Proof: Let $f:(M,d)to(Y,rho)$ be a continuous function. If $Ksubset M$ is compact,then $f(K)subset Y$ is also...
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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 2 days ago
HUANG Fuzhe
61
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HUANG Fuzhe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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0
down vote
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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 2 days ago
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
2 days ago
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
2 days ago
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
2 days ago
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
2 days ago
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 2 days ago
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 2 days ago
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 2 days ago
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 2 days ago
asked 2 days ago
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 2 days ago
HUANG Fuzhe
61
asked 2 days ago
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
2 days ago
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
2 days ago
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
2 days ago
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
2 days ago
add a comment |
1
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
2 days ago
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
2 days ago
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
2 days ago
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
2 days ago
1
1
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
2 days ago
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
2 days ago
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
2 days ago
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
2 days ago
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
2 days ago
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
2 days ago
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
2 days ago
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
2 days ago
add a comment |
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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.
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1
What have you tried? Please add your attempts to your post, otherwise your post would be closed soon.
– xbh
2 days ago
Welcome to MSE. Please read this text about how to ask a good question.
– José Carlos Santos
2 days ago
Possible duplicate of Proving continuous image of compact sets are compact
– José Carlos Santos
2 days ago
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
2 days ago