If A complement is the union of two separated sets, prove that the union of those separated sets with A is...
$begingroup$
Let $A$ be a connected subset of a connected metric space $(X,d)$.
Assume $A^{c}$ is the union of two separated sets $B$ and $C$.
Prove that $A cup B$ and $A cup C$ are connected.
Attempt
Proving $A cup B$ is connected is sufficient.
Assume towards a contradiction that $A cup B$ is not connected. (So it is disconnected).
There exists open sets $G_{1}$ and $G_{2}$ such that $A cup B subseteq G_{1} cup G_{2}$, $(Acup B) cap G_{1} neq phi$, $(Acup B) cap G_{2} neq phi$ and $(Acup B)cap G_{1} cap G_{2} = phi$.
Since $A subseteq A cup B$ and $A$ is connected, then either $A$ lies in $G_{1}$ or $G_{2}$.
WLOG, assume $A$ lies in $G_{1}$.
I do not know how to proceed from here. I need to obtain a contradiction to finish my proof.
Edit
I have corrected my proof.
real-analysis proof-verification metric-spaces
edited Dec 5 '18 at 7:45
Fluffy Skye
1018
asked Dec 4 '18 at 9:00
Snop D.Snop D.
285
$endgroup$
add a comment |
$begingroup$
Let $A$ be a connected subset of a connected metric space $(X,d)$.
Assume $A^{c}$ is the union of two separated sets $B$ and $C$.
Prove that $A cup B$ and $A cup C$ are connected.
Attempt
Proving $A cup B$ is connected is sufficient.
Assume towards a contradiction that $A cup B$ is not connected. (So it is disconnected).
There exists open sets $G_{1}$ and $G_{2}$ such that $A cup B subseteq G_{1} cup G_{2}$, $(Acup B) cap G_{1} neq phi$, $(Acup B) cap G_{2} neq phi$ and $(Acup B)cap G_{1} cap G_{2} = phi$.
Since $A subseteq A cup B$ and $A$ is connected, then either $A$ lies in $G_{1}$ or $G_{2}$.
WLOG, assume $A$ lies in $G_{1}$.
I do not know how to proceed from here. I need to obtain a contradiction to finish my proof.
Edit
I have corrected my proof.
real-analysis proof-verification metric-spaces
edited Dec 5 '18 at 7:45
Fluffy Skye
1018
asked Dec 4 '18 at 9:00
Snop D.Snop D.
285
$endgroup$
$begingroup$
You cannot prove that $A cup B$ is connected for any subset $B$ of $A^{c}$. You don't use the fact that $B$ and $C$ are separated anywhere in your proof so this cannot be correct. @Snop D.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:25
add a comment |
$begingroup$
Let $A$ be a connected subset of a connected metric space $(X,d)$.
Assume $A^{c}$ is the union of two separated sets $B$ and $C$.
Prove that $A cup B$ and $A cup C$ are connected.
Attempt
Proving $A cup B$ is connected is sufficient.
Assume towards a contradiction that $A cup B$ is not connected. (So it is disconnected).
There exists open sets $G_{1}$ and $G_{2}$ such that $A cup B subseteq G_{1} cup G_{2}$, $(Acup B) cap G_{1} neq phi$, $(Acup B) cap G_{2} neq phi$ and $(Acup B)cap G_{1} cap G_{2} = phi$.
Since $A subseteq A cup B$ and $A$ is connected, then either $A$ lies in $G_{1}$ or $G_{2}$.
WLOG, assume $A$ lies in $G_{1}$.
I do not know how to proceed from here. I need to obtain a contradiction to finish my proof.
Edit
I have corrected my proof.
real-analysis proof-verification metric-spaces
edited Dec 5 '18 at 7:45
Fluffy Skye
1018
asked Dec 4 '18 at 9:00
Snop D.Snop D.
285
$endgroup$
Let $A$ be a connected subset of a connected metric space $(X,d)$.
Assume $A^{c}$ is the union of two separated sets $B$ and $C$.
Prove that $A cup B$ and $A cup C$ are connected.
Attempt
Proving $A cup B$ is connected is sufficient.
Assume towards a contradiction that $A cup B$ is not connected. (So it is disconnected).
There exists open sets $G_{1}$ and $G_{2}$ such that $A cup B subseteq G_{1} cup G_{2}$, $(Acup B) cap G_{1} neq phi$, $(Acup B) cap G_{2} neq phi$ and $(Acup B)cap G_{1} cap G_{2} = phi$.
Since $A subseteq A cup B$ and $A$ is connected, then either $A$ lies in $G_{1}$ or $G_{2}$.
WLOG, assume $A$ lies in $G_{1}$.
I do not know how to proceed from here. I need to obtain a contradiction to finish my proof.
Edit
I have corrected my proof.
real-analysis proof-verification metric-spaces
real-analysis proof-verification metric-spaces
edited Dec 5 '18 at 7:45
Fluffy Skye
1018
asked Dec 4 '18 at 9:00
Snop D.Snop D.
285
edited Dec 5 '18 at 7:45
Fluffy Skye
1018
asked Dec 4 '18 at 9:00
Snop D.Snop D.
285
edited Dec 5 '18 at 7:45
Fluffy Skye
1018
edited Dec 5 '18 at 7:45
Fluffy Skye
1018
edited Dec 5 '18 at 7:45
Fluffy Skye
1018
1018
asked Dec 4 '18 at 9:00
Snop D.Snop D.
285
asked Dec 4 '18 at 9:00
Snop D.Snop D.
285
asked Dec 4 '18 at 9:00
Snop D.Snop D.
285
285
$begingroup$
You cannot prove that $A cup B$ is connected for any subset $B$ of $A^{c}$. You don't use the fact that $B$ and $C$ are separated anywhere in your proof so this cannot be correct. @Snop D.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:25
add a comment |
$begingroup$
You cannot prove that $A cup B$ is connected for any subset $B$ of $A^{c}$. You don't use the fact that $B$ and $C$ are separated anywhere in your proof so this cannot be correct. @Snop D.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:25
$begingroup$
You cannot prove that $A cup B$ is connected for any subset $B$ of $A^{c}$. You don't use the fact that $B$ and $C$ are separated anywhere in your proof so this cannot be correct. @Snop D.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:25
$begingroup$
You cannot prove that $A cup B$ is connected for any subset $B$ of $A^{c}$. You don't use the fact that $B$ and $C$ are separated anywhere in your proof so this cannot be correct. @Snop D.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are almost there. If $A$ lies entirely in $G_1$, and $Acap G_2neq emptyset$, then you can easily conclude that $G_1cap G_2$ cannot be an empty set, therefore, a contradiction follows.
You can conclude this because you know that $Acap G_2subseteq G_1cap G_2$, and $emptyset$ has no non-empty subsets.
answered Dec 4 '18 at 9:07
5xum5xum
90.2k393161
$endgroup$
$begingroup$
The hypothesis that $B$ and $C$ are separated and their union is $A^{c}$ is necessary for this result. Where is this used in this proof?
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:10
$begingroup$
@KaviRamaMurthy OP already reached a point where he proved that $A$ lies in either $G_1$ or $G_2$. I just continued from there.
$endgroup$
– 5xum
Dec 4 '18 at 9:14
$begingroup$
@5xum That seems to work! Perhaps I need the $A^{c}$ part to prove the condition in my Edit.
$endgroup$
– Snop D.
Dec 4 '18 at 9:19
$begingroup$
I had made a mistake in my proof. I have corrected it now.
$endgroup$
– Snop D.
Dec 4 '18 at 22:19
add a comment |
$begingroup$
Isn't the complement of $A cup B$ just C? And I think it's easy to see that C is open.
answered Dec 4 '18 at 9:24
modest_mildewmodest_mildew
112
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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oldest
votes
$begingroup$
You are almost there. If $A$ lies entirely in $G_1$, and $Acap G_2neq emptyset$, then you can easily conclude that $G_1cap G_2$ cannot be an empty set, therefore, a contradiction follows.
You can conclude this because you know that $Acap G_2subseteq G_1cap G_2$, and $emptyset$ has no non-empty subsets.
answered Dec 4 '18 at 9:07
5xum5xum
90.2k393161
$endgroup$
$begingroup$
The hypothesis that $B$ and $C$ are separated and their union is $A^{c}$ is necessary for this result. Where is this used in this proof?
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:10
$begingroup$
@KaviRamaMurthy OP already reached a point where he proved that $A$ lies in either $G_1$ or $G_2$. I just continued from there.
$endgroup$
– 5xum
Dec 4 '18 at 9:14
$begingroup$
@5xum That seems to work! Perhaps I need the $A^{c}$ part to prove the condition in my Edit.
$endgroup$
– Snop D.
Dec 4 '18 at 9:19
$begingroup$
I had made a mistake in my proof. I have corrected it now.
$endgroup$
– Snop D.
Dec 4 '18 at 22:19
add a comment |
$begingroup$
You are almost there. If $A$ lies entirely in $G_1$, and $Acap G_2neq emptyset$, then you can easily conclude that $G_1cap G_2$ cannot be an empty set, therefore, a contradiction follows.
You can conclude this because you know that $Acap G_2subseteq G_1cap G_2$, and $emptyset$ has no non-empty subsets.
answered Dec 4 '18 at 9:07
5xum5xum
90.2k393161
$endgroup$
$begingroup$
The hypothesis that $B$ and $C$ are separated and their union is $A^{c}$ is necessary for this result. Where is this used in this proof?
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:10
$begingroup$
@KaviRamaMurthy OP already reached a point where he proved that $A$ lies in either $G_1$ or $G_2$. I just continued from there.
$endgroup$
– 5xum
Dec 4 '18 at 9:14
$begingroup$
@5xum That seems to work! Perhaps I need the $A^{c}$ part to prove the condition in my Edit.
$endgroup$
– Snop D.
Dec 4 '18 at 9:19
$begingroup$
I had made a mistake in my proof. I have corrected it now.
$endgroup$
– Snop D.
Dec 4 '18 at 22:19
add a comment |
$begingroup$
You are almost there. If $A$ lies entirely in $G_1$, and $Acap G_2neq emptyset$, then you can easily conclude that $G_1cap G_2$ cannot be an empty set, therefore, a contradiction follows.
You can conclude this because you know that $Acap G_2subseteq G_1cap G_2$, and $emptyset$ has no non-empty subsets.
answered Dec 4 '18 at 9:07
5xum5xum
90.2k393161
$endgroup$
You are almost there. If $A$ lies entirely in $G_1$, and $Acap G_2neq emptyset$, then you can easily conclude that $G_1cap G_2$ cannot be an empty set, therefore, a contradiction follows.
You can conclude this because you know that $Acap G_2subseteq G_1cap G_2$, and $emptyset$ has no non-empty subsets.
answered Dec 4 '18 at 9:07
5xum5xum
90.2k393161
answered Dec 4 '18 at 9:07
5xum5xum
90.2k393161
answered Dec 4 '18 at 9:07
5xum5xum
90.2k393161
answered Dec 4 '18 at 9:07
5xum5xum
90.2k393161
90.2k393161
$begingroup$
The hypothesis that $B$ and $C$ are separated and their union is $A^{c}$ is necessary for this result. Where is this used in this proof?
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:10
$begingroup$
@KaviRamaMurthy OP already reached a point where he proved that $A$ lies in either $G_1$ or $G_2$. I just continued from there.
$endgroup$
– 5xum
Dec 4 '18 at 9:14
$begingroup$
@5xum That seems to work! Perhaps I need the $A^{c}$ part to prove the condition in my Edit.
$endgroup$
– Snop D.
Dec 4 '18 at 9:19
$begingroup$
I had made a mistake in my proof. I have corrected it now.
$endgroup$
– Snop D.
Dec 4 '18 at 22:19
add a comment |
$begingroup$
The hypothesis that $B$ and $C$ are separated and their union is $A^{c}$ is necessary for this result. Where is this used in this proof?
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:10
$begingroup$
@KaviRamaMurthy OP already reached a point where he proved that $A$ lies in either $G_1$ or $G_2$. I just continued from there.
$endgroup$
– 5xum
Dec 4 '18 at 9:14
$begingroup$
@5xum That seems to work! Perhaps I need the $A^{c}$ part to prove the condition in my Edit.
$endgroup$
– Snop D.
Dec 4 '18 at 9:19
$begingroup$
I had made a mistake in my proof. I have corrected it now.
$endgroup$
– Snop D.
Dec 4 '18 at 22:19
$begingroup$
The hypothesis that $B$ and $C$ are separated and their union is $A^{c}$ is necessary for this result. Where is this used in this proof?
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:10
$begingroup$
The hypothesis that $B$ and $C$ are separated and their union is $A^{c}$ is necessary for this result. Where is this used in this proof?
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:10
$begingroup$
@KaviRamaMurthy OP already reached a point where he proved that $A$ lies in either $G_1$ or $G_2$. I just continued from there.
$endgroup$
– 5xum
Dec 4 '18 at 9:14
$begingroup$
@KaviRamaMurthy OP already reached a point where he proved that $A$ lies in either $G_1$ or $G_2$. I just continued from there.
$endgroup$
– 5xum
Dec 4 '18 at 9:14
$begingroup$
@5xum That seems to work! Perhaps I need the $A^{c}$ part to prove the condition in my Edit.
$endgroup$
– Snop D.
Dec 4 '18 at 9:19
$begingroup$
@5xum That seems to work! Perhaps I need the $A^{c}$ part to prove the condition in my Edit.
$endgroup$
– Snop D.
Dec 4 '18 at 9:19
$begingroup$
I had made a mistake in my proof. I have corrected it now.
$endgroup$
– Snop D.
Dec 4 '18 at 22:19
$begingroup$
I had made a mistake in my proof. I have corrected it now.
$endgroup$
– Snop D.
Dec 4 '18 at 22:19
add a comment |
$begingroup$
Isn't the complement of $A cup B$ just C? And I think it's easy to see that C is open.
answered Dec 4 '18 at 9:24
modest_mildewmodest_mildew
112
$endgroup$
add a comment |
$begingroup$
Isn't the complement of $A cup B$ just C? And I think it's easy to see that C is open.
answered Dec 4 '18 at 9:24
modest_mildewmodest_mildew
112
$endgroup$
add a comment |
$begingroup$
Isn't the complement of $A cup B$ just C? And I think it's easy to see that C is open.
answered Dec 4 '18 at 9:24
modest_mildewmodest_mildew
112
$endgroup$
Isn't the complement of $A cup B$ just C? And I think it's easy to see that C is open.
answered Dec 4 '18 at 9:24
modest_mildewmodest_mildew
112
edited Dec 4 '18 at 9:31
answered Dec 4 '18 at 9:24
modest_mildewmodest_mildew
112
answered Dec 4 '18 at 9:24
modest_mildewmodest_mildew
112
answered Dec 4 '18 at 9:24
modest_mildewmodest_mildew
112
112
add a comment |
add a comment |
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$begingroup$
You cannot prove that $A cup B$ is connected for any subset $B$ of $A^{c}$. You don't use the fact that $B$ and $C$ are separated anywhere in your proof so this cannot be correct. @Snop D.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 9:25