If $C$ is connected subset of a disconnected metric space $X=Acup B$ then either $Csubseteq A $ or $C...
If $C$ is a connected subset of a disconnected metric space $X=Acup
B$ where $overline{A}cap B = Acap overline{B} = phi$ then either
$Csubseteq A $ or $C subseteq B$
If $Csubseteq A $ or $C subseteq B$ then we're done.
Assume neither $Csubseteq A $ or $C subseteq B$
$Rightarrow ; ; exists ; x,y in X $ such that
$xin Ccap A$ and $y in Ccap B$, obviously $xneq y $ since $Acap B= phi$
We have $Ccap A subseteq C$ and $Ccap B subseteq C$ and both are non empty.
C=$Ccap X = Ccap (Acup B) = (Ccap A)cup (Ccap B)$
Also,
$Ccap A subseteq A$ and
$Ccap B subseteq B$ and since $Acap B = phi $
$(Ccap A)cap (Ccap B) = phi$
Here is where I am having problem :-
Can I directly say that $(Ccap A)$ and $(Ccap B)$ are open in C?
Because that will immediately show that C is disconnected and we'd have the required contradiction.
general-topology proof-verification metric-spaces proof-writing
asked Nov 26 '18 at 14:01
Vishweshwar Tyagi
381211
add a comment |
If $C$ is a connected subset of a disconnected metric space $X=Acup
B$ where $overline{A}cap B = Acap overline{B} = phi$ then either
$Csubseteq A $ or $C subseteq B$
If $Csubseteq A $ or $C subseteq B$ then we're done.
Assume neither $Csubseteq A $ or $C subseteq B$
$Rightarrow ; ; exists ; x,y in X $ such that
$xin Ccap A$ and $y in Ccap B$, obviously $xneq y $ since $Acap B= phi$
We have $Ccap A subseteq C$ and $Ccap B subseteq C$ and both are non empty.
C=$Ccap X = Ccap (Acup B) = (Ccap A)cup (Ccap B)$
Also,
$Ccap A subseteq A$ and
$Ccap B subseteq B$ and since $Acap B = phi $
$(Ccap A)cap (Ccap B) = phi$
Here is where I am having problem :-
Can I directly say that $(Ccap A)$ and $(Ccap B)$ are open in C?
Because that will immediately show that C is disconnected and we'd have the required contradiction.
general-topology proof-verification metric-spaces proof-writing
asked Nov 26 '18 at 14:01
Vishweshwar Tyagi
381211
It might be helpful to recall what it means for a set to be open in $C$ (in the subspace topology).
– Xander Henderson
Nov 26 '18 at 14:03
add a comment |
If $C$ is a connected subset of a disconnected metric space $X=Acup
B$ where $overline{A}cap B = Acap overline{B} = phi$ then either
$Csubseteq A $ or $C subseteq B$
If $Csubseteq A $ or $C subseteq B$ then we're done.
Assume neither $Csubseteq A $ or $C subseteq B$
$Rightarrow ; ; exists ; x,y in X $ such that
$xin Ccap A$ and $y in Ccap B$, obviously $xneq y $ since $Acap B= phi$
We have $Ccap A subseteq C$ and $Ccap B subseteq C$ and both are non empty.
C=$Ccap X = Ccap (Acup B) = (Ccap A)cup (Ccap B)$
Also,
$Ccap A subseteq A$ and
$Ccap B subseteq B$ and since $Acap B = phi $
$(Ccap A)cap (Ccap B) = phi$
Here is where I am having problem :-
Can I directly say that $(Ccap A)$ and $(Ccap B)$ are open in C?
Because that will immediately show that C is disconnected and we'd have the required contradiction.
general-topology proof-verification metric-spaces proof-writing
asked Nov 26 '18 at 14:01
Vishweshwar Tyagi
381211
If $C$ is a connected subset of a disconnected metric space $X=Acup
B$ where $overline{A}cap B = Acap overline{B} = phi$ then either
$Csubseteq A $ or $C subseteq B$
If $Csubseteq A $ or $C subseteq B$ then we're done.
Assume neither $Csubseteq A $ or $C subseteq B$
$Rightarrow ; ; exists ; x,y in X $ such that
$xin Ccap A$ and $y in Ccap B$, obviously $xneq y $ since $Acap B= phi$
We have $Ccap A subseteq C$ and $Ccap B subseteq C$ and both are non empty.
C=$Ccap X = Ccap (Acup B) = (Ccap A)cup (Ccap B)$
Also,
$Ccap A subseteq A$ and
$Ccap B subseteq B$ and since $Acap B = phi $
$(Ccap A)cap (Ccap B) = phi$
Here is where I am having problem :-
Can I directly say that $(Ccap A)$ and $(Ccap B)$ are open in C?
Because that will immediately show that C is disconnected and we'd have the required contradiction.
general-topology proof-verification metric-spaces proof-writing
general-topology proof-verification metric-spaces proof-writing
asked Nov 26 '18 at 14:01
Vishweshwar Tyagi
381211
asked Nov 26 '18 at 14:01
Vishweshwar Tyagi
381211
asked Nov 26 '18 at 14:01
Vishweshwar Tyagi
381211
asked Nov 26 '18 at 14:01
Vishweshwar Tyagi
381211
asked Nov 26 '18 at 14:01
Vishweshwar Tyagi
381211
381211
It might be helpful to recall what it means for a set to be open in $C$ (in the subspace topology).
– Xander Henderson
Nov 26 '18 at 14:03
add a comment |
It might be helpful to recall what it means for a set to be open in $C$ (in the subspace topology).
– Xander Henderson
Nov 26 '18 at 14:03
It might be helpful to recall what it means for a set to be open in $C$ (in the subspace topology).
– Xander Henderson
Nov 26 '18 at 14:03
It might be helpful to recall what it means for a set to be open in $C$ (in the subspace topology).
– Xander Henderson
Nov 26 '18 at 14:03
add a comment |
2 Answers
2
active
oldest
votes
Since $Acap overline B=emptyset$ and $X=Acup B$, then $B=overline B$ is closed and $A$ is open. Therefore $Ccap A$ is open in $C$ (because open subsets of $C$ are just intersection of $C$ with open subset of $X$.)
The same reasoning shows that $Ccap B$ is open.
answered Nov 26 '18 at 14:36
user126154
5,378716
add a comment |
Observe that $B$ is the complement of closed set $overline A$ hence is open.
Similarly $A$ is open.
Then $Ccap A$ and $Ccap B$ are open in $C$.
You can now finish your proof by stating that one of them must be empty (because $C$ is connected) so that $Csubseteq A$ or $Csubseteq B$.
answered Nov 26 '18 at 14:36
drhab
98.1k544129
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since $Acap overline B=emptyset$ and $X=Acup B$, then $B=overline B$ is closed and $A$ is open. Therefore $Ccap A$ is open in $C$ (because open subsets of $C$ are just intersection of $C$ with open subset of $X$.)
The same reasoning shows that $Ccap B$ is open.
answered Nov 26 '18 at 14:36
user126154
5,378716
add a comment |
Since $Acap overline B=emptyset$ and $X=Acup B$, then $B=overline B$ is closed and $A$ is open. Therefore $Ccap A$ is open in $C$ (because open subsets of $C$ are just intersection of $C$ with open subset of $X$.)
The same reasoning shows that $Ccap B$ is open.
answered Nov 26 '18 at 14:36
user126154
5,378716
add a comment |
Since $Acap overline B=emptyset$ and $X=Acup B$, then $B=overline B$ is closed and $A$ is open. Therefore $Ccap A$ is open in $C$ (because open subsets of $C$ are just intersection of $C$ with open subset of $X$.)
The same reasoning shows that $Ccap B$ is open.
answered Nov 26 '18 at 14:36
user126154
5,378716
Since $Acap overline B=emptyset$ and $X=Acup B$, then $B=overline B$ is closed and $A$ is open. Therefore $Ccap A$ is open in $C$ (because open subsets of $C$ are just intersection of $C$ with open subset of $X$.)
The same reasoning shows that $Ccap B$ is open.
answered Nov 26 '18 at 14:36
user126154
5,378716
answered Nov 26 '18 at 14:36
user126154
5,378716
answered Nov 26 '18 at 14:36
user126154
5,378716
answered Nov 26 '18 at 14:36
user126154
5,378716
5,378716
add a comment |
add a comment |
Observe that $B$ is the complement of closed set $overline A$ hence is open.
Similarly $A$ is open.
Then $Ccap A$ and $Ccap B$ are open in $C$.
You can now finish your proof by stating that one of them must be empty (because $C$ is connected) so that $Csubseteq A$ or $Csubseteq B$.
answered Nov 26 '18 at 14:36
drhab
98.1k544129
add a comment |
Observe that $B$ is the complement of closed set $overline A$ hence is open.
Similarly $A$ is open.
Then $Ccap A$ and $Ccap B$ are open in $C$.
You can now finish your proof by stating that one of them must be empty (because $C$ is connected) so that $Csubseteq A$ or $Csubseteq B$.
answered Nov 26 '18 at 14:36
drhab
98.1k544129
add a comment |
Observe that $B$ is the complement of closed set $overline A$ hence is open.
Similarly $A$ is open.
Then $Ccap A$ and $Ccap B$ are open in $C$.
You can now finish your proof by stating that one of them must be empty (because $C$ is connected) so that $Csubseteq A$ or $Csubseteq B$.
answered Nov 26 '18 at 14:36
drhab
98.1k544129
Observe that $B$ is the complement of closed set $overline A$ hence is open.
Similarly $A$ is open.
Then $Ccap A$ and $Ccap B$ are open in $C$.
You can now finish your proof by stating that one of them must be empty (because $C$ is connected) so that $Csubseteq A$ or $Csubseteq B$.
answered Nov 26 '18 at 14:36
drhab
98.1k544129
edited Nov 26 '18 at 14:45
answered Nov 26 '18 at 14:36
drhab
98.1k544129
answered Nov 26 '18 at 14:36
drhab
98.1k544129
answered Nov 26 '18 at 14:36
drhab
98.1k544129
98.1k544129
add a comment |
add a comment |
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It might be helpful to recall what it means for a set to be open in $C$ (in the subspace topology).
– Xander Henderson
Nov 26 '18 at 14:03