Prove that $G$ is an open interval for two-valued continuous function $f$
Suppose $Gsubset mathbb R$ is a non-empty open set such that the function $f:G rightarrow {0,1}$ is a two-valued function and is continuous. Show that any two-valued function on $G$ is a constant if and only if the set $G$ is an open interval.
I am concerning about the construction of open set for set $G$ and have no idea to start off with.
general-topology continuity metric-spaces
asked Dec 1 at 3:07
weilam06
11910
add a comment |
Suppose $Gsubset mathbb R$ is a non-empty open set such that the function $f:G rightarrow {0,1}$ is a two-valued function and is continuous. Show that any two-valued function on $G$ is a constant if and only if the set $G$ is an open interval.
I am concerning about the construction of open set for set $G$ and have no idea to start off with.
general-topology continuity metric-spaces
asked Dec 1 at 3:07
weilam06
11910
Do you know about this fact? If $f:Xrightarrow Y$ is continuous and $X$ is connected, then $f(X)$ is connected
– user25959
Dec 1 at 3:11
I don't know this fact, though.
– weilam06
Dec 1 at 3:13
So I have to prove the subset $G$ is connected in order to proceed?
– weilam06
Dec 1 at 3:36
1
You can also use the intermediate value theorem.
– user25959
Dec 1 at 3:46
I would like to listen to the idea that IVT works in the proof here.
– weilam06
Dec 1 at 3:52
add a comment |
Suppose $Gsubset mathbb R$ is a non-empty open set such that the function $f:G rightarrow {0,1}$ is a two-valued function and is continuous. Show that any two-valued function on $G$ is a constant if and only if the set $G$ is an open interval.
I am concerning about the construction of open set for set $G$ and have no idea to start off with.
general-topology continuity metric-spaces
asked Dec 1 at 3:07
weilam06
11910
Suppose $Gsubset mathbb R$ is a non-empty open set such that the function $f:G rightarrow {0,1}$ is a two-valued function and is continuous. Show that any two-valued function on $G$ is a constant if and only if the set $G$ is an open interval.
I am concerning about the construction of open set for set $G$ and have no idea to start off with.
general-topology continuity metric-spaces
general-topology continuity metric-spaces
asked Dec 1 at 3:07
weilam06
11910
asked Dec 1 at 3:07
weilam06
11910
edited Dec 1 at 3:31
asked Dec 1 at 3:07
weilam06
11910
asked Dec 1 at 3:07
weilam06
11910
asked Dec 1 at 3:07
weilam06
11910
11910
Do you know about this fact? If $f:Xrightarrow Y$ is continuous and $X$ is connected, then $f(X)$ is connected
– user25959
Dec 1 at 3:11
I don't know this fact, though.
– weilam06
Dec 1 at 3:13
So I have to prove the subset $G$ is connected in order to proceed?
– weilam06
Dec 1 at 3:36
1
You can also use the intermediate value theorem.
– user25959
Dec 1 at 3:46
I would like to listen to the idea that IVT works in the proof here.
– weilam06
Dec 1 at 3:52
add a comment |
Do you know about this fact? If $f:Xrightarrow Y$ is continuous and $X$ is connected, then $f(X)$ is connected
– user25959
Dec 1 at 3:11
I don't know this fact, though.
– weilam06
Dec 1 at 3:13
So I have to prove the subset $G$ is connected in order to proceed?
– weilam06
Dec 1 at 3:36
1
You can also use the intermediate value theorem.
– user25959
Dec 1 at 3:46
I would like to listen to the idea that IVT works in the proof here.
– weilam06
Dec 1 at 3:52
Do you know about this fact? If $f:Xrightarrow Y$ is continuous and $X$ is connected, then $f(X)$ is connected
– user25959
Dec 1 at 3:11
Do you know about this fact? If $f:Xrightarrow Y$ is continuous and $X$ is connected, then $f(X)$ is connected
– user25959
Dec 1 at 3:11
I don't know this fact, though.
– weilam06
Dec 1 at 3:13
I don't know this fact, though.
– weilam06
Dec 1 at 3:13
So I have to prove the subset $G$ is connected in order to proceed?
– weilam06
Dec 1 at 3:36
So I have to prove the subset $G$ is connected in order to proceed?
– weilam06
Dec 1 at 3:36
1
1
You can also use the intermediate value theorem.
– user25959
Dec 1 at 3:46
You can also use the intermediate value theorem.
– user25959
Dec 1 at 3:46
I would like to listen to the idea that IVT works in the proof here.
– weilam06
Dec 1 at 3:52
I would like to listen to the idea that IVT works in the proof here.
– weilam06
Dec 1 at 3:52
add a comment |
2 Answers
2
active
oldest
votes
Theorem: A metric space $X$ is connected if and only if any continuous function $f:Xto {0,1}$ is constant.
Proof:Suppose $X$ is connected and $f:Xto {0,1}$ is continuous. If $f$ is not a constant function, then $f$ is onto. Let $A=f^{-1}(0)$ and $B=f^{-1}(1)$. Then $Acup B=X$,and $A,Bneq emptyset$. Also note that both are proper subsets of $X$ and are open and closed in $X$, a contradiction.
Suppose $X$ is not connected. Let $A$ and $B$ be the disconnection. Then define $f:Xto {0,1}$ such that $$f(x)=begin{cases}
0, &text{if $xin A$}\
1,&text{if $xin B$}
end{cases}$$
$f$ is a non-constant continuous function(verify).
Another useful theorem is
A subset $I$ of $Bbb{R}$ is connected if and only if $I$ is an
interval.
You can find a proof here.
I hope now you can complete your answer on your own.
answered Dec 1 at 4:08
Thomas Shelby
1,443216
add a comment |
If $G$ is an open interval then IVP tells you that if $f$ if takes both the values $0$ and $1$ then it must take all values in between. This proves one way.
For the other way use the fact any open set is a disjoint union of open intervals. If $G$ is a union of two or mote intervals define $f$ to be $1$ on one of them and $0$ on all the others. You will get a continuous function taking both the values $0$ and $1$ (and no other value).
answered Dec 1 at 4:57
Kavi Rama Murthy
49.9k31854
add a comment |
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2 Answers
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2 Answers
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oldest
votes
Theorem: A metric space $X$ is connected if and only if any continuous function $f:Xto {0,1}$ is constant.
Proof:Suppose $X$ is connected and $f:Xto {0,1}$ is continuous. If $f$ is not a constant function, then $f$ is onto. Let $A=f^{-1}(0)$ and $B=f^{-1}(1)$. Then $Acup B=X$,and $A,Bneq emptyset$. Also note that both are proper subsets of $X$ and are open and closed in $X$, a contradiction.
Suppose $X$ is not connected. Let $A$ and $B$ be the disconnection. Then define $f:Xto {0,1}$ such that $$f(x)=begin{cases}
0, &text{if $xin A$}\
1,&text{if $xin B$}
end{cases}$$
$f$ is a non-constant continuous function(verify).
Another useful theorem is
A subset $I$ of $Bbb{R}$ is connected if and only if $I$ is an
interval.
You can find a proof here.
I hope now you can complete your answer on your own.
answered Dec 1 at 4:08
Thomas Shelby
1,443216
add a comment |
Theorem: A metric space $X$ is connected if and only if any continuous function $f:Xto {0,1}$ is constant.
Proof:Suppose $X$ is connected and $f:Xto {0,1}$ is continuous. If $f$ is not a constant function, then $f$ is onto. Let $A=f^{-1}(0)$ and $B=f^{-1}(1)$. Then $Acup B=X$,and $A,Bneq emptyset$. Also note that both are proper subsets of $X$ and are open and closed in $X$, a contradiction.
Suppose $X$ is not connected. Let $A$ and $B$ be the disconnection. Then define $f:Xto {0,1}$ such that $$f(x)=begin{cases}
0, &text{if $xin A$}\
1,&text{if $xin B$}
end{cases}$$
$f$ is a non-constant continuous function(verify).
Another useful theorem is
A subset $I$ of $Bbb{R}$ is connected if and only if $I$ is an
interval.
You can find a proof here.
I hope now you can complete your answer on your own.
answered Dec 1 at 4:08
Thomas Shelby
1,443216
add a comment |
Theorem: A metric space $X$ is connected if and only if any continuous function $f:Xto {0,1}$ is constant.
Proof:Suppose $X$ is connected and $f:Xto {0,1}$ is continuous. If $f$ is not a constant function, then $f$ is onto. Let $A=f^{-1}(0)$ and $B=f^{-1}(1)$. Then $Acup B=X$,and $A,Bneq emptyset$. Also note that both are proper subsets of $X$ and are open and closed in $X$, a contradiction.
Suppose $X$ is not connected. Let $A$ and $B$ be the disconnection. Then define $f:Xto {0,1}$ such that $$f(x)=begin{cases}
0, &text{if $xin A$}\
1,&text{if $xin B$}
end{cases}$$
$f$ is a non-constant continuous function(verify).
Another useful theorem is
A subset $I$ of $Bbb{R}$ is connected if and only if $I$ is an
interval.
You can find a proof here.
I hope now you can complete your answer on your own.
answered Dec 1 at 4:08
Thomas Shelby
1,443216
Theorem: A metric space $X$ is connected if and only if any continuous function $f:Xto {0,1}$ is constant.
Proof:Suppose $X$ is connected and $f:Xto {0,1}$ is continuous. If $f$ is not a constant function, then $f$ is onto. Let $A=f^{-1}(0)$ and $B=f^{-1}(1)$. Then $Acup B=X$,and $A,Bneq emptyset$. Also note that both are proper subsets of $X$ and are open and closed in $X$, a contradiction.
Suppose $X$ is not connected. Let $A$ and $B$ be the disconnection. Then define $f:Xto {0,1}$ such that $$f(x)=begin{cases}
0, &text{if $xin A$}\
1,&text{if $xin B$}
end{cases}$$
$f$ is a non-constant continuous function(verify).
Another useful theorem is
A subset $I$ of $Bbb{R}$ is connected if and only if $I$ is an
interval.
You can find a proof here.
I hope now you can complete your answer on your own.
answered Dec 1 at 4:08
Thomas Shelby
1,443216
edited Dec 1 at 4:19
answered Dec 1 at 4:08
Thomas Shelby
1,443216
answered Dec 1 at 4:08
Thomas Shelby
1,443216
answered Dec 1 at 4:08
Thomas Shelby
1,443216
1,443216
add a comment |
add a comment |
If $G$ is an open interval then IVP tells you that if $f$ if takes both the values $0$ and $1$ then it must take all values in between. This proves one way.
For the other way use the fact any open set is a disjoint union of open intervals. If $G$ is a union of two or mote intervals define $f$ to be $1$ on one of them and $0$ on all the others. You will get a continuous function taking both the values $0$ and $1$ (and no other value).
answered Dec 1 at 4:57
Kavi Rama Murthy
49.9k31854
add a comment |
If $G$ is an open interval then IVP tells you that if $f$ if takes both the values $0$ and $1$ then it must take all values in between. This proves one way.
For the other way use the fact any open set is a disjoint union of open intervals. If $G$ is a union of two or mote intervals define $f$ to be $1$ on one of them and $0$ on all the others. You will get a continuous function taking both the values $0$ and $1$ (and no other value).
answered Dec 1 at 4:57
Kavi Rama Murthy
49.9k31854
add a comment |
If $G$ is an open interval then IVP tells you that if $f$ if takes both the values $0$ and $1$ then it must take all values in between. This proves one way.
For the other way use the fact any open set is a disjoint union of open intervals. If $G$ is a union of two or mote intervals define $f$ to be $1$ on one of them and $0$ on all the others. You will get a continuous function taking both the values $0$ and $1$ (and no other value).
answered Dec 1 at 4:57
Kavi Rama Murthy
49.9k31854
If $G$ is an open interval then IVP tells you that if $f$ if takes both the values $0$ and $1$ then it must take all values in between. This proves one way.
For the other way use the fact any open set is a disjoint union of open intervals. If $G$ is a union of two or mote intervals define $f$ to be $1$ on one of them and $0$ on all the others. You will get a continuous function taking both the values $0$ and $1$ (and no other value).
answered Dec 1 at 4:57
Kavi Rama Murthy
49.9k31854
answered Dec 1 at 4:57
Kavi Rama Murthy
49.9k31854
answered Dec 1 at 4:57
Kavi Rama Murthy
49.9k31854
answered Dec 1 at 4:57
Kavi Rama Murthy
49.9k31854
49.9k31854
add a comment |
add a comment |
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Do you know about this fact? If $f:Xrightarrow Y$ is continuous and $X$ is connected, then $f(X)$ is connected
– user25959
Dec 1 at 3:11
I don't know this fact, though.
– weilam06
Dec 1 at 3:13
So I have to prove the subset $G$ is connected in order to proceed?
– weilam06
Dec 1 at 3:36
1
You can also use the intermediate value theorem.
– user25959
Dec 1 at 3:46
I would like to listen to the idea that IVT works in the proof here.
– weilam06
Dec 1 at 3:52