Exterior Covering Number of $epsilon /2$ is greater of equal than Covering Number of $epsilon$
$begingroup$
Definition $(epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K subset T$ and let $epsilon >0$, A subset $N subset K $ is called $epsilon -Net$ of $K$ if every point in $K$ is within a distance $epsilon$ if every point in K is within a distance $epsilon$ of some point of $N$,i.e.
$$forall x in K exists x_0 in N : d (x,x_0) leq epsilon$$
Definition(Covering Number): For metric space $(T,d)$ The covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N(K,d,epsilon )$, is the smallest possible cardinarity an $epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers in $K$ and radii $epsilon$ whose union covers $K$
Definition(Exterior Covering Number): For metric space $(T,d)$ The exterior covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N^{ext}(K,d,epsilon )$, is the smallest number of closed balls with centers not necessary in $K$ and radii $epsilon$ whose union covers $K$
then I was asked to prove that:
$$N(K,d,epsilon) leq N^{ext} (K,d,epsilon /2) $$
how to see that ?
here is my attempt: since each $epsilon -ball$ in $N(K,d, epsilon) $ should intersect at least one $epsilon /2 -ball$ in $N^{ext}(N,d,epsilon/2)$ ,thus I am trying to show that by contradiction: for each two distinct $epsilon -ball$ in $N(K,d, epsilon)$ ,the $epsilon /2 -balls$ they intersect must contain a distinct one.
real-analysis probability general-topology functional-analysis
asked Dec 28 '18 at 0:43
ShaoyuPeiShaoyuPei
1778
$endgroup$
add a comment |
$begingroup$
Definition $(epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K subset T$ and let $epsilon >0$, A subset $N subset K $ is called $epsilon -Net$ of $K$ if every point in $K$ is within a distance $epsilon$ if every point in K is within a distance $epsilon$ of some point of $N$,i.e.
$$forall x in K exists x_0 in N : d (x,x_0) leq epsilon$$
Definition(Covering Number): For metric space $(T,d)$ The covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N(K,d,epsilon )$, is the smallest possible cardinarity an $epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers in $K$ and radii $epsilon$ whose union covers $K$
Definition(Exterior Covering Number): For metric space $(T,d)$ The exterior covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N^{ext}(K,d,epsilon )$, is the smallest number of closed balls with centers not necessary in $K$ and radii $epsilon$ whose union covers $K$
then I was asked to prove that:
$$N(K,d,epsilon) leq N^{ext} (K,d,epsilon /2) $$
how to see that ?
here is my attempt: since each $epsilon -ball$ in $N(K,d, epsilon) $ should intersect at least one $epsilon /2 -ball$ in $N^{ext}(N,d,epsilon/2)$ ,thus I am trying to show that by contradiction: for each two distinct $epsilon -ball$ in $N(K,d, epsilon)$ ,the $epsilon /2 -balls$ they intersect must contain a distinct one.
real-analysis probability general-topology functional-analysis
asked Dec 28 '18 at 0:43
ShaoyuPeiShaoyuPei
1778
$endgroup$
add a comment |
$begingroup$
Definition $(epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K subset T$ and let $epsilon >0$, A subset $N subset K $ is called $epsilon -Net$ of $K$ if every point in $K$ is within a distance $epsilon$ if every point in K is within a distance $epsilon$ of some point of $N$,i.e.
$$forall x in K exists x_0 in N : d (x,x_0) leq epsilon$$
Definition(Covering Number): For metric space $(T,d)$ The covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N(K,d,epsilon )$, is the smallest possible cardinarity an $epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers in $K$ and radii $epsilon$ whose union covers $K$
Definition(Exterior Covering Number): For metric space $(T,d)$ The exterior covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N^{ext}(K,d,epsilon )$, is the smallest number of closed balls with centers not necessary in $K$ and radii $epsilon$ whose union covers $K$
then I was asked to prove that:
$$N(K,d,epsilon) leq N^{ext} (K,d,epsilon /2) $$
how to see that ?
here is my attempt: since each $epsilon -ball$ in $N(K,d, epsilon) $ should intersect at least one $epsilon /2 -ball$ in $N^{ext}(N,d,epsilon/2)$ ,thus I am trying to show that by contradiction: for each two distinct $epsilon -ball$ in $N(K,d, epsilon)$ ,the $epsilon /2 -balls$ they intersect must contain a distinct one.
real-analysis probability general-topology functional-analysis
asked Dec 28 '18 at 0:43
ShaoyuPeiShaoyuPei
1778
$endgroup$
Definition $(epsilon -Net)$ :let $(T,d)$ be a metric space .Consider a subset $ K subset T$ and let $epsilon >0$, A subset $N subset K $ is called $epsilon -Net$ of $K$ if every point in $K$ is within a distance $epsilon$ if every point in K is within a distance $epsilon$ of some point of $N$,i.e.
$$forall x in K exists x_0 in N : d (x,x_0) leq epsilon$$
Definition(Covering Number): For metric space $(T,d)$ The covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N(K,d,epsilon )$, is the smallest possible cardinarity an $epsilon -Net$ of K , or equivalently ,is the smallest number of closed balls with centers in $K$ and radii $epsilon$ whose union covers $K$
Definition(Exterior Covering Number): For metric space $(T,d)$ The exterior covering number of $K subset T$ respect to a given $epsilon geq 0$ ,denotes as $N^{ext}(K,d,epsilon )$, is the smallest number of closed balls with centers not necessary in $K$ and radii $epsilon$ whose union covers $K$
then I was asked to prove that:
$$N(K,d,epsilon) leq N^{ext} (K,d,epsilon /2) $$
how to see that ?
here is my attempt: since each $epsilon -ball$ in $N(K,d, epsilon) $ should intersect at least one $epsilon /2 -ball$ in $N^{ext}(N,d,epsilon/2)$ ,thus I am trying to show that by contradiction: for each two distinct $epsilon -ball$ in $N(K,d, epsilon)$ ,the $epsilon /2 -balls$ they intersect must contain a distinct one.
real-analysis probability general-topology functional-analysis
real-analysis probability general-topology functional-analysis
asked Dec 28 '18 at 0:43
ShaoyuPeiShaoyuPei
1778
asked Dec 28 '18 at 0:43
ShaoyuPeiShaoyuPei
1778
edited Dec 28 '18 at 0:51
ShaoyuPei
asked Dec 28 '18 at 0:43
ShaoyuPeiShaoyuPei
1778
asked Dec 28 '18 at 0:43
ShaoyuPeiShaoyuPei
1778
asked Dec 28 '18 at 0:43
ShaoyuPeiShaoyuPei
1778
1778
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.
For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).
By the triangle inequality,
$$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
for each $i$, which means that
$$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.
answered Dec 28 '18 at 1:01
Kenny WongKenny Wong
19k21440
$endgroup$
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%2f3054478%2fexterior-covering-number-of-epsilon-2-is-greater-of-equal-than-covering-numb%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 that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.
For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).
By the triangle inequality,
$$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
for each $i$, which means that
$$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.
answered Dec 28 '18 at 1:01
Kenny WongKenny Wong
19k21440
$endgroup$
add a comment |
$begingroup$
Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.
For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).
By the triangle inequality,
$$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
for each $i$, which means that
$$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.
answered Dec 28 '18 at 1:01
Kenny WongKenny Wong
19k21440
$endgroup$
add a comment |
$begingroup$
Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.
For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).
By the triangle inequality,
$$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
for each $i$, which means that
$$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.
answered Dec 28 '18 at 1:01
Kenny WongKenny Wong
19k21440
$endgroup$
Suppose that $$bar B(x_1, epsilon / 2), dots, bar B(x_{N^{rm ext}}, epsilon / 2)$$ is an external covering of $K$ of minimal size.
For each $i in { 1, dots, N^{rm ext}}$, there exists a $k_i in K$ that is contained in $bar B(x_i, epsilon / 2)$. (Otherwise $bar B(x_i, epsilon / 2)$ would be redundant, contradicting the minimality of size of the covering).
By the triangle inequality,
$$ bar B(x_i , epsilon / 2) subseteq bar B(k_i, epsilon), $$
for each $i$, which means that
$$bar B(k_1, epsilon ), dots, bar B(k_{N^{rm ext}}, epsilon)$$
is an internal covering of size $N^{rm ext}$. Hence the smallest internal covering has size at most $N^{rm ext}$.
answered Dec 28 '18 at 1:01
Kenny WongKenny Wong
19k21440
edited Dec 28 '18 at 1:33
answered Dec 28 '18 at 1:01
Kenny WongKenny Wong
19k21440
answered Dec 28 '18 at 1:01
Kenny WongKenny Wong
19k21440
answered Dec 28 '18 at 1:01
Kenny WongKenny Wong
19k21440
19k21440
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%2f3054478%2fexterior-covering-number-of-epsilon-2-is-greater-of-equal-than-covering-numb%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
