Is space $C[0,1]$ is Hausdorff?
$begingroup$
Is space $C[0,1]$ is Hausdorff?
I just started studying metric spaces. I encountered the Hausdroff space definition. I wanted to construct a space which does not have that property.
I thought of $C[0,1]$ as candidate. Because we can define 2 distant function which agree at single point.
SO It is not possible to haved isjoint open balls.
Is my argument correct?
Any Help will be appreciated
general-topology analysis metric-spaces
edited Jan 12 at 15:09
klirk
2,631530
asked Jan 12 at 12:24
MathLoverMathLover
49710
$endgroup$
add a comment |
$begingroup$
Is space $C[0,1]$ is Hausdorff?
I just started studying metric spaces. I encountered the Hausdroff space definition. I wanted to construct a space which does not have that property.
I thought of $C[0,1]$ as candidate. Because we can define 2 distant function which agree at single point.
SO It is not possible to haved isjoint open balls.
Is my argument correct?
Any Help will be appreciated
general-topology analysis metric-spaces
edited Jan 12 at 15:09
klirk
2,631530
asked Jan 12 at 12:24
MathLoverMathLover
49710
$endgroup$
$begingroup$
Note that $mathcal{C}[0,1]$ is just a set. Which topology do you have in mind?
$endgroup$
– José Carlos Santos
Jan 12 at 12:28
$begingroup$
Sir ,topology generated by supremum metric f
$endgroup$
– MathLover
Jan 12 at 12:29
$begingroup$
Any metric space is Hausdorff. You can take it as an easy exercise to prove that.
$endgroup$
– Mark
Jan 12 at 12:30
$begingroup$
Ohh I got for distantness sup(f,g)>$epsilon$ which is not here.Thanks a lot
$endgroup$
– MathLover
Jan 12 at 12:31
1
$begingroup$
Now if you want to find a space which is not Hausdorff then the easiest example is a set $X$ with cardinality at least $2$ with the trivial topology. If you want a more interesting example then look at an infinite set $X$ with the co-finite topology.
$endgroup$
– Mark
Jan 12 at 12:33
add a comment |
$begingroup$
Is space $C[0,1]$ is Hausdorff?
I just started studying metric spaces. I encountered the Hausdroff space definition. I wanted to construct a space which does not have that property.
I thought of $C[0,1]$ as candidate. Because we can define 2 distant function which agree at single point.
SO It is not possible to haved isjoint open balls.
Is my argument correct?
Any Help will be appreciated
general-topology analysis metric-spaces
edited Jan 12 at 15:09
klirk
2,631530
asked Jan 12 at 12:24
MathLoverMathLover
49710
$endgroup$
Is space $C[0,1]$ is Hausdorff?
I just started studying metric spaces. I encountered the Hausdroff space definition. I wanted to construct a space which does not have that property.
I thought of $C[0,1]$ as candidate. Because we can define 2 distant function which agree at single point.
SO It is not possible to haved isjoint open balls.
Is my argument correct?
Any Help will be appreciated
general-topology analysis metric-spaces
general-topology analysis metric-spaces
edited Jan 12 at 15:09
klirk
2,631530
asked Jan 12 at 12:24
MathLoverMathLover
49710
edited Jan 12 at 15:09
klirk
2,631530
asked Jan 12 at 12:24
MathLoverMathLover
49710
edited Jan 12 at 15:09
klirk
2,631530
edited Jan 12 at 15:09
klirk
2,631530
edited Jan 12 at 15:09
klirk
2,631530
2,631530
asked Jan 12 at 12:24
MathLoverMathLover
49710
asked Jan 12 at 12:24
MathLoverMathLover
49710
asked Jan 12 at 12:24
MathLoverMathLover
49710
49710
$begingroup$
Note that $mathcal{C}[0,1]$ is just a set. Which topology do you have in mind?
$endgroup$
– José Carlos Santos
Jan 12 at 12:28
$begingroup$
Sir ,topology generated by supremum metric f
$endgroup$
– MathLover
Jan 12 at 12:29
$begingroup$
Any metric space is Hausdorff. You can take it as an easy exercise to prove that.
$endgroup$
– Mark
Jan 12 at 12:30
$begingroup$
Ohh I got for distantness sup(f,g)>$epsilon$ which is not here.Thanks a lot
$endgroup$
– MathLover
Jan 12 at 12:31
1
$begingroup$
Now if you want to find a space which is not Hausdorff then the easiest example is a set $X$ with cardinality at least $2$ with the trivial topology. If you want a more interesting example then look at an infinite set $X$ with the co-finite topology.
$endgroup$
– Mark
Jan 12 at 12:33
add a comment |
$begingroup$
Note that $mathcal{C}[0,1]$ is just a set. Which topology do you have in mind?
$endgroup$
– José Carlos Santos
Jan 12 at 12:28
$begingroup$
Sir ,topology generated by supremum metric f
$endgroup$
– MathLover
Jan 12 at 12:29
$begingroup$
Any metric space is Hausdorff. You can take it as an easy exercise to prove that.
$endgroup$
– Mark
Jan 12 at 12:30
$begingroup$
Ohh I got for distantness sup(f,g)>$epsilon$ which is not here.Thanks a lot
$endgroup$
– MathLover
Jan 12 at 12:31
1
$begingroup$
Now if you want to find a space which is not Hausdorff then the easiest example is a set $X$ with cardinality at least $2$ with the trivial topology. If you want a more interesting example then look at an infinite set $X$ with the co-finite topology.
$endgroup$
– Mark
Jan 12 at 12:33
$begingroup$
Note that $mathcal{C}[0,1]$ is just a set. Which topology do you have in mind?
$endgroup$
– José Carlos Santos
Jan 12 at 12:28
$begingroup$
Note that $mathcal{C}[0,1]$ is just a set. Which topology do you have in mind?
$endgroup$
– José Carlos Santos
Jan 12 at 12:28
$begingroup$
Sir ,topology generated by supremum metric f
$endgroup$
– MathLover
Jan 12 at 12:29
$begingroup$
Sir ,topology generated by supremum metric f
$endgroup$
– MathLover
Jan 12 at 12:29
$begingroup$
Any metric space is Hausdorff. You can take it as an easy exercise to prove that.
$endgroup$
– Mark
Jan 12 at 12:30
$begingroup$
Any metric space is Hausdorff. You can take it as an easy exercise to prove that.
$endgroup$
– Mark
Jan 12 at 12:30
$begingroup$
Ohh I got for distantness sup(f,g)>$epsilon$ which is not here.Thanks a lot
$endgroup$
– MathLover
Jan 12 at 12:31
$begingroup$
Ohh I got for distantness sup(f,g)>$epsilon$ which is not here.Thanks a lot
$endgroup$
– MathLover
Jan 12 at 12:31
1
1
$begingroup$
Now if you want to find a space which is not Hausdorff then the easiest example is a set $X$ with cardinality at least $2$ with the trivial topology. If you want a more interesting example then look at an infinite set $X$ with the co-finite topology.
$endgroup$
– Mark
Jan 12 at 12:33
$begingroup$
Now if you want to find a space which is not Hausdorff then the easiest example is a set $X$ with cardinality at least $2$ with the trivial topology. If you want a more interesting example then look at an infinite set $X$ with the co-finite topology.
$endgroup$
– Mark
Jan 12 at 12:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The set $C[0,1]$ by itself is not a topological space. If you are taking about this set with the standard metric $d(f,g)=sup {|f(x)-g(x)|:0leq x leq 1}$ then it is Hausdorff because any metric space is Hausdorff. [ If $x neq y$ and $r=d(x,y)$ then $B(x,frac r 2)$ and $B(y,frac r 2)$ are disjoint open sets containing $x$ and $y$ respectively.
edited Jan 12 at 12:33
José Carlos Santos
157k22126227
answered Jan 12 at 12:30
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
$endgroup$
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%2f3070848%2fis-space-c0-1-is-hausdorff%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
$begingroup$
The set $C[0,1]$ by itself is not a topological space. If you are taking about this set with the standard metric $d(f,g)=sup {|f(x)-g(x)|:0leq x leq 1}$ then it is Hausdorff because any metric space is Hausdorff. [ If $x neq y$ and $r=d(x,y)$ then $B(x,frac r 2)$ and $B(y,frac r 2)$ are disjoint open sets containing $x$ and $y$ respectively.
edited Jan 12 at 12:33
José Carlos Santos
157k22126227
answered Jan 12 at 12:30
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
$endgroup$
add a comment |
$begingroup$
The set $C[0,1]$ by itself is not a topological space. If you are taking about this set with the standard metric $d(f,g)=sup {|f(x)-g(x)|:0leq x leq 1}$ then it is Hausdorff because any metric space is Hausdorff. [ If $x neq y$ and $r=d(x,y)$ then $B(x,frac r 2)$ and $B(y,frac r 2)$ are disjoint open sets containing $x$ and $y$ respectively.
edited Jan 12 at 12:33
José Carlos Santos
157k22126227
answered Jan 12 at 12:30
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
$endgroup$
add a comment |
$begingroup$
The set $C[0,1]$ by itself is not a topological space. If you are taking about this set with the standard metric $d(f,g)=sup {|f(x)-g(x)|:0leq x leq 1}$ then it is Hausdorff because any metric space is Hausdorff. [ If $x neq y$ and $r=d(x,y)$ then $B(x,frac r 2)$ and $B(y,frac r 2)$ are disjoint open sets containing $x$ and $y$ respectively.
edited Jan 12 at 12:33
José Carlos Santos
157k22126227
answered Jan 12 at 12:30
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
$endgroup$
The set $C[0,1]$ by itself is not a topological space. If you are taking about this set with the standard metric $d(f,g)=sup {|f(x)-g(x)|:0leq x leq 1}$ then it is Hausdorff because any metric space is Hausdorff. [ If $x neq y$ and $r=d(x,y)$ then $B(x,frac r 2)$ and $B(y,frac r 2)$ are disjoint open sets containing $x$ and $y$ respectively.
edited Jan 12 at 12:33
José Carlos Santos
157k22126227
answered Jan 12 at 12:30
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
edited Jan 12 at 12:33
José Carlos Santos
157k22126227
edited Jan 12 at 12:33
José Carlos Santos
157k22126227
edited Jan 12 at 12:33
José Carlos Santos
157k22126227
157k22126227
answered Jan 12 at 12:30
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
answered Jan 12 at 12:30
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
answered Jan 12 at 12:30
Kavi Rama MurthyKavi Rama Murthy
56.1k42158
56.1k42158
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%2f3070848%2fis-space-c0-1-is-hausdorff%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
$begingroup$
Note that $mathcal{C}[0,1]$ is just a set. Which topology do you have in mind?
$endgroup$
– José Carlos Santos
Jan 12 at 12:28
$begingroup$
Sir ,topology generated by supremum metric f
$endgroup$
– MathLover
Jan 12 at 12:29
$begingroup$
Any metric space is Hausdorff. You can take it as an easy exercise to prove that.
$endgroup$
– Mark
Jan 12 at 12:30
$begingroup$
Ohh I got for distantness sup(f,g)>$epsilon$ which is not here.Thanks a lot
$endgroup$
– MathLover
Jan 12 at 12:31
1
$begingroup$
Now if you want to find a space which is not Hausdorff then the easiest example is a set $X$ with cardinality at least $2$ with the trivial topology. If you want a more interesting example then look at an infinite set $X$ with the co-finite topology.
$endgroup$
– Mark
Jan 12 at 12:33