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
473211
$endgroup$
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
473211
$endgroup$
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
473211
$endgroup$
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
473211
asked Dec 14 '18 at 1:51
William SunWilliam Sun
473211
edited Dec 14 '18 at 2:02
William Sun
asked Dec 14 '18 at 1:51
William SunWilliam Sun
473211
asked Dec 14 '18 at 1:51
William SunWilliam Sun
473211
asked Dec 14 '18 at 1:51
William SunWilliam Sun
473211
473211
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
1
active
oldest
votes
$begingroup$
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
111k348118
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038818%2fis-t-1-condition-necessary-in-the-definition-of-completely-normality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
111k348118
$endgroup$
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
111k348118
$endgroup$
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
111k348118
$endgroup$
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
111k348118
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 14 '18 at 4:58
Henno BrandsmaHenno Brandsma
111k348118
111k348118
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038818%2fis-t-1-condition-necessary-in-the-definition-of-completely-normality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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