Suppose $Ysubset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, prove that $Ycup A$ and...
Question is :
Suppose $Ysubset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, Prove that $Ycup A$ and $Ycup B$ are connected.
What i have tried is :
Suppose $Ycup A$ has separation say $Ccup D$ then all i could see is that
Either $Ysubset C$ or $Ysubset D$ as $Y$ is connected and $
Ccup D$ is separation for a subset that contains $Y$.
Without loss of generality we could assume $Ysubset C$
As $Ycup A=Ccup D$ and $Ysubset C$ i can say $Dsubset A$ (I do not know how does this help)
I have $X-Y=Acup B$ with $Acap B=emptyset$
I have not used connectedness of $X$ till now, So i thought of using that and end up with following :
$X-Y=Acup BRightarrow X=Ycup Acup B=(Ycup A)cup (Ycup B)$
I do not know what to conclude from this...
I would be thankful if some one can help me by giving some "hints"
Thank you
general-topology connectedness
edited Jun 27 '16 at 10:47
Irregular User
2,86251843
add a comment |
Question is :
Suppose $Ysubset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, Prove that $Ycup A$ and $Ycup B$ are connected.
What i have tried is :
Suppose $Ycup A$ has separation say $Ccup D$ then all i could see is that
Either $Ysubset C$ or $Ysubset D$ as $Y$ is connected and $
Ccup D$ is separation for a subset that contains $Y$.
Without loss of generality we could assume $Ysubset C$
As $Ycup A=Ccup D$ and $Ysubset C$ i can say $Dsubset A$ (I do not know how does this help)
I have $X-Y=Acup B$ with $Acap B=emptyset$
I have not used connectedness of $X$ till now, So i thought of using that and end up with following :
$X-Y=Acup BRightarrow X=Ycup Acup B=(Ycup A)cup (Ycup B)$
I do not know what to conclude from this...
I would be thankful if some one can help me by giving some "hints"
Thank you
general-topology connectedness
edited Jun 27 '16 at 10:47
Irregular User
2,86251843
add a comment |
Question is :
Suppose $Ysubset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, Prove that $Ycup A$ and $Ycup B$ are connected.
What i have tried is :
Suppose $Ycup A$ has separation say $Ccup D$ then all i could see is that
Either $Ysubset C$ or $Ysubset D$ as $Y$ is connected and $
Ccup D$ is separation for a subset that contains $Y$.
Without loss of generality we could assume $Ysubset C$
As $Ycup A=Ccup D$ and $Ysubset C$ i can say $Dsubset A$ (I do not know how does this help)
I have $X-Y=Acup B$ with $Acap B=emptyset$
I have not used connectedness of $X$ till now, So i thought of using that and end up with following :
$X-Y=Acup BRightarrow X=Ycup Acup B=(Ycup A)cup (Ycup B)$
I do not know what to conclude from this...
I would be thankful if some one can help me by giving some "hints"
Thank you
general-topology connectedness
edited Jun 27 '16 at 10:47
Irregular User
2,86251843
Question is :
Suppose $Ysubset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, Prove that $Ycup A$ and $Ycup B$ are connected.
What i have tried is :
Suppose $Ycup A$ has separation say $Ccup D$ then all i could see is that
Either $Ysubset C$ or $Ysubset D$ as $Y$ is connected and $
Ccup D$ is separation for a subset that contains $Y$.
Without loss of generality we could assume $Ysubset C$
As $Ycup A=Ccup D$ and $Ysubset C$ i can say $Dsubset A$ (I do not know how does this help)
I have $X-Y=Acup B$ with $Acap B=emptyset$
I have not used connectedness of $X$ till now, So i thought of using that and end up with following :
$X-Y=Acup BRightarrow X=Ycup Acup B=(Ycup A)cup (Ycup B)$
I do not know what to conclude from this...
I would be thankful if some one can help me by giving some "hints"
Thank you
general-topology connectedness
general-topology connectedness
edited Jun 27 '16 at 10:47
Irregular User
2,86251843
edited Jun 27 '16 at 10:47
Irregular User
2,86251843
edited Jun 27 '16 at 10:47
Irregular User
2,86251843
edited Jun 27 '16 at 10:47
Irregular User
2,86251843
edited Jun 27 '16 at 10:47
Irregular User
2,86251843
2,86251843
asked Apr 2 '14 at 11:34
user87543
add a comment |
add a comment |
2 Answers
2
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oldest
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As $A,B$ form a separation for $X-Y$ and $X$ is connected, $Y$ must be non-empty.
You suppose that $Ycup A$ has a separation $Ccup D$. Then $Y$, being connected, is a subset of $C$
Now $A$ is clopen in $X-Y$, that means there are sets $U$ open and $K$ closed, such that $U-Y=K-Y=A$. You know that $Dsubseteq A$ is clopen in $Ycup A$. Can you show that $D$ is open in $U$ and closed in $K$?
answered Apr 2 '14 at 14:09
Stefan HamckeStefan Hamcke
21.6k42877
Second paragraph is confusing me... could you please edit it a bit...
– user87543
Apr 3 '14 at 4:40
@PraphullaKoushik: Which sentence confuses you?
– Stefan Hamcke
Apr 3 '14 at 14:14
1
Well, I should say I had already solved this problem some time ago, then forgot the proof so I had to solve it again, but it wasn"t entirely new to me. When I deal with such a problem I usually draw a picture and draw into it the sets that I can think of, in order to see how they relate. Here it was a large disk $X$, a smaller disk $Y$ within $X$, and two lines from the outer circle to the inner circle which you can think of separating $X-Y$ into $A$ and $B$. Then I separated $Acup Y$ into $D$ and $C$ where $D$ was just a subset of $A$ ... (to be continued)
– Stefan Hamcke
Apr 3 '14 at 14:54
2
@MathsLover: Actually, I don't define anything here. $A$ is clopen in $X-Y$ if it's both open and closed in $X-Y$. And this implies the existence of the sets $U$ and $K$.
– Stefan Hamcke
Aug 21 '14 at 16:31
1
@MathsLover: Yes, that argument is right :)
– Stefan Hamcke
Aug 21 '14 at 19:12
|
show 9 more comments
Continuing with your argument, We can prove that $B cup C$ and $D$ form the partition of $X$.(Assuming your conclusion that $D subset A$).
We have $X = (B cup C) cup D$.
$C$ and $D$ form separation of $Y cup A$. So, using theorem (23.1), no limit point of $C$ is in $D$. Similarly, no limit point of $B$ is in $D$ since $D subset A$ and $A$ and $B$ form separation of $X-Y$. From this, we can easily say that no limit point of $B cup C$ is in $D$ i.e. Every limit point of $B cup C$ wrt $X$ is in $B cup C$. So, $B cup C$ is closed in $X$ and $D$ is open in $X$.
Now, use a similar argument that, no limit point of $D$ is in $B cup C$ implying that $D$ is closed in $X$ and $Bcup C$ is open in X. Hence, $B cup C$ and $D$ are $textit{clopen}$ in $X$ and form partition of $X$.
answered Jan 4 at 8:46
MUHMUH
395216
add a comment |
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2 Answers
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2 Answers
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As $A,B$ form a separation for $X-Y$ and $X$ is connected, $Y$ must be non-empty.
You suppose that $Ycup A$ has a separation $Ccup D$. Then $Y$, being connected, is a subset of $C$
Now $A$ is clopen in $X-Y$, that means there are sets $U$ open and $K$ closed, such that $U-Y=K-Y=A$. You know that $Dsubseteq A$ is clopen in $Ycup A$. Can you show that $D$ is open in $U$ and closed in $K$?
answered Apr 2 '14 at 14:09
Stefan HamckeStefan Hamcke
21.6k42877
Second paragraph is confusing me... could you please edit it a bit...
– user87543
Apr 3 '14 at 4:40
@PraphullaKoushik: Which sentence confuses you?
– Stefan Hamcke
Apr 3 '14 at 14:14
1
Well, I should say I had already solved this problem some time ago, then forgot the proof so I had to solve it again, but it wasn"t entirely new to me. When I deal with such a problem I usually draw a picture and draw into it the sets that I can think of, in order to see how they relate. Here it was a large disk $X$, a smaller disk $Y$ within $X$, and two lines from the outer circle to the inner circle which you can think of separating $X-Y$ into $A$ and $B$. Then I separated $Acup Y$ into $D$ and $C$ where $D$ was just a subset of $A$ ... (to be continued)
– Stefan Hamcke
Apr 3 '14 at 14:54
2
@MathsLover: Actually, I don't define anything here. $A$ is clopen in $X-Y$ if it's both open and closed in $X-Y$. And this implies the existence of the sets $U$ and $K$.
– Stefan Hamcke
Aug 21 '14 at 16:31
1
@MathsLover: Yes, that argument is right :)
– Stefan Hamcke
Aug 21 '14 at 19:12
|
show 9 more comments
As $A,B$ form a separation for $X-Y$ and $X$ is connected, $Y$ must be non-empty.
You suppose that $Ycup A$ has a separation $Ccup D$. Then $Y$, being connected, is a subset of $C$
Now $A$ is clopen in $X-Y$, that means there are sets $U$ open and $K$ closed, such that $U-Y=K-Y=A$. You know that $Dsubseteq A$ is clopen in $Ycup A$. Can you show that $D$ is open in $U$ and closed in $K$?
answered Apr 2 '14 at 14:09
Stefan HamckeStefan Hamcke
21.6k42877
Second paragraph is confusing me... could you please edit it a bit...
– user87543
Apr 3 '14 at 4:40
@PraphullaKoushik: Which sentence confuses you?
– Stefan Hamcke
Apr 3 '14 at 14:14
1
Well, I should say I had already solved this problem some time ago, then forgot the proof so I had to solve it again, but it wasn"t entirely new to me. When I deal with such a problem I usually draw a picture and draw into it the sets that I can think of, in order to see how they relate. Here it was a large disk $X$, a smaller disk $Y$ within $X$, and two lines from the outer circle to the inner circle which you can think of separating $X-Y$ into $A$ and $B$. Then I separated $Acup Y$ into $D$ and $C$ where $D$ was just a subset of $A$ ... (to be continued)
– Stefan Hamcke
Apr 3 '14 at 14:54
2
@MathsLover: Actually, I don't define anything here. $A$ is clopen in $X-Y$ if it's both open and closed in $X-Y$. And this implies the existence of the sets $U$ and $K$.
– Stefan Hamcke
Aug 21 '14 at 16:31
1
@MathsLover: Yes, that argument is right :)
– Stefan Hamcke
Aug 21 '14 at 19:12
|
show 9 more comments
As $A,B$ form a separation for $X-Y$ and $X$ is connected, $Y$ must be non-empty.
You suppose that $Ycup A$ has a separation $Ccup D$. Then $Y$, being connected, is a subset of $C$
Now $A$ is clopen in $X-Y$, that means there are sets $U$ open and $K$ closed, such that $U-Y=K-Y=A$. You know that $Dsubseteq A$ is clopen in $Ycup A$. Can you show that $D$ is open in $U$ and closed in $K$?
answered Apr 2 '14 at 14:09
Stefan HamckeStefan Hamcke
21.6k42877
As $A,B$ form a separation for $X-Y$ and $X$ is connected, $Y$ must be non-empty.
You suppose that $Ycup A$ has a separation $Ccup D$. Then $Y$, being connected, is a subset of $C$
Now $A$ is clopen in $X-Y$, that means there are sets $U$ open and $K$ closed, such that $U-Y=K-Y=A$. You know that $Dsubseteq A$ is clopen in $Ycup A$. Can you show that $D$ is open in $U$ and closed in $K$?
answered Apr 2 '14 at 14:09
Stefan HamckeStefan Hamcke
21.6k42877
edited Jun 29 '14 at 20:16
answered Apr 2 '14 at 14:09
Stefan HamckeStefan Hamcke
21.6k42877
answered Apr 2 '14 at 14:09
Stefan HamckeStefan Hamcke
21.6k42877
answered Apr 2 '14 at 14:09
Stefan HamckeStefan Hamcke
21.6k42877
21.6k42877
Second paragraph is confusing me... could you please edit it a bit...
– user87543
Apr 3 '14 at 4:40
@PraphullaKoushik: Which sentence confuses you?
– Stefan Hamcke
Apr 3 '14 at 14:14
1
Well, I should say I had already solved this problem some time ago, then forgot the proof so I had to solve it again, but it wasn"t entirely new to me. When I deal with such a problem I usually draw a picture and draw into it the sets that I can think of, in order to see how they relate. Here it was a large disk $X$, a smaller disk $Y$ within $X$, and two lines from the outer circle to the inner circle which you can think of separating $X-Y$ into $A$ and $B$. Then I separated $Acup Y$ into $D$ and $C$ where $D$ was just a subset of $A$ ... (to be continued)
– Stefan Hamcke
Apr 3 '14 at 14:54
2
@MathsLover: Actually, I don't define anything here. $A$ is clopen in $X-Y$ if it's both open and closed in $X-Y$. And this implies the existence of the sets $U$ and $K$.
– Stefan Hamcke
Aug 21 '14 at 16:31
1
@MathsLover: Yes, that argument is right :)
– Stefan Hamcke
Aug 21 '14 at 19:12
|
show 9 more comments
Second paragraph is confusing me... could you please edit it a bit...
– user87543
Apr 3 '14 at 4:40
@PraphullaKoushik: Which sentence confuses you?
– Stefan Hamcke
Apr 3 '14 at 14:14
1
Well, I should say I had already solved this problem some time ago, then forgot the proof so I had to solve it again, but it wasn"t entirely new to me. When I deal with such a problem I usually draw a picture and draw into it the sets that I can think of, in order to see how they relate. Here it was a large disk $X$, a smaller disk $Y$ within $X$, and two lines from the outer circle to the inner circle which you can think of separating $X-Y$ into $A$ and $B$. Then I separated $Acup Y$ into $D$ and $C$ where $D$ was just a subset of $A$ ... (to be continued)
– Stefan Hamcke
Apr 3 '14 at 14:54
2
@MathsLover: Actually, I don't define anything here. $A$ is clopen in $X-Y$ if it's both open and closed in $X-Y$. And this implies the existence of the sets $U$ and $K$.
– Stefan Hamcke
Aug 21 '14 at 16:31
1
@MathsLover: Yes, that argument is right :)
– Stefan Hamcke
Aug 21 '14 at 19:12
Second paragraph is confusing me... could you please edit it a bit...
– user87543
Apr 3 '14 at 4:40
Second paragraph is confusing me... could you please edit it a bit...
– user87543
Apr 3 '14 at 4:40
@PraphullaKoushik: Which sentence confuses you?
– Stefan Hamcke
Apr 3 '14 at 14:14
@PraphullaKoushik: Which sentence confuses you?
– Stefan Hamcke
Apr 3 '14 at 14:14
1
1
Well, I should say I had already solved this problem some time ago, then forgot the proof so I had to solve it again, but it wasn"t entirely new to me. When I deal with such a problem I usually draw a picture and draw into it the sets that I can think of, in order to see how they relate. Here it was a large disk $X$, a smaller disk $Y$ within $X$, and two lines from the outer circle to the inner circle which you can think of separating $X-Y$ into $A$ and $B$. Then I separated $Acup Y$ into $D$ and $C$ where $D$ was just a subset of $A$ ... (to be continued)
– Stefan Hamcke
Apr 3 '14 at 14:54
Well, I should say I had already solved this problem some time ago, then forgot the proof so I had to solve it again, but it wasn"t entirely new to me. When I deal with such a problem I usually draw a picture and draw into it the sets that I can think of, in order to see how they relate. Here it was a large disk $X$, a smaller disk $Y$ within $X$, and two lines from the outer circle to the inner circle which you can think of separating $X-Y$ into $A$ and $B$. Then I separated $Acup Y$ into $D$ and $C$ where $D$ was just a subset of $A$ ... (to be continued)
– Stefan Hamcke
Apr 3 '14 at 14:54
2
2
@MathsLover: Actually, I don't define anything here. $A$ is clopen in $X-Y$ if it's both open and closed in $X-Y$. And this implies the existence of the sets $U$ and $K$.
– Stefan Hamcke
Aug 21 '14 at 16:31
@MathsLover: Actually, I don't define anything here. $A$ is clopen in $X-Y$ if it's both open and closed in $X-Y$. And this implies the existence of the sets $U$ and $K$.
– Stefan Hamcke
Aug 21 '14 at 16:31
1
1
@MathsLover: Yes, that argument is right :)
– Stefan Hamcke
Aug 21 '14 at 19:12
@MathsLover: Yes, that argument is right :)
– Stefan Hamcke
Aug 21 '14 at 19:12
|
show 9 more comments
Continuing with your argument, We can prove that $B cup C$ and $D$ form the partition of $X$.(Assuming your conclusion that $D subset A$).
We have $X = (B cup C) cup D$.
$C$ and $D$ form separation of $Y cup A$. So, using theorem (23.1), no limit point of $C$ is in $D$. Similarly, no limit point of $B$ is in $D$ since $D subset A$ and $A$ and $B$ form separation of $X-Y$. From this, we can easily say that no limit point of $B cup C$ is in $D$ i.e. Every limit point of $B cup C$ wrt $X$ is in $B cup C$. So, $B cup C$ is closed in $X$ and $D$ is open in $X$.
Now, use a similar argument that, no limit point of $D$ is in $B cup C$ implying that $D$ is closed in $X$ and $Bcup C$ is open in X. Hence, $B cup C$ and $D$ are $textit{clopen}$ in $X$ and form partition of $X$.
answered Jan 4 at 8:46
MUHMUH
395216
add a comment |
Continuing with your argument, We can prove that $B cup C$ and $D$ form the partition of $X$.(Assuming your conclusion that $D subset A$).
We have $X = (B cup C) cup D$.
$C$ and $D$ form separation of $Y cup A$. So, using theorem (23.1), no limit point of $C$ is in $D$. Similarly, no limit point of $B$ is in $D$ since $D subset A$ and $A$ and $B$ form separation of $X-Y$. From this, we can easily say that no limit point of $B cup C$ is in $D$ i.e. Every limit point of $B cup C$ wrt $X$ is in $B cup C$. So, $B cup C$ is closed in $X$ and $D$ is open in $X$.
Now, use a similar argument that, no limit point of $D$ is in $B cup C$ implying that $D$ is closed in $X$ and $Bcup C$ is open in X. Hence, $B cup C$ and $D$ are $textit{clopen}$ in $X$ and form partition of $X$.
answered Jan 4 at 8:46
MUHMUH
395216
add a comment |
Continuing with your argument, We can prove that $B cup C$ and $D$ form the partition of $X$.(Assuming your conclusion that $D subset A$).
We have $X = (B cup C) cup D$.
$C$ and $D$ form separation of $Y cup A$. So, using theorem (23.1), no limit point of $C$ is in $D$. Similarly, no limit point of $B$ is in $D$ since $D subset A$ and $A$ and $B$ form separation of $X-Y$. From this, we can easily say that no limit point of $B cup C$ is in $D$ i.e. Every limit point of $B cup C$ wrt $X$ is in $B cup C$. So, $B cup C$ is closed in $X$ and $D$ is open in $X$.
Now, use a similar argument that, no limit point of $D$ is in $B cup C$ implying that $D$ is closed in $X$ and $Bcup C$ is open in X. Hence, $B cup C$ and $D$ are $textit{clopen}$ in $X$ and form partition of $X$.
answered Jan 4 at 8:46
MUHMUH
395216
Continuing with your argument, We can prove that $B cup C$ and $D$ form the partition of $X$.(Assuming your conclusion that $D subset A$).
We have $X = (B cup C) cup D$.
$C$ and $D$ form separation of $Y cup A$. So, using theorem (23.1), no limit point of $C$ is in $D$. Similarly, no limit point of $B$ is in $D$ since $D subset A$ and $A$ and $B$ form separation of $X-Y$. From this, we can easily say that no limit point of $B cup C$ is in $D$ i.e. Every limit point of $B cup C$ wrt $X$ is in $B cup C$. So, $B cup C$ is closed in $X$ and $D$ is open in $X$.
Now, use a similar argument that, no limit point of $D$ is in $B cup C$ implying that $D$ is closed in $X$ and $Bcup C$ is open in X. Hence, $B cup C$ and $D$ are $textit{clopen}$ in $X$ and form partition of $X$.
answered Jan 4 at 8:46
MUHMUH
395216
answered Jan 4 at 8:46
MUHMUH
395216
answered Jan 4 at 8:46
MUHMUH
395216
answered Jan 4 at 8:46
MUHMUH
395216
395216
add a comment |
add a comment |
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