Mixing
Sufficient condition for onto.
Clash Royale CLAN TAG#URR8PPP
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I want to ask you for following statement
f is continuous from X to Y, where X is compact Hausdorff, connected and Y is ordered set in order topology, then f is onto?
My attempt
For given $y in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff,
$B=f^-1(bar N_y$) is compact in X , Where $bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* in B$ such that $f(x*)=y$ . Thus $f$ is onto
First, I want to ask whether this proof is correct or not.
Second, Is there any counter example if X has only connect condition ?
Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$
Thank you!
general-topology
asked 1 hour ago
fivestar
31519
add a comment |Â
up vote
1
down vote
favorite
I want to ask you for following statement
f is continuous from X to Y, where X is compact Hausdorff, connected and Y is ordered set in order topology, then f is onto?
My attempt
For given $y in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff,
$B=f^-1(bar N_y$) is compact in X , Where $bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* in B$ such that $f(x*)=y$ . Thus $f$ is onto
First, I want to ask whether this proof is correct or not.
Second, Is there any counter example if X has only connect condition ?
Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$
Thank you!
general-topology
asked 1 hour ago
fivestar
31519
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to ask you for following statement
f is continuous from X to Y, where X is compact Hausdorff, connected and Y is ordered set in order topology, then f is onto?
My attempt
For given $y in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff,
$B=f^-1(bar N_y$) is compact in X , Where $bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* in B$ such that $f(x*)=y$ . Thus $f$ is onto
First, I want to ask whether this proof is correct or not.
Second, Is there any counter example if X has only connect condition ?
Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$
Thank you!
general-topology
asked 1 hour ago
fivestar
31519
I want to ask you for following statement
f is continuous from X to Y, where X is compact Hausdorff, connected and Y is ordered set in order topology, then f is onto?
My attempt
For given $y in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff,
$B=f^-1(bar N_y$) is compact in X , Where $bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* in B$ such that $f(x*)=y$ . Thus $f$ is onto
First, I want to ask whether this proof is correct or not.
Second, Is there any counter example if X has only connect condition ?
Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$
Thank you!
general-topology
general-topology
asked 1 hour ago
fivestar
31519
asked 1 hour ago
fivestar
31519
asked 1 hour ago
fivestar
31519
asked 1 hour ago
fivestar
31519
asked 1 hour ago
fivestar
31519
31519
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.
A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).
answered 1 hour ago
Mees de Vries
15k12450
Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
– fivestar
1 hour ago
add a comment |Â
up vote
2
down vote
Your proof is not correct as pointed out by Mees de Vries.
However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.
answered 1 hour ago
sqtrat
1,888518
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.
A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).
answered 1 hour ago
Mees de Vries
15k12450
Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
– fivestar
1 hour ago
add a comment |Â
up vote
3
down vote
Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.
A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).
answered 1 hour ago
Mees de Vries
15k12450
Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
– fivestar
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.
A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).
answered 1 hour ago
Mees de Vries
15k12450
Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.
A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).
answered 1 hour ago
Mees de Vries
15k12450
answered 1 hour ago
Mees de Vries
15k12450
answered 1 hour ago
Mees de Vries
15k12450
answered 1 hour ago
Mees de Vries
15k12450
15k12450
Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
– fivestar
1 hour ago
add a comment |Â
Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
– fivestar
1 hour ago
Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
– fivestar
1 hour ago
Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
– fivestar
1 hour ago
add a comment |Â
up vote
2
down vote
Your proof is not correct as pointed out by Mees de Vries.
However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.
answered 1 hour ago
sqtrat
1,888518
add a comment |Â
up vote
2
down vote
Your proof is not correct as pointed out by Mees de Vries.
However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.
answered 1 hour ago
sqtrat
1,888518
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your proof is not correct as pointed out by Mees de Vries.
However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.
answered 1 hour ago
sqtrat
1,888518
Your proof is not correct as pointed out by Mees de Vries.
However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.
answered 1 hour ago
sqtrat
1,888518
answered 1 hour ago
sqtrat
1,888518
answered 1 hour ago
sqtrat
1,888518
answered 1 hour ago
sqtrat
1,888518
1,888518
add a comment |Â
add a comment |Â
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