The existence of a vector arbitrarily separated from a close subspace
$begingroup$
Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$|z| = 1$$ satisfying $$|z-y| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not
belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to
$Y$: $$inf_{y in Y} |x-y| = d > 0$$ There is then a $y_0 in Y$ such that $$|x - y_0| < 2d quad (*)$$
Denote $z' = x - y_0$; we can then write $(*)$ as $|z'| < 2d$. It follows from definition of $d$ above that $$|z' - y | geq d quad (**)$$ for all $y$ in $Y$. Setting $z = frac{z'}{|z'|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
functional-analysis convex-analysis
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
$endgroup$
add a comment |
$begingroup$
Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$|z| = 1$$ satisfying $$|z-y| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not
belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to
$Y$: $$inf_{y in Y} |x-y| = d > 0$$ There is then a $y_0 in Y$ such that $$|x - y_0| < 2d quad (*)$$
Denote $z' = x - y_0$; we can then write $(*)$ as $|z'| < 2d$. It follows from definition of $d$ above that $$|z' - y | geq d quad (**)$$ for all $y$ in $Y$. Setting $z = frac{z'}{|z'|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
functional-analysis convex-analysis
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
$endgroup$
add a comment |
$begingroup$
Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$|z| = 1$$ satisfying $$|z-y| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not
belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to
$Y$: $$inf_{y in Y} |x-y| = d > 0$$ There is then a $y_0 in Y$ such that $$|x - y_0| < 2d quad (*)$$
Denote $z' = x - y_0$; we can then write $(*)$ as $|z'| < 2d$. It follows from definition of $d$ above that $$|z' - y | geq d quad (**)$$ for all $y$ in $Y$. Setting $z = frac{z'}{|z'|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
functional-analysis convex-analysis
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
$endgroup$
Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$|z| = 1$$ satisfying $$|z-y| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not
belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to
$Y$: $$inf_{y in Y} |x-y| = d > 0$$ There is then a $y_0 in Y$ such that $$|x - y_0| < 2d quad (*)$$
Denote $z' = x - y_0$; we can then write $(*)$ as $|z'| < 2d$. It follows from definition of $d$ above that $$|z' - y | geq d quad (**)$$ for all $y$ in $Y$. Setting $z = frac{z'}{|z'|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
functional-analysis convex-analysis
functional-analysis convex-analysis
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
asked Dec 28 '18 at 16:56
yoshiyoshi
1,242917
1,242917
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.5k11446
$endgroup$
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
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%2f3055067%2fthe-existence-of-a-vector-arbitrarily-separated-from-a-close-subspace%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
$begingroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.5k11446
$endgroup$
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
add a comment |
$begingroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.5k11446
$endgroup$
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
add a comment |
$begingroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.5k11446
$endgroup$
$|z'-|z'|y|geq d$ since $|z'|yin Y$, this implies that $|{{z'-|z'|y}over {|z'|}}||geq {dover{|z'|}}$ Which implies that $|{z'over{|z'|}}-y|geq 1/2$
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.5k11446
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.5k11446
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.5k11446
answered Dec 28 '18 at 17:27
Tsemo AristideTsemo Aristide
59.5k11446
59.5k11446
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
add a comment |
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
$begingroup$
thanks! quick question, how did you see this? it wasn't obvious to me to choose this particular value of $y in Y$
$endgroup$
– yoshi
Dec 28 '18 at 17:31
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%2f3055067%2fthe-existence-of-a-vector-arbitrarily-separated-from-a-close-subspace%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
