$A$ and $B$ connected implies atleast one of $A cup B$ or $A cap B$ is connected The Next CEO of Stack OverflowEmpty space connected? If so, then what are its components?Union of connected sets also connected?Prove $Y$ is connectedProperty of normal spacesStep Connected if and only if Connected$A cup B$ and $A cap B$ connected $implies A$ and $B$ are connectedif $V_1cong U_1, V_2cong U_2$, is $(V_1cup V_2 cong U_1cup U_2)$? Pasting homeomorphismsIf $overline Acap B = Acap overline B = varnothing$, $Acup B$ is disconnected.Question about compact set contained in an union of open setsSuppose $H$ and $K$ are closed such that $H cup K$ and $H cap K$ are connected. Prove that $H$ and $K$ are connected.Is $[0,1]cup [2,3]$ connected?Suppose that the sets $A_1,A_2 subset mathbbR^n $ are connected and that they are not disjoint. Prove that $A_1 cup A_2$ is connected.
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$A$ and $B$ connected implies atleast one of $A cup B$ or $A cap B$ is connected
The Next CEO of Stack OverflowEmpty space connected? If so, then what are its components?Union of connected sets also connected?Prove $Y$ is connectedProperty of normal spacesStep Connected if and only if Connected$A cup B$ and $A cap B$ connected $implies A$ and $B$ are connectedif $V_1cong U_1, V_2cong U_2$, is $(V_1cup V_2 cong U_1cup U_2)$? Pasting homeomorphismsIf $overline Acap B = Acap overline B = varnothing$, $Acup B$ is disconnected.Question about compact set contained in an union of open setsSuppose $H$ and $K$ are closed such that $H cup K$ and $H cap K$ are connected. Prove that $H$ and $K$ are connected.Is $[0,1]cup [2,3]$ connected?Suppose that the sets $A_1,A_2 subset mathbbR^n $ are connected and that they are not disjoint. Prove that $A_1 cup A_2$ is connected.
1
$begingroup$
Let $A$ and $B$ be connected subsets of a metric space. Prove that at
least one of $A cup B$ or $A cap B$ is connected
My two attempts:
- Assume neither $A cup B$ nor $Acap B$ are connected (hence they are disconnected, then we can partition them $$Acup B rightarrow U_1,U_2quad A cap B rightarrow V_1,V_2 $$
for some open, disjoint sets $U_1,U_2,V_1,V_2$. From here I'm not sure where to continue, and I suppose I need to arrive to a contradiction showing that either $A$ or $B$ is disconnected, which can't be as they are given to be connected but I'm not sure how to reach there. - I tried to go about it by showing that if one of $A cup B$ or $A cap B$ is disconnected the other one must be connceted. Let $A cup B$ be disconnected, then it can be partionted into two open disjoint sets $U_1,U_2$. Now fix $xin A cap B$ and then see that $x in U_1$ or $x in U_2$. WLOG choose the former. So we have $x in A$ and $xin B$ and $x in U_1$. But once again I'm not sure how to use this to arrive to the fact to show that $A cap B$ is connected
Any help would be appreciated.
general-topology connectedness
asked Mar 20 at 9:53
Hushus46Hushus46
513314
$endgroup$
add a comment |
1
$begingroup$
Let $A$ and $B$ be connected subsets of a metric space. Prove that at
least one of $A cup B$ or $A cap B$ is connected
My two attempts:
- Assume neither $A cup B$ nor $Acap B$ are connected (hence they are disconnected, then we can partition them $$Acup B rightarrow U_1,U_2quad A cap B rightarrow V_1,V_2 $$
for some open, disjoint sets $U_1,U_2,V_1,V_2$. From here I'm not sure where to continue, and I suppose I need to arrive to a contradiction showing that either $A$ or $B$ is disconnected, which can't be as they are given to be connected but I'm not sure how to reach there. - I tried to go about it by showing that if one of $A cup B$ or $A cap B$ is disconnected the other one must be connceted. Let $A cup B$ be disconnected, then it can be partionted into two open disjoint sets $U_1,U_2$. Now fix $xin A cap B$ and then see that $x in U_1$ or $x in U_2$. WLOG choose the former. So we have $x in A$ and $xin B$ and $x in U_1$. But once again I'm not sure how to use this to arrive to the fact to show that $A cap B$ is connected
Any help would be appreciated.
general-topology connectedness
asked Mar 20 at 9:53
Hushus46Hushus46
513314
$endgroup$
$begingroup$
That depends on your definition of connectedness (does it allow the empty set to be connected?).
$endgroup$
– YuiTo Cheng
Mar 20 at 10:00
$begingroup$
@YuiToCheng Yes this is taken with the usual notion that $emptyset$ is connected
$endgroup$
– Hushus46
Mar 20 at 10:02
$begingroup$
related question and another related question.
$endgroup$
– drhab
Mar 20 at 10:20
add a comment |
1
1
1
$begingroup$
Let $A$ and $B$ be connected subsets of a metric space. Prove that at
least one of $A cup B$ or $A cap B$ is connected
My two attempts:
- Assume neither $A cup B$ nor $Acap B$ are connected (hence they are disconnected, then we can partition them $$Acup B rightarrow U_1,U_2quad A cap B rightarrow V_1,V_2 $$
for some open, disjoint sets $U_1,U_2,V_1,V_2$. From here I'm not sure where to continue, and I suppose I need to arrive to a contradiction showing that either $A$ or $B$ is disconnected, which can't be as they are given to be connected but I'm not sure how to reach there. - I tried to go about it by showing that if one of $A cup B$ or $A cap B$ is disconnected the other one must be connceted. Let $A cup B$ be disconnected, then it can be partionted into two open disjoint sets $U_1,U_2$. Now fix $xin A cap B$ and then see that $x in U_1$ or $x in U_2$. WLOG choose the former. So we have $x in A$ and $xin B$ and $x in U_1$. But once again I'm not sure how to use this to arrive to the fact to show that $A cap B$ is connected
Any help would be appreciated.
general-topology connectedness
asked Mar 20 at 9:53
Hushus46Hushus46
513314
$endgroup$
Let $A$ and $B$ be connected subsets of a metric space. Prove that at
least one of $A cup B$ or $A cap B$ is connected
My two attempts:
- Assume neither $A cup B$ nor $Acap B$ are connected (hence they are disconnected, then we can partition them $$Acup B rightarrow U_1,U_2quad A cap B rightarrow V_1,V_2 $$
for some open, disjoint sets $U_1,U_2,V_1,V_2$. From here I'm not sure where to continue, and I suppose I need to arrive to a contradiction showing that either $A$ or $B$ is disconnected, which can't be as they are given to be connected but I'm not sure how to reach there. - I tried to go about it by showing that if one of $A cup B$ or $A cap B$ is disconnected the other one must be connceted. Let $A cup B$ be disconnected, then it can be partionted into two open disjoint sets $U_1,U_2$. Now fix $xin A cap B$ and then see that $x in U_1$ or $x in U_2$. WLOG choose the former. So we have $x in A$ and $xin B$ and $x in U_1$. But once again I'm not sure how to use this to arrive to the fact to show that $A cap B$ is connected
Any help would be appreciated.
general-topology connectedness
general-topology connectedness
asked Mar 20 at 9:53
Hushus46Hushus46
513314
asked Mar 20 at 9:53
Hushus46Hushus46
513314
edited Mar 20 at 10:00
Hushus46
asked Mar 20 at 9:53
Hushus46Hushus46
513314
asked Mar 20 at 9:53
Hushus46Hushus46
513314
asked Mar 20 at 9:53
Hushus46Hushus46
513314
513314
$begingroup$
That depends on your definition of connectedness (does it allow the empty set to be connected?).
$endgroup$
– YuiTo Cheng
Mar 20 at 10:00
$begingroup$
@YuiToCheng Yes this is taken with the usual notion that $emptyset$ is connected
$endgroup$
– Hushus46
Mar 20 at 10:02
$begingroup$
related question and another related question.
$endgroup$
– drhab
Mar 20 at 10:20
add a comment |
$begingroup$
That depends on your definition of connectedness (does it allow the empty set to be connected?).
$endgroup$
– YuiTo Cheng
Mar 20 at 10:00
$begingroup$
@YuiToCheng Yes this is taken with the usual notion that $emptyset$ is connected
$endgroup$
– Hushus46
Mar 20 at 10:02
$begingroup$
related question and another related question.
$endgroup$
– drhab
Mar 20 at 10:20
$begingroup$
That depends on your definition of connectedness (does it allow the empty set to be connected?).
$endgroup$
– YuiTo Cheng
Mar 20 at 10:00
$begingroup$
That depends on your definition of connectedness (does it allow the empty set to be connected?).
$endgroup$
– YuiTo Cheng
Mar 20 at 10:00
$begingroup$
@YuiToCheng Yes this is taken with the usual notion that $emptyset$ is connected
$endgroup$
– Hushus46
Mar 20 at 10:02
$begingroup$
@YuiToCheng Yes this is taken with the usual notion that $emptyset$ is connected
$endgroup$
– Hushus46
Mar 20 at 10:02
$begingroup$
related question and another related question.
$endgroup$
– drhab
Mar 20 at 10:20
$begingroup$
related question and another related question.
$endgroup$
– drhab
Mar 20 at 10:20
add a comment |
1 Answer
1
active
oldest
votes
8
$begingroup$
There are two possibilities:
$Acap B=emptyset$: then $Acap B$ is connected.
$Acap Bneqemptyset$: then $Acup B$ is connected, because if $fcolon Acup Blongrightarrow0,1$ (with $0,1$ endowed with the discrete topology) is continuous then, if $pin Acap B$, $f(A)=biglf(p)bigr=f(B)$, and therefore $f$ is constant.
answered Mar 20 at 9:59
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
2
$begingroup$
I was about to type the very same answer but you are superfast!
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:01
2
$begingroup$
Same story for me: again you are superfast.
$endgroup$
– drhab
Mar 20 at 10:04
1
$begingroup$
This answer is perfect, just waiting for a couple of minutes before I'm allowed to accept it. Thanks
$endgroup$
– Hushus46
Mar 20 at 10:04
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
8
$begingroup$
There are two possibilities:
$Acap B=emptyset$: then $Acap B$ is connected.
$Acap Bneqemptyset$: then $Acup B$ is connected, because if $fcolon Acup Blongrightarrow0,1$ (with $0,1$ endowed with the discrete topology) is continuous then, if $pin Acap B$, $f(A)=biglf(p)bigr=f(B)$, and therefore $f$ is constant.
answered Mar 20 at 9:59
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
2
$begingroup$
I was about to type the very same answer but you are superfast!
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:01
2
$begingroup$
Same story for me: again you are superfast.
$endgroup$
– drhab
Mar 20 at 10:04
1
$begingroup$
This answer is perfect, just waiting for a couple of minutes before I'm allowed to accept it. Thanks
$endgroup$
– Hushus46
Mar 20 at 10:04
add a comment |
8
$begingroup$
There are two possibilities:
$Acap B=emptyset$: then $Acap B$ is connected.
$Acap Bneqemptyset$: then $Acup B$ is connected, because if $fcolon Acup Blongrightarrow0,1$ (with $0,1$ endowed with the discrete topology) is continuous then, if $pin Acap B$, $f(A)=biglf(p)bigr=f(B)$, and therefore $f$ is constant.
answered Mar 20 at 9:59
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
2
$begingroup$
I was about to type the very same answer but you are superfast!
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:01
2
$begingroup$
Same story for me: again you are superfast.
$endgroup$
– drhab
Mar 20 at 10:04
1
$begingroup$
This answer is perfect, just waiting for a couple of minutes before I'm allowed to accept it. Thanks
$endgroup$
– Hushus46
Mar 20 at 10:04
add a comment |
8
8
8
$begingroup$
There are two possibilities:
$Acap B=emptyset$: then $Acap B$ is connected.
$Acap Bneqemptyset$: then $Acup B$ is connected, because if $fcolon Acup Blongrightarrow0,1$ (with $0,1$ endowed with the discrete topology) is continuous then, if $pin Acap B$, $f(A)=biglf(p)bigr=f(B)$, and therefore $f$ is constant.
answered Mar 20 at 9:59
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
There are two possibilities:
$Acap B=emptyset$: then $Acap B$ is connected.
$Acap Bneqemptyset$: then $Acup B$ is connected, because if $fcolon Acup Blongrightarrow0,1$ (with $0,1$ endowed with the discrete topology) is continuous then, if $pin Acap B$, $f(A)=biglf(p)bigr=f(B)$, and therefore $f$ is constant.
answered Mar 20 at 9:59
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 20 at 9:59
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 20 at 9:59
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 20 at 9:59
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
2
$begingroup$
I was about to type the very same answer but you are superfast!
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:01
2
$begingroup$
Same story for me: again you are superfast.
$endgroup$
– drhab
Mar 20 at 10:04
1
$begingroup$
This answer is perfect, just waiting for a couple of minutes before I'm allowed to accept it. Thanks
$endgroup$
– Hushus46
Mar 20 at 10:04
add a comment |
2
$begingroup$
I was about to type the very same answer but you are superfast!
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:01
2
$begingroup$
Same story for me: again you are superfast.
$endgroup$
– drhab
Mar 20 at 10:04
1
$begingroup$
This answer is perfect, just waiting for a couple of minutes before I'm allowed to accept it. Thanks
$endgroup$
– Hushus46
Mar 20 at 10:04
2
2
$begingroup$
I was about to type the very same answer but you are superfast!
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:01
$begingroup$
I was about to type the very same answer but you are superfast!
$endgroup$
– Kavi Rama Murthy
Mar 20 at 10:01
2
2
$begingroup$
Same story for me: again you are superfast.
$endgroup$
– drhab
Mar 20 at 10:04
$begingroup$
Same story for me: again you are superfast.
$endgroup$
– drhab
Mar 20 at 10:04
1
1
$begingroup$
This answer is perfect, just waiting for a couple of minutes before I'm allowed to accept it. Thanks
$endgroup$
– Hushus46
Mar 20 at 10:04
$begingroup$
This answer is perfect, just waiting for a couple of minutes before I'm allowed to accept it. Thanks
$endgroup$
– Hushus46
Mar 20 at 10:04
add a comment |
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$begingroup$
That depends on your definition of connectedness (does it allow the empty set to be connected?).
$endgroup$
– YuiTo Cheng
Mar 20 at 10:00
$begingroup$
@YuiToCheng Yes this is taken with the usual notion that $emptyset$ is connected
$endgroup$
– Hushus46
Mar 20 at 10:02
$begingroup$
related question and another related question.
$endgroup$
– drhab
Mar 20 at 10:20