming one-point sets are closed allow regular (normal) space to be Hausdorff(regular) ?
general-topology
asked Jan 3 at 6:48
onurcanbektasonurcanbektas
3,36611036
$endgroup$
In the book of Topology by Munkres, at page 193, it is given that
Definition. Suppose that one-point sets are closed in X. Then X is said to be regular if for each pair consisting of a point x and a
closed set B disjoint from x, there exist disjoint open sets
containing x and B, respectively. The space X is said to be normal if
for each pair A, B of disjoint closed sets of X, there exist disjoint
open sets containing A and B, respectively.
It is clear that a regular space is Hausdorff, and that a normal space
is regular. (We need to include the condition that one-point sets be
closed as part of the definition of regularity and normality in order
for this to be the case. A two-point space in the indiscrete topology
satisfies the other part of the definitions of regularity and
normality, even though it is not Hausdorff.)
But, how does assuming one-point sets are closed allow regular (normal) space to be Hausdorff(regular) ?
general-topology
general-topology
asked Jan 3 at 6:48
onurcanbektasonurcanbektas
3,36611036
asked Jan 3 at 6:48
onurcanbektasonurcanbektas
3,36611036
edited Jan 3 at 6:52
onurcanbektas
asked Jan 3 at 6:48
onurcanbektasonurcanbektas
3,36611036
asked Jan 3 at 6:48
onurcanbektasonurcanbektas
3,36611036
asked Jan 3 at 6:48
onurcanbektasonurcanbektas
3,36611036
3,36611036
$begingroup$
Because you can take B={y}, a different point, in your definition of regular.
$endgroup$
– Gaffney
Jan 3 at 6:51
1
$begingroup$
@Gaffney In case when $B= {y}$ is open ? I didn't full get it .
$endgroup$
– onurcanbektas
Jan 3 at 6:53
$begingroup$
If it's open then you can't conclude that the space is Hausdorff. You don't conclude anything.
$endgroup$
– Gaffney
Jan 3 at 7:04
$begingroup$
@Gaffney Yes, I know that, but I still don't understand what are you trying to say in your first comment.
$endgroup$
– onurcanbektas
Jan 3 at 7:05
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@Gaffney Why did you deleted your answer ?
$endgroup$
– onurcanbektas
Jan 3 at 7:20
add a comment |
$begingroup$
Because you can take B={y}, a different point, in your definition of regular.
$endgroup$
– Gaffney
Jan 3 at 6:51
1
$begingroup$
@Gaffney In case when $B= {y}$ is open ? I didn't full get it .
$endgroup$
– onurcanbektas
Jan 3 at 6:53
$begingroup$
If it's open then you can't conclude that the space is Hausdorff. You don't conclude anything.
$endgroup$
– Gaffney
Jan 3 at 7:04
$begingroup$
@Gaffney Yes, I know that, but I still don't understand what are you trying to say in your first comment.
$endgroup$
– onurcanbektas
Jan 3 at 7:05
$begingroup$
@Gaffney Why did you deleted your answer ?
$endgroup$
– onurcanbektas
Jan 3 at 7:20
$begingroup$
Because you can take B={y}, a different point, in your definition of regular.
$endgroup$
– Gaffney
Jan 3 at 6:51
$begingroup$
Because you can take B={y}, a different point, in your definition of regular.
$endgroup$
– Gaffney
Jan 3 at 6:51
1
1
$begingroup$
@Gaffney In case when $B= {y}$ is open ? I didn't full get it .
$endgroup$
– onurcanbektas
Jan 3 at 6:53
$begingroup$
@Gaffney In case when $B= {y}$ is open ? I didn't full get it .
$endgroup$
– onurcanbektas
Jan 3 at 6:53
$begingroup$
If it's open then you can't conclude that the space is Hausdorff. You don't conclude anything.
$endgroup$
– Gaffney
Jan 3 at 7:04
$begingroup$
If it's open then you can't conclude that the space is Hausdorff. You don't conclude anything.
$endgroup$
– Gaffney
Jan 3 at 7:04
$begingroup$
@Gaffney Yes, I know that, but I still don't understand what are you trying to say in your first comment.
$endgroup$
– onurcanbektas
Jan 3 at 7:05
$begingroup$
@Gaffney Yes, I know that, but I still don't understand what are you trying to say in your first comment.
$endgroup$
– onurcanbektas
Jan 3 at 7:05
$begingroup$
@Gaffney Why did you deleted your answer ?
$endgroup$
– onurcanbektas
Jan 3 at 7:20
$begingroup$
@Gaffney Why did you deleted your answer ?
$endgroup$
– onurcanbektas
Jan 3 at 7:20
add a comment |
2 Answers
2
active
oldest
votes
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Further considerations:
(i). Consider Sierpinski space $S={x,y}$ with $xne y,$ where $S,emptyset,$ and ${x}$ are open but ${y}$ is not open. If $A, B$ are disjoint closed subsets of $S$ then at least one of $A, B$ is empty so $A,B$ are covered by disjoint open sets. But $S$ does not satisfy the condition for regularity: $x$ does not belong to the closed subset ${y}$ but the only open set covering ${y}$ is the whole space $S$.
(ii). Let $X$ be a normal space. Let $pin X$ and let $Y$ be a closed subset of $X$ with $pnot in Y.$ Since $X$ is a $T_1$ space (one-point subsets are closed), the sets ${p}, Y$ are closed and disjoint. So, since $X$ is normal, there are disjoint open sets with ${p}subset U$ and $Ysubset V. $ That is, $pin U$ and $Ysubset V.$ So $X$ is regular.
answered Jan 3 at 8:11
DanielWainfleetDanielWainfleet
34.6k31648
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Sierpinski space is useful for many examples and counter-examples.
$endgroup$
– DanielWainfleet
Jan 5 at 0:58
add a comment |
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If $x neq y$ then ${y}$ is a closed set adn $x notin {y}$ so there exist disjoint open sets $U,V$ such that $x in U$ and ${y} subset V$. This shows that the space is Hausdorff.
answered Jan 3 at 7:19
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Further considerations:
(i). Consider Sierpinski space $S={x,y}$ with $xne y,$ where $S,emptyset,$ and ${x}$ are open but ${y}$ is not open. If $A, B$ are disjoint closed subsets of $S$ then at least one of $A, B$ is empty so $A,B$ are covered by disjoint open sets. But $S$ does not satisfy the condition for regularity: $x$ does not belong to the closed subset ${y}$ but the only open set covering ${y}$ is the whole space $S$.
(ii). Let $X$ be a normal space. Let $pin X$ and let $Y$ be a closed subset of $X$ with $pnot in Y.$ Since $X$ is a $T_1$ space (one-point subsets are closed), the sets ${p}, Y$ are closed and disjoint. So, since $X$ is normal, there are disjoint open sets with ${p}subset U$ and $Ysubset V. $ That is, $pin U$ and $Ysubset V.$ So $X$ is regular.
answered Jan 3 at 8:11
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
$begingroup$
Sierpinski space is useful for many examples and counter-examples.
$endgroup$
– DanielWainfleet
Jan 5 at 0:58
add a comment |
$begingroup$
Further considerations:
(i). Consider Sierpinski space $S={x,y}$ with $xne y,$ where $S,emptyset,$ and ${x}$ are open but ${y}$ is not open. If $A, B$ are disjoint closed subsets of $S$ then at least one of $A, B$ is empty so $A,B$ are covered by disjoint open sets. But $S$ does not satisfy the condition for regularity: $x$ does not belong to the closed subset ${y}$ but the only open set covering ${y}$ is the whole space $S$.
(ii). Let $X$ be a normal space. Let $pin X$ and let $Y$ be a closed subset of $X$ with $pnot in Y.$ Since $X$ is a $T_1$ space (one-point subsets are closed), the sets ${p}, Y$ are closed and disjoint. So, since $X$ is normal, there are disjoint open sets with ${p}subset U$ and $Ysubset V. $ That is, $pin U$ and $Ysubset V.$ So $X$ is regular.
answered Jan 3 at 8:11
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
$begingroup$
Sierpinski space is useful for many examples and counter-examples.
$endgroup$
– DanielWainfleet
Jan 5 at 0:58
add a comment |
$begingroup$
Further considerations:
(i). Consider Sierpinski space $S={x,y}$ with $xne y,$ where $S,emptyset,$ and ${x}$ are open but ${y}$ is not open. If $A, B$ are disjoint closed subsets of $S$ then at least one of $A, B$ is empty so $A,B$ are covered by disjoint open sets. But $S$ does not satisfy the condition for regularity: $x$ does not belong to the closed subset ${y}$ but the only open set covering ${y}$ is the whole space $S$.
(ii). Let $X$ be a normal space. Let $pin X$ and let $Y$ be a closed subset of $X$ with $pnot in Y.$ Since $X$ is a $T_1$ space (one-point subsets are closed), the sets ${p}, Y$ are closed and disjoint. So, since $X$ is normal, there are disjoint open sets with ${p}subset U$ and $Ysubset V. $ That is, $pin U$ and $Ysubset V.$ So $X$ is regular.
answered Jan 3 at 8:11
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
Further considerations:
(i). Consider Sierpinski space $S={x,y}$ with $xne y,$ where $S,emptyset,$ and ${x}$ are open but ${y}$ is not open. If $A, B$ are disjoint closed subsets of $S$ then at least one of $A, B$ is empty so $A,B$ are covered by disjoint open sets. But $S$ does not satisfy the condition for regularity: $x$ does not belong to the closed subset ${y}$ but the only open set covering ${y}$ is the whole space $S$.
(ii). Let $X$ be a normal space. Let $pin X$ and let $Y$ be a closed subset of $X$ with $pnot in Y.$ Since $X$ is a $T_1$ space (one-point subsets are closed), the sets ${p}, Y$ are closed and disjoint. So, since $X$ is normal, there are disjoint open sets with ${p}subset U$ and $Ysubset V. $ That is, $pin U$ and $Ysubset V.$ So $X$ is regular.
answered Jan 3 at 8:11
DanielWainfleetDanielWainfleet
34.6k31648
edited Jan 3 at 8:25
answered Jan 3 at 8:11
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 8:11
DanielWainfleetDanielWainfleet
34.6k31648
answered Jan 3 at 8:11
DanielWainfleetDanielWainfleet
34.6k31648
34.6k31648
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Sierpinski space is useful for many examples and counter-examples.
$endgroup$
– DanielWainfleet
Jan 5 at 0:58
add a comment |
$begingroup$
Sierpinski space is useful for many examples and counter-examples.
$endgroup$
– DanielWainfleet
Jan 5 at 0:58
$begingroup$
Sierpinski space is useful for many examples and counter-examples.
$endgroup$
– DanielWainfleet
Jan 5 at 0:58
$begingroup$
Sierpinski space is useful for many examples and counter-examples.
$endgroup$
– DanielWainfleet
Jan 5 at 0:58
add a comment |
$begingroup$
If $x neq y$ then ${y}$ is a closed set adn $x notin {y}$ so there exist disjoint open sets $U,V$ such that $x in U$ and ${y} subset V$. This shows that the space is Hausdorff.
answered Jan 3 at 7:19
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
add a comment |
$begingroup$
If $x neq y$ then ${y}$ is a closed set adn $x notin {y}$ so there exist disjoint open sets $U,V$ such that $x in U$ and ${y} subset V$. This shows that the space is Hausdorff.
answered Jan 3 at 7:19
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
add a comment |
$begingroup$
If $x neq y$ then ${y}$ is a closed set adn $x notin {y}$ so there exist disjoint open sets $U,V$ such that $x in U$ and ${y} subset V$. This shows that the space is Hausdorff.
answered Jan 3 at 7:19
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
$endgroup$
If $x neq y$ then ${y}$ is a closed set adn $x notin {y}$ so there exist disjoint open sets $U,V$ such that $x in U$ and ${y} subset V$. This shows that the space is Hausdorff.
answered Jan 3 at 7:19
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Jan 3 at 7:19
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Jan 3 at 7:19
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
answered Jan 3 at 7:19
Kavi Rama MurthyKavi Rama Murthy
53.8k32055
53.8k32055
add a comment |
add a comment |
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$begingroup$
Because you can take B={y}, a different point, in your definition of regular.
$endgroup$
– Gaffney
Jan 3 at 6:51
1
$begingroup$
@Gaffney In case when $B= {y}$ is open ? I didn't full get it .
$endgroup$
– onurcanbektas
Jan 3 at 6:53
$begingroup$
If it's open then you can't conclude that the space is Hausdorff. You don't conclude anything.
$endgroup$
– Gaffney
Jan 3 at 7:04
$begingroup$
@Gaffney Yes, I know that, but I still don't understand what are you trying to say in your first comment.
$endgroup$
– onurcanbektas
Jan 3 at 7:05
$begingroup$
@Gaffney Why did you deleted your answer ?
$endgroup$
– onurcanbektas
Jan 3 at 7:20