Compact spaces and Urysohn Lemma
$begingroup$
Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x in X$ a point. We define the distance $d(x, A)$ via
$$
d(x,A)=inf({ d(x,a)mid a in A})
$$
(a) Show that $dcolon X to mathbb{R}$ is a continuous function and that $xin overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f colon X to mathbb{R}$
defined by
$$
f(x)=frac{d(x, A)}{d(x, A) + d(x, B)}
$$
Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as
in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for
all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a)
Can we say that the function is continuous if we have $x,yin X$ so, that
$d(x,A)leq d(x,a)leq d(x,y) + d(y,a)$
for $a in A$
$d(x,A)-d(x,y) leq inf{d(y,a)}=d(y,A)$
so
$d(x,A)-d(y,A)leq d(x,y)$
Is this right or am I missing something? And how can I show that $xinoverline{A}$ if and only if $d(x,A)=0$?
Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
general-topology
edited Jan 10 at 5:14
Henno Brandsma
117k349127
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
$endgroup$
add a comment |
$begingroup$
Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x in X$ a point. We define the distance $d(x, A)$ via
$$
d(x,A)=inf({ d(x,a)mid a in A})
$$
(a) Show that $dcolon X to mathbb{R}$ is a continuous function and that $xin overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f colon X to mathbb{R}$
defined by
$$
f(x)=frac{d(x, A)}{d(x, A) + d(x, B)}
$$
Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as
in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for
all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a)
Can we say that the function is continuous if we have $x,yin X$ so, that
$d(x,A)leq d(x,a)leq d(x,y) + d(y,a)$
for $a in A$
$d(x,A)-d(x,y) leq inf{d(y,a)}=d(y,A)$
so
$d(x,A)-d(y,A)leq d(x,y)$
Is this right or am I missing something? And how can I show that $xinoverline{A}$ if and only if $d(x,A)=0$?
Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
general-topology
edited Jan 10 at 5:14
Henno Brandsma
117k349127
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
$endgroup$
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
add a comment |
$begingroup$
Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x in X$ a point. We define the distance $d(x, A)$ via
$$
d(x,A)=inf({ d(x,a)mid a in A})
$$
(a) Show that $dcolon X to mathbb{R}$ is a continuous function and that $xin overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f colon X to mathbb{R}$
defined by
$$
f(x)=frac{d(x, A)}{d(x, A) + d(x, B)}
$$
Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as
in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for
all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a)
Can we say that the function is continuous if we have $x,yin X$ so, that
$d(x,A)leq d(x,a)leq d(x,y) + d(y,a)$
for $a in A$
$d(x,A)-d(x,y) leq inf{d(y,a)}=d(y,A)$
so
$d(x,A)-d(y,A)leq d(x,y)$
Is this right or am I missing something? And how can I show that $xinoverline{A}$ if and only if $d(x,A)=0$?
Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
general-topology
edited Jan 10 at 5:14
Henno Brandsma
117k349127
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
$endgroup$
Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x in X$ a point. We define the distance $d(x, A)$ via
$$
d(x,A)=inf({ d(x,a)mid a in A})
$$
(a) Show that $dcolon X to mathbb{R}$ is a continuous function and that $xin overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f colon X to mathbb{R}$
defined by
$$
f(x)=frac{d(x, A)}{d(x, A) + d(x, B)}
$$
Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as
in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for
all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a)
Can we say that the function is continuous if we have $x,yin X$ so, that
$d(x,A)leq d(x,a)leq d(x,y) + d(y,a)$
for $a in A$
$d(x,A)-d(x,y) leq inf{d(y,a)}=d(y,A)$
so
$d(x,A)-d(y,A)leq d(x,y)$
Is this right or am I missing something? And how can I show that $xinoverline{A}$ if and only if $d(x,A)=0$?
Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
general-topology
general-topology
edited Jan 10 at 5:14
Henno Brandsma
117k349127
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
edited Jan 10 at 5:14
Henno Brandsma
117k349127
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
edited Jan 10 at 5:14
Henno Brandsma
117k349127
edited Jan 10 at 5:14
Henno Brandsma
117k349127
edited Jan 10 at 5:14
Henno Brandsma
117k349127
117k349127
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
111
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
add a comment |
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
9,1962820
$endgroup$
add a comment |
Your Answer
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%2f3067323%2fcompact-spaces-and-urysohn-lemma%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
$begingroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
9,1962820
$endgroup$
add a comment |
$begingroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
9,1962820
$endgroup$
add a comment |
$begingroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
9,1962820
$endgroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
9,1962820
edited Jan 9 at 13:01
answered Jan 9 at 12:52
William ElliotWilliam Elliot
9,1962820
answered Jan 9 at 12:52
William ElliotWilliam Elliot
9,1962820
answered Jan 9 at 12:52
William ElliotWilliam Elliot
9,1962820
9,1962820
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%2f3067323%2fcompact-spaces-and-urysohn-lemma%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
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12