Is $T_1$ condition necessary in the definition of completely normality?
$begingroup$
Engelking states a theorem(2.1.7) in the book General Topology as follows:
For every $T_1$ spaces the following are equivalent:
(1) Every subspace of $X$ is normal.
(2) Every open subspace of $X$ is normal.
(3) Two separated sets have disjoint open neighborhoods.
I was asked to show that (1) and (3) are equivalent, however, I don’t know where the $T_1$ condition is used in the proof. My proof goes as follows:
If $X$ satisfies (3), then let $Y$ be a subspace and $A,Bsubset Y$ disjoint closed sets in the subspace $Y$. Denote the closure operator in $X$ by $Cl$ and in $Y$ by $Cl^*$. We claim that $A, B$ are separated in $X$: $$Cl(A)cap B= Cl(A)cap Cl^*(B) = Cl(A)cap Cl(B)cap Y = Cl^*(A)cap Cl^*(B)$$ But this is empty by the fact that $A, B$ are closed. Then we can take disjoint open neighborhood of $A,B$ in $X$, and intersect both by $Y$.
If $X$ satisfies (1), then take two separated sets $A, B$ in $X$. The open subspace $Y=X-(Cl(A)cap Cl(B))$ contains each of $A $ and $B$ by the fact that they are separated. The closure of them in $Y$ is empty: $Cl^*(A)cap Cl^*(B)= Cl(A)cap Cl(B)cap Y$ is empty because $Y$ does not contain $Cl(A)cap Cl(B)$. Take disjoint open neighborhood of $Cl^*(A)$ and $Cl^*(B)$ in $Y$, then by the fact that $Y$ is open, we are done with the proof.
So where exactly did I use $T_1$ condition in my proof?
general-topology
asked Dec 14 '18 at 1:51
William SunWilliam Sun
471211
$endgroup$
add a comment |
$begingroup$
Engelking states a theorem(2.1.7) in the book General Topology as follows:
For every $T_1$ spaces the following are equivalent:
(1) Every subspace of $X$ is normal.
(2) Every open subspace of $X$ is normal.
(3) Two separated sets have disjoint open neighborhoods.
I was asked to show that (1) and (3) are equivalent, however, I don’t know where the $T_1$ condition is used in the proof. My proof goes as follows:
If $X$ satisfies (3), then let $Y$ be a subspace and $A,Bsubset Y$ disjoint closed sets in the subspace $Y$. Denote the closure operator in $X$ by $Cl$ and in $Y$ by $Cl^*$. We claim that $A, B$ are separated in $X$: $$Cl(A)cap B= Cl(A)cap Cl^*(B) = Cl(A)cap Cl(B)cap Y = Cl^*(A)cap Cl^*(B)$$ But this is empty by the fact that $A, B$ are closed. Then we can take disjoint open neighborhood of $A,B$ in $X$, and intersect both by $Y$.
If $X$ satisfies (1), then take two separated sets $A, B$ in $X$. The open subspace $Y=X-(Cl(A)cap Cl(B))$ contains each of $A $ and $B$ by the fact that they are separated. The closure of them in $Y$ is empty: $Cl^*(A)cap Cl^*(B)= Cl(A)cap Cl(B)cap Y$ is empty because $Y$ does not contain $Cl(A)cap Cl(B)$. Take disjoint open neighborhood of $Cl^*(A)$ and $Cl^*(B)$ in $Y$, then by the fact that $Y$ is open, we are done with the proof.
So where exactly did I use $T_1$ condition in my proof?
general-topology
asked Dec 14 '18 at 1:51
William SunWilliam Sun
471211
$endgroup$
1
$begingroup$
Does author require normal spaces to be Hausdorff?
$endgroup$
– William Elliot
Dec 14 '18 at 3:31
$begingroup$
@WilliamElliot The author requires normal spaces to be both $T_1$ and $T_4$. I understand it now-I use this book just for a reference so I was not aware it has an alternate definition for normal space. Thank you for the comment.
$endgroup$
– William Sun
Dec 14 '18 at 3:35
add a comment |
$begingroup$
Engelking states a theorem(2.1.7) in the book General Topology as follows:
For every $T_1$ spaces the following are equivalent:
(1) Every subspace of $X$ is normal.
(2) Every open subspace of $X$ is normal.
(3) Two separated sets have disjoint open neighborhoods.
I was asked to show that (1) and (3) are equivalent, however, I don’t know where the $T_1$ condition is used in the proof. My proof goes as follows:
If $X$ satisfies (3), then let $Y$ be a subspace and $A,Bsubset Y$ disjoint closed sets in the subspace $Y$. Denote the closure operator in $X$ by $Cl$ and in $Y$ by $Cl^*$. We claim that $A, B$ are separated in $X$: $$Cl(A)cap B= Cl(A)cap Cl^*(B) = Cl(A)cap Cl(B)cap Y = Cl^*(A)cap Cl^*(B)$$ But this is empty by the fact that $A, B$ are closed. Then we can take disjoint open neighborhood of $A,B$ in $X$, and intersect both by $Y$.
If $X$ satisfies (1), then take two separated sets $A, B$ in $X$. The open subspace $Y=X-(Cl(A)cap Cl(B))$ contains each of $A $ and $B$ by the fact that they are separated. The closure of them in $Y$ is empty: $Cl^*(A)cap Cl^*(B)= Cl(A)cap Cl(B)cap Y$ is empty because $Y$ does not contain $Cl(A)cap Cl(B)$. Take disjoint open neighborhood of $Cl^*(A)$ and $Cl^*(B)$ in $Y$, then by the fact that $Y$ is open, we are done with the proof.
So where exactly did I use $T_1$ condition in my proof?
general-topology
asked Dec 14 '18 at 1:51
William SunWilliam Sun
471211
$endgroup$
Engelking states a theorem(2.1.7) in the book General Topology as follows:
For every $T_1$ spaces the following are equivalent:
(1) Every subspace of $X$ is normal.
(2) Every open subspace of $X$ is normal.
(3) Two separated sets have disjoint open neighborhoods.
I was asked to show that (1) and (3) are equivalent, however, I don’t know where the $T_1$ condition is used in the proof. My proof goes as follows:
If $X$ satisfies (3), then let $Y$ be a subspace and $A,Bsubset Y$ disjoint closed sets in the subspace $Y$. Denote the closure operator in $X$ by $Cl$ and in $Y$ by $Cl^*$. We claim that $A, B$ are separated in $X$: $$Cl(A)cap B= Cl(A)cap Cl^*(B) = Cl(A)cap Cl(B)cap Y = Cl^*(A)cap Cl^*(B)$$ But this is empty by the fact that $A, B$ are closed. Then we can take disjoint open neighborhood of $A,B$ in $X$, and intersect both by $Y$.
If $X$ satisfies (1), then take two separated sets $A, B$ in $X$. The open subspace $Y=X-(Cl(A)cap Cl(B))$ contains each of $A $ and $B$ by the fact that they are separated. The closure of them in $Y$ is empty: $Cl^*(A)cap Cl^*(B)= Cl(A)cap Cl(B)cap Y$ is empty because $Y$ does not contain $Cl(A)cap Cl(B)$. Take disjoint open neighborhood of $Cl^*(A)$ and $Cl^*(B)$ in $Y$, then by the fact that $Y$ is open, we are done with the proof.
So where exactly did I use $T_1$ condition in my proof?
general-topology
general-topology
asked Dec 14 '18 at 1:51
William SunWilliam Sun
471211
asked Dec 14 '18 at 1:51
William SunWilliam Sun
471211
edited Dec 14 '18 at 2:02
William Sun
asked Dec 14 '18 at 1:51
William SunWilliam Sun
471211
asked Dec 14 '18 at 1:51
William SunWilliam Sun
471211
asked Dec 14 '18 at 1:51
William SunWilliam Sun
471211
471211
1
$begingroup$
Does author require normal spaces to be Hausdorff?
$endgroup$
– William Elliot
Dec 14 '18 at 3:31
$begingroup$
@WilliamElliot The author requires normal spaces to be both $T_1$ and $T_4$. I understand it now-I use this book just for a reference so I was not aware it has an alternate definition for normal space. Thank you for the comment.
$endgroup$
– William Sun
Dec 14 '18 at 3:35
add a comment |
1
$begingroup$
Does author require normal spaces to be Hausdorff?
$endgroup$
– William Elliot
Dec 14 '18 at 3:31
$begingroup$
@WilliamElliot The author requires normal spaces to be both $T_1$ and $T_4$. I understand it now-I use this book just for a reference so I was not aware it has an alternate definition for normal space. Thank you for the comment.
$endgroup$
– William Sun
Dec 14 '18 at 3:35
1
1
$begingroup$
Does author require normal spaces to be Hausdorff?
$endgroup$
– William Elliot
Dec 14 '18 at 3:31
$begingroup$
Does author require normal spaces to be Hausdorff?
$endgroup$
– William Elliot
Dec 14 '18 at 3:31
$begingroup$
@WilliamElliot The author requires normal spaces to be both $T_1$ and $T_4$. I understand it now-I use this book just for a reference so I was not aware it has an alternate definition for normal space. Thank you for the comment.
$endgroup$
– William Sun
Dec 14 '18 at 3:35
$begingroup$
@WilliamElliot The author requires normal spaces to be both $T_1$ and $T_4$. I understand it now-I use this book just for a reference so I was not aware it has an alternate definition for normal space. Thank you for the comment.
$endgroup$
– William Sun
Dec 14 '18 at 3:35
add a comment |
1 Answer
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$begingroup$
Be aware that Engelking has the tendency to assume extra separation axioms in some definitions: compact/paracompact includes Hausdorff, normal and regular includes $T_1$, perfectly normal includes normal (which includes $T_1$) etc.
So in order to have a space all of whose subspaces are normal, we can only consider $T_1$ spaces to begin with. It's the price of admission, as it were.
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
109k347114
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add a comment |
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$begingroup$
Be aware that Engelking has the tendency to assume extra separation axioms in some definitions: compact/paracompact includes Hausdorff, normal and regular includes $T_1$, perfectly normal includes normal (which includes $T_1$) etc.
So in order to have a space all of whose subspaces are normal, we can only consider $T_1$ spaces to begin with. It's the price of admission, as it were.
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
add a comment |
$begingroup$
Be aware that Engelking has the tendency to assume extra separation axioms in some definitions: compact/paracompact includes Hausdorff, normal and regular includes $T_1$, perfectly normal includes normal (which includes $T_1$) etc.
So in order to have a space all of whose subspaces are normal, we can only consider $T_1$ spaces to begin with. It's the price of admission, as it were.
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
add a comment |
$begingroup$
Be aware that Engelking has the tendency to assume extra separation axioms in some definitions: compact/paracompact includes Hausdorff, normal and regular includes $T_1$, perfectly normal includes normal (which includes $T_1$) etc.
So in order to have a space all of whose subspaces are normal, we can only consider $T_1$ spaces to begin with. It's the price of admission, as it were.
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
Be aware that Engelking has the tendency to assume extra separation axioms in some definitions: compact/paracompact includes Hausdorff, normal and regular includes $T_1$, perfectly normal includes normal (which includes $T_1$) etc.
So in order to have a space all of whose subspaces are normal, we can only consider $T_1$ spaces to begin with. It's the price of admission, as it were.
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
109k347114
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
109k347114
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
109k347114
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
109k347114
109k347114
add a comment |
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1
$begingroup$
Does author require normal spaces to be Hausdorff?
$endgroup$
– William Elliot
Dec 14 '18 at 3:31
$begingroup$
@WilliamElliot The author requires normal spaces to be both $T_1$ and $T_4$. I understand it now-I use this book just for a reference so I was not aware it has an alternate definition for normal space. Thank you for the comment.
$endgroup$
– William Sun
Dec 14 '18 at 3:35