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
9,07421653
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1 Answer
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1 Answer
1
active
oldest
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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
9,07421653
$endgroup$
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
9,07421653
$endgroup$
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
9,07421653
$endgroup$
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
9,07421653
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
9,07421653
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
9,07421653
answered Dec 21 '18 at 18:43
BigbearZzzBigbearZzz
9,07421653
9,07421653
add a comment |
add a comment |
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Hint: image of compact space is compact
$endgroup$
– Wojowu
Dec 21 '18 at 18:09