Proof: Let $f:(M,d)to(Y,rho)$ be a continuous function. If $Ksubset M$ is compact,then $f(K)subset Y$ is also...
up vote
0
down vote
favorite
Let $f:(M,d)to(Y,rho)$ be a continuous function.If $Ksubset M$ is compact,then $f(K)subset Y$is also compact.
My attempt:
I know that f is continuous function, so there is a $xin X$, if for any $epsilon>0$ ,there is a $delta>0$,s.t. $d(x,x')<delta$,so $rho(f(x),f(x'))<epsilon$.
Then $Ksubset M$ is compact. so there is a collection {$X_alpha$} is an open cover of $K$,$Ksubset cup_alpha X_alpha$.
Thus, if for any $x_alphain X_alpha$, $d(x_alpha,x_alpha')<delta$,so $rho(f(x_alpha),f(x_alpha'))<epsilon$. So,$Ksubsetcup_alpha X_alpha$,and $y=f(x)in Y$,so $f(K)subsetcup_alpha f(X_alpha)$,$f(K)subset Y$ is also compact.
This proof may have some wrong...
continuity metric-spaces compactness
asked yesterday
HUANG Fuzhe
61
New contributor
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
Let $f:(M,d)to(Y,rho)$ be a continuous function.If $Ksubset M$ is compact,then $f(K)subset Y$is also compact.
My attempt:
I know that f is continuous function, so there is a $xin X$, if for any $epsilon>0$ ,there is a $delta>0$,s.t. $d(x,x')<delta$,so $rho(f(x),f(x'))<epsilon$.
Then $Ksubset M$ is compact. so there is a collection {$X_alpha$} is an open cover of $K$,$Ksubset cup_alpha X_alpha$.
Thus, if for any $x_alphain X_alpha$, $d(x_alpha,x_alpha')<delta$,so $rho(f(x_alpha),f(x_alpha'))<epsilon$. So,$Ksubsetcup_alpha X_alpha$,and $y=f(x)in Y$,so $f(K)subsetcup_alpha f(X_alpha)$,$f(K)subset Y$ is also compact.
This proof may have some wrong...
continuity metric-spaces compactness
asked yesterday
HUANG Fuzhe
61
New contributor
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
yesterday
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
yesterday
Regarding your works, that is not the definition of compactness. A subset $K$ is compact iff for every open cover for $K$, there is a finite subcover of the open cover. To show $f(K)$ is compact, you need to show that any open cover ${U_{alpha} }$ has a finite subcover ${U_{alpha_1},dots,U_{alpha_n}}$.
– Kelvin Lois
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:(M,d)to(Y,rho)$ be a continuous function.If $Ksubset M$ is compact,then $f(K)subset Y$is also compact.
My attempt:
I know that f is continuous function, so there is a $xin X$, if for any $epsilon>0$ ,there is a $delta>0$,s.t. $d(x,x')<delta$,so $rho(f(x),f(x'))<epsilon$.
Then $Ksubset M$ is compact. so there is a collection {$X_alpha$} is an open cover of $K$,$Ksubset cup_alpha X_alpha$.
Thus, if for any $x_alphain X_alpha$, $d(x_alpha,x_alpha')<delta$,so $rho(f(x_alpha),f(x_alpha'))<epsilon$. So,$Ksubsetcup_alpha X_alpha$,and $y=f(x)in Y$,so $f(K)subsetcup_alpha f(X_alpha)$,$f(K)subset Y$ is also compact.
This proof may have some wrong...
continuity metric-spaces compactness
asked yesterday
HUANG Fuzhe
61
New contributor
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $f:(M,d)to(Y,rho)$ be a continuous function.If $Ksubset M$ is compact,then $f(K)subset Y$is also compact.
My attempt:
I know that f is continuous function, so there is a $xin X$, if for any $epsilon>0$ ,there is a $delta>0$,s.t. $d(x,x')<delta$,so $rho(f(x),f(x'))<epsilon$.
Then $Ksubset M$ is compact. so there is a collection {$X_alpha$} is an open cover of $K$,$Ksubset cup_alpha X_alpha$.
Thus, if for any $x_alphain X_alpha$, $d(x_alpha,x_alpha')<delta$,so $rho(f(x_alpha),f(x_alpha'))<epsilon$. So,$Ksubsetcup_alpha X_alpha$,and $y=f(x)in Y$,so $f(K)subsetcup_alpha f(X_alpha)$,$f(K)subset Y$ is also compact.
This proof may have some wrong...
continuity metric-spaces compactness
continuity metric-spaces compactness
asked yesterday
HUANG Fuzhe
61
New contributor
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
HUANG Fuzhe
61
New contributor
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
HUANG Fuzhe
61
New contributor
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
HUANG Fuzhe
61
asked yesterday
HUANG Fuzhe
61
61
New contributor
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
yesterday
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
yesterday
Regarding your works, that is not the definition of compactness. A subset $K$ is compact iff for every open cover for $K$, there is a finite subcover of the open cover. To show $f(K)$ is compact, you need to show that any open cover ${U_{alpha} }$ has a finite subcover ${U_{alpha_1},dots,U_{alpha_n}}$.
– Kelvin Lois
yesterday
add a comment |
1
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
yesterday
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
yesterday
Regarding your works, that is not the definition of compactness. A subset $K$ is compact iff for every open cover for $K$, there is a finite subcover of the open cover. To show $f(K)$ is compact, you need to show that any open cover ${U_{alpha} }$ has a finite subcover ${U_{alpha_1},dots,U_{alpha_n}}$.
– Kelvin Lois
yesterday
1
1
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
yesterday
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
yesterday
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
yesterday
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
yesterday
Regarding your works, that is not the definition of compactness. A subset $K$ is compact iff for every open cover for $K$, there is a finite subcover of the open cover. To show $f(K)$ is compact, you need to show that any open cover ${U_{alpha} }$ has a finite subcover ${U_{alpha_1},dots,U_{alpha_n}}$.
– Kelvin Lois
yesterday
Regarding your works, that is not the definition of compactness. A subset $K$ is compact iff for every open cover for $K$, there is a finite subcover of the open cover. To show $f(K)$ is compact, you need to show that any open cover ${U_{alpha} }$ has a finite subcover ${U_{alpha_1},dots,U_{alpha_n}}$.
– Kelvin Lois
yesterday
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
HUANG Fuzhe is a new contributor. Be nice, and check out our Code of Conduct.
HUANG Fuzhe is a new contributor. Be nice, and check out our Code of Conduct.
HUANG Fuzhe is a new contributor. Be nice, and check out our Code of Conduct.
HUANG Fuzhe is a new contributor. Be nice, and check out our Code of Conduct.
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%2f2999335%2fproof-let-fm-d-toy-rho-be-a-continuous-function-if-k-subset-m-is-com%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
1
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
yesterday
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
yesterday
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
yesterday
Regarding your works, that is not the definition of compactness. A subset $K$ is compact iff for every open cover for $K$, there is a finite subcover of the open cover. To show $f(K)$ is compact, you need to show that any open cover ${U_{alpha} }$ has a finite subcover ${U_{alpha_1},dots,U_{alpha_n}}$.
– Kelvin Lois
yesterday