Choosing a correct option related to compactness and completeness about image of map.
$begingroup$
Let $f : X to Y$ be a continuous map between metric spaces. Then $f(X)$
is a complete subset of $Y$ if
A. the space $X$ is compact
B. the space $Y$ is compact
C. the space $X$ is complete
D. the space $Y$ is complete.
My Attempt.
If I want to show completeness then I assume a cauchy sequence of $f(X)$ say $y_1,y_2,y_3,y_4 dots dots$ be a cauchy sequence in $f(X)$ let $y_1=f(x_1),y_2=f(x_2),y_3=f(x_3),y_4=f(x_4) dots dots$
Then how do we proceed futher systematically?
I have following ideas but I cannot use/implement properly.
we want $y=f(x)$ where $(y_m) to y$ and $(x_m) to x$ but I wonder how to use them because I don't know $(x_m)$ converge or not.
metric-spaces compactness complete-spaces
$endgroup$
add a comment |
$begingroup$
Let $f : X to Y$ be a continuous map between metric spaces. Then $f(X)$
is a complete subset of $Y$ if
A. the space $X$ is compact
B. the space $Y$ is compact
C. the space $X$ is complete
D. the space $Y$ is complete.
My Attempt.
If I want to show completeness then I assume a cauchy sequence of $f(X)$ say $y_1,y_2,y_3,y_4 dots dots$ be a cauchy sequence in $f(X)$ let $y_1=f(x_1),y_2=f(x_2),y_3=f(x_3),y_4=f(x_4) dots dots$
Then how do we proceed futher systematically?
I have following ideas but I cannot use/implement properly.
we want $y=f(x)$ where $(y_m) to y$ and $(x_m) to x$ but I wonder how to use them because I don't know $(x_m)$ converge or not.
metric-spaces compactness complete-spaces
$endgroup$
$begingroup$
Hint: image of compact space is compact
$endgroup$
– Wojowu
Dec 21 '18 at 18:09
add a comment |
$begingroup$
Let $f : X to Y$ be a continuous map between metric spaces. Then $f(X)$
is a complete subset of $Y$ if
A. the space $X$ is compact
B. the space $Y$ is compact
C. the space $X$ is complete
D. the space $Y$ is complete.
My Attempt.
If I want to show completeness then I assume a cauchy sequence of $f(X)$ say $y_1,y_2,y_3,y_4 dots dots$ be a cauchy sequence in $f(X)$ let $y_1=f(x_1),y_2=f(x_2),y_3=f(x_3),y_4=f(x_4) dots dots$
Then how do we proceed futher systematically?
I have following ideas but I cannot use/implement properly.
we want $y=f(x)$ where $(y_m) to y$ and $(x_m) to x$ but I wonder how to use them because I don't know $(x_m)$ converge or not.
metric-spaces compactness complete-spaces
$endgroup$
Let $f : X to Y$ be a continuous map between metric spaces. Then $f(X)$
is a complete subset of $Y$ if
A. the space $X$ is compact
B. the space $Y$ is compact
C. the space $X$ is complete
D. the space $Y$ is complete.
My Attempt.
If I want to show completeness then I assume a cauchy sequence of $f(X)$ say $y_1,y_2,y_3,y_4 dots dots$ be a cauchy sequence in $f(X)$ let $y_1=f(x_1),y_2=f(x_2),y_3=f(x_3),y_4=f(x_4) dots dots$
Then how do we proceed futher systematically?
I have following ideas but I cannot use/implement properly.
we want $y=f(x)$ where $(y_m) to y$ and $(x_m) to x$ but I wonder how to use them because I don't know $(x_m)$ converge or not.
metric-spaces compactness complete-spaces
metric-spaces compactness complete-spaces
asked Dec 21 '18 at 18:06
user408906
$begingroup$
Hint: image of compact space is compact
$endgroup$
– Wojowu
Dec 21 '18 at 18:09
add a comment |
$begingroup$
Hint: image of compact space is compact
$endgroup$
– Wojowu
Dec 21 '18 at 18:09
$begingroup$
Hint: image of compact space is compact
$endgroup$
– Wojowu
Dec 21 '18 at 18:09
$begingroup$
Hint: image of compact space is compact
$endgroup$
– Wojowu
Dec 21 '18 at 18:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A. is the correct answer since the continuous image $f(X)$ of a compact set $X$ is always compact, and hence complete.
B. C. D. are all false. Indeed, we can take $X=Bbb R$ which is complete and take $Y=[-pi/2,pi/2]$ which is both compact amd complete. However, for $f(x)=arctan(x)$ we have $f(X)=(-pi/2,pi/2)$ which is not complete.
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
8,94521652
$endgroup$
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%2f3048757%2fchoosing-a-correct-option-related-to-compactness-and-completeness-about-image-of%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 the correct answer since the continuous image $f(X)$ of a compact set $X$ is always compact, and hence complete.
B. C. D. are all false. Indeed, we can take $X=Bbb R$ which is complete and take $Y=[-pi/2,pi/2]$ which is both compact amd complete. However, for $f(x)=arctan(x)$ we have $f(X)=(-pi/2,pi/2)$ which is not complete.
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
8,94521652
$endgroup$
add a comment |
$begingroup$
A. is the correct answer since the continuous image $f(X)$ of a compact set $X$ is always compact, and hence complete.
B. C. D. are all false. Indeed, we can take $X=Bbb R$ which is complete and take $Y=[-pi/2,pi/2]$ which is both compact amd complete. However, for $f(x)=arctan(x)$ we have $f(X)=(-pi/2,pi/2)$ which is not complete.
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
8,94521652
$endgroup$
add a comment |
$begingroup$
A. is the correct answer since the continuous image $f(X)$ of a compact set $X$ is always compact, and hence complete.
B. C. D. are all false. Indeed, we can take $X=Bbb R$ which is complete and take $Y=[-pi/2,pi/2]$ which is both compact amd complete. However, for $f(x)=arctan(x)$ we have $f(X)=(-pi/2,pi/2)$ which is not complete.
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
8,94521652
$endgroup$
A. is the correct answer since the continuous image $f(X)$ of a compact set $X$ is always compact, and hence complete.
B. C. D. are all false. Indeed, we can take $X=Bbb R$ which is complete and take $Y=[-pi/2,pi/2]$ which is both compact amd complete. However, for $f(x)=arctan(x)$ we have $f(X)=(-pi/2,pi/2)$ which is not complete.
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
8,94521652
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
8,94521652
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
8,94521652
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
8,94521652
8,94521652
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%2f3048757%2fchoosing-a-correct-option-related-to-compactness-and-completeness-about-image-of%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
$begingroup$
Hint: image of compact space is compact
$endgroup$
– Wojowu
Dec 21 '18 at 18:09