Locally Convex tvs closure of ${0}$
$begingroup$
Let $E$ be a topological vector space locally convex, defined by the family of seminorms $mathcal{F}=(p_j)_{jin J}$.
I can't prove that $underset{jin J}bigcap Ker(p_j)=overline{{0}}$
functional-analysis topological-vector-spaces locally-convex-spaces
asked Dec 16 '18 at 8:10
Anas BOUALIIAnas BOUALII
1397
$endgroup$
add a comment |
$begingroup$
Let $E$ be a topological vector space locally convex, defined by the family of seminorms $mathcal{F}=(p_j)_{jin J}$.
I can't prove that $underset{jin J}bigcap Ker(p_j)=overline{{0}}$
functional-analysis topological-vector-spaces locally-convex-spaces
asked Dec 16 '18 at 8:10
Anas BOUALIIAnas BOUALII
1397
$endgroup$
$begingroup$
This condition is equivalent to your topology being Hausdorff. If you didn't start with enough seminorms, then your TVS won't have this property. However, if you require your TVS to be Hausdorff, distinct points can be separated by seminorms. What would that tell you about the intersection of the kernels of the seminorms?
$endgroup$
– Ashwin Trisal
Dec 16 '18 at 8:13
1
$begingroup$
@AshwinTrisal No, this is not true, this identity always holds. If $X$ is $T_1$ we get that the right hand closure is just ${0}$ and we do get the separation property.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 9:29
add a comment |
$begingroup$
Let $E$ be a topological vector space locally convex, defined by the family of seminorms $mathcal{F}=(p_j)_{jin J}$.
I can't prove that $underset{jin J}bigcap Ker(p_j)=overline{{0}}$
functional-analysis topological-vector-spaces locally-convex-spaces
asked Dec 16 '18 at 8:10
Anas BOUALIIAnas BOUALII
1397
$endgroup$
Let $E$ be a topological vector space locally convex, defined by the family of seminorms $mathcal{F}=(p_j)_{jin J}$.
I can't prove that $underset{jin J}bigcap Ker(p_j)=overline{{0}}$
functional-analysis topological-vector-spaces locally-convex-spaces
functional-analysis topological-vector-spaces locally-convex-spaces
asked Dec 16 '18 at 8:10
Anas BOUALIIAnas BOUALII
1397
asked Dec 16 '18 at 8:10
Anas BOUALIIAnas BOUALII
1397
asked Dec 16 '18 at 8:10
Anas BOUALIIAnas BOUALII
1397
asked Dec 16 '18 at 8:10
Anas BOUALIIAnas BOUALII
1397
asked Dec 16 '18 at 8:10
Anas BOUALIIAnas BOUALII
1397
1397
$begingroup$
This condition is equivalent to your topology being Hausdorff. If you didn't start with enough seminorms, then your TVS won't have this property. However, if you require your TVS to be Hausdorff, distinct points can be separated by seminorms. What would that tell you about the intersection of the kernels of the seminorms?
$endgroup$
– Ashwin Trisal
Dec 16 '18 at 8:13
1
$begingroup$
@AshwinTrisal No, this is not true, this identity always holds. If $X$ is $T_1$ we get that the right hand closure is just ${0}$ and we do get the separation property.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 9:29
add a comment |
$begingroup$
This condition is equivalent to your topology being Hausdorff. If you didn't start with enough seminorms, then your TVS won't have this property. However, if you require your TVS to be Hausdorff, distinct points can be separated by seminorms. What would that tell you about the intersection of the kernels of the seminorms?
$endgroup$
– Ashwin Trisal
Dec 16 '18 at 8:13
1
$begingroup$
@AshwinTrisal No, this is not true, this identity always holds. If $X$ is $T_1$ we get that the right hand closure is just ${0}$ and we do get the separation property.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 9:29
$begingroup$
This condition is equivalent to your topology being Hausdorff. If you didn't start with enough seminorms, then your TVS won't have this property. However, if you require your TVS to be Hausdorff, distinct points can be separated by seminorms. What would that tell you about the intersection of the kernels of the seminorms?
$endgroup$
– Ashwin Trisal
Dec 16 '18 at 8:13
$begingroup$
This condition is equivalent to your topology being Hausdorff. If you didn't start with enough seminorms, then your TVS won't have this property. However, if you require your TVS to be Hausdorff, distinct points can be separated by seminorms. What would that tell you about the intersection of the kernels of the seminorms?
$endgroup$
– Ashwin Trisal
Dec 16 '18 at 8:13
1
1
$begingroup$
@AshwinTrisal No, this is not true, this identity always holds. If $X$ is $T_1$ we get that the right hand closure is just ${0}$ and we do get the separation property.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 9:29
$begingroup$
@AshwinTrisal No, this is not true, this identity always holds. If $X$ is $T_1$ we get that the right hand closure is just ${0}$ and we do get the separation property.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 9:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $x in overline{{0}}$.
We know that all $p_j$ are by definition continuous in the topology generated by the semininorms, so for any $j in J$: $$p_j(x) in p_j[overline{{0}}] subseteq overline{p_j[{0}]} = overline{{0}} = {0}$$ as the last closure is taken in the reals (or complex numbers) where singletons are closed, and so $x in operatorname{Ker}(p_j)$ for all $j$, also using that $p_j(0)=0$ of course.
This shows $$overline{{0}} subseteq bigcap_{j in J} operatorname{Ker}(p_j)$$
To see the other inclusion, let $x$ be in $bigcap_{j in J} operatorname{Ker}(p_j)$ and let $O$ be a basic neighbourhood of $x$. This means that there are finitely many $p_{j_1},ldots,p_{j_n}$ and an $epsilon >0$ such that $O = bigcap_{i=1}^n B_{j_i}(x,varepsilon)$, where $B_{j_i}(x,varepsilon)= {y in X: p_{j_i}(x - y) < varepsilon}$ is the open ball around $x$ induced by the seminorm $p_{j_i}$.
The fact that $x in operatorname{Ker}(p_{j_i})$ implies that $0 in B_{j_i}(x,varepsilon)$, and so $0 in O$. As $O$ is an arbitary base element containing $x$ we have shown that $x in overline{{0}}$ as required.
answered Dec 16 '18 at 9:28
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
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%2f3042358%2flocally-convex-tvs-closure-of-0%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
$begingroup$
Suppose $x in overline{{0}}$.
We know that all $p_j$ are by definition continuous in the topology generated by the semininorms, so for any $j in J$: $$p_j(x) in p_j[overline{{0}}] subseteq overline{p_j[{0}]} = overline{{0}} = {0}$$ as the last closure is taken in the reals (or complex numbers) where singletons are closed, and so $x in operatorname{Ker}(p_j)$ for all $j$, also using that $p_j(0)=0$ of course.
This shows $$overline{{0}} subseteq bigcap_{j in J} operatorname{Ker}(p_j)$$
To see the other inclusion, let $x$ be in $bigcap_{j in J} operatorname{Ker}(p_j)$ and let $O$ be a basic neighbourhood of $x$. This means that there are finitely many $p_{j_1},ldots,p_{j_n}$ and an $epsilon >0$ such that $O = bigcap_{i=1}^n B_{j_i}(x,varepsilon)$, where $B_{j_i}(x,varepsilon)= {y in X: p_{j_i}(x - y) < varepsilon}$ is the open ball around $x$ induced by the seminorm $p_{j_i}$.
The fact that $x in operatorname{Ker}(p_{j_i})$ implies that $0 in B_{j_i}(x,varepsilon)$, and so $0 in O$. As $O$ is an arbitary base element containing $x$ we have shown that $x in overline{{0}}$ as required.
answered Dec 16 '18 at 9:28
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
add a comment |
$begingroup$
Suppose $x in overline{{0}}$.
We know that all $p_j$ are by definition continuous in the topology generated by the semininorms, so for any $j in J$: $$p_j(x) in p_j[overline{{0}}] subseteq overline{p_j[{0}]} = overline{{0}} = {0}$$ as the last closure is taken in the reals (or complex numbers) where singletons are closed, and so $x in operatorname{Ker}(p_j)$ for all $j$, also using that $p_j(0)=0$ of course.
This shows $$overline{{0}} subseteq bigcap_{j in J} operatorname{Ker}(p_j)$$
To see the other inclusion, let $x$ be in $bigcap_{j in J} operatorname{Ker}(p_j)$ and let $O$ be a basic neighbourhood of $x$. This means that there are finitely many $p_{j_1},ldots,p_{j_n}$ and an $epsilon >0$ such that $O = bigcap_{i=1}^n B_{j_i}(x,varepsilon)$, where $B_{j_i}(x,varepsilon)= {y in X: p_{j_i}(x - y) < varepsilon}$ is the open ball around $x$ induced by the seminorm $p_{j_i}$.
The fact that $x in operatorname{Ker}(p_{j_i})$ implies that $0 in B_{j_i}(x,varepsilon)$, and so $0 in O$. As $O$ is an arbitary base element containing $x$ we have shown that $x in overline{{0}}$ as required.
answered Dec 16 '18 at 9:28
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
add a comment |
$begingroup$
Suppose $x in overline{{0}}$.
We know that all $p_j$ are by definition continuous in the topology generated by the semininorms, so for any $j in J$: $$p_j(x) in p_j[overline{{0}}] subseteq overline{p_j[{0}]} = overline{{0}} = {0}$$ as the last closure is taken in the reals (or complex numbers) where singletons are closed, and so $x in operatorname{Ker}(p_j)$ for all $j$, also using that $p_j(0)=0$ of course.
This shows $$overline{{0}} subseteq bigcap_{j in J} operatorname{Ker}(p_j)$$
To see the other inclusion, let $x$ be in $bigcap_{j in J} operatorname{Ker}(p_j)$ and let $O$ be a basic neighbourhood of $x$. This means that there are finitely many $p_{j_1},ldots,p_{j_n}$ and an $epsilon >0$ such that $O = bigcap_{i=1}^n B_{j_i}(x,varepsilon)$, where $B_{j_i}(x,varepsilon)= {y in X: p_{j_i}(x - y) < varepsilon}$ is the open ball around $x$ induced by the seminorm $p_{j_i}$.
The fact that $x in operatorname{Ker}(p_{j_i})$ implies that $0 in B_{j_i}(x,varepsilon)$, and so $0 in O$. As $O$ is an arbitary base element containing $x$ we have shown that $x in overline{{0}}$ as required.
answered Dec 16 '18 at 9:28
Henno BrandsmaHenno Brandsma
109k347115
$endgroup$
Suppose $x in overline{{0}}$.
We know that all $p_j$ are by definition continuous in the topology generated by the semininorms, so for any $j in J$: $$p_j(x) in p_j[overline{{0}}] subseteq overline{p_j[{0}]} = overline{{0}} = {0}$$ as the last closure is taken in the reals (or complex numbers) where singletons are closed, and so $x in operatorname{Ker}(p_j)$ for all $j$, also using that $p_j(0)=0$ of course.
This shows $$overline{{0}} subseteq bigcap_{j in J} operatorname{Ker}(p_j)$$
To see the other inclusion, let $x$ be in $bigcap_{j in J} operatorname{Ker}(p_j)$ and let $O$ be a basic neighbourhood of $x$. This means that there are finitely many $p_{j_1},ldots,p_{j_n}$ and an $epsilon >0$ such that $O = bigcap_{i=1}^n B_{j_i}(x,varepsilon)$, where $B_{j_i}(x,varepsilon)= {y in X: p_{j_i}(x - y) < varepsilon}$ is the open ball around $x$ induced by the seminorm $p_{j_i}$.
The fact that $x in operatorname{Ker}(p_{j_i})$ implies that $0 in B_{j_i}(x,varepsilon)$, and so $0 in O$. As $O$ is an arbitary base element containing $x$ we have shown that $x in overline{{0}}$ as required.
answered Dec 16 '18 at 9:28
Henno BrandsmaHenno Brandsma
109k347115
answered Dec 16 '18 at 9:28
Henno BrandsmaHenno Brandsma
109k347115
answered Dec 16 '18 at 9:28
Henno BrandsmaHenno Brandsma
109k347115
answered Dec 16 '18 at 9:28
Henno BrandsmaHenno Brandsma
109k347115
109k347115
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%2f3042358%2flocally-convex-tvs-closure-of-0%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
$begingroup$
This condition is equivalent to your topology being Hausdorff. If you didn't start with enough seminorms, then your TVS won't have this property. However, if you require your TVS to be Hausdorff, distinct points can be separated by seminorms. What would that tell you about the intersection of the kernels of the seminorms?
$endgroup$
– Ashwin Trisal
Dec 16 '18 at 8:13
1
$begingroup$
@AshwinTrisal No, this is not true, this identity always holds. If $X$ is $T_1$ we get that the right hand closure is just ${0}$ and we do get the separation property.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 9:29