How to show that the given set is open?
$begingroup$
Consider the following subsets of the plane $mathbb{R}^2$:
$$X={(x,y)|y=0}cup {(x,y)|x>0text{ and}; y=1/x}$$
How to show that $A$ and $B$ are open in $X$ under subspace topology.
Efforts:
Let's define $A={(x,y)|y=0}$ and $B={(x,y)|x>0text{ and}; y=1/x}$.
To show that $A$ is open I need to find an open set $N$ of $mathbb{R}^2$ such that $Xcap N=A$. I am not able to proceed further.
I welcome any hints.
Thanks for reading.
general-topology
asked Jan 6 at 10:59
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
add a comment |
$begingroup$
Consider the following subsets of the plane $mathbb{R}^2$:
$$X={(x,y)|y=0}cup {(x,y)|x>0text{ and}; y=1/x}$$
How to show that $A$ and $B$ are open in $X$ under subspace topology.
Efforts:
Let's define $A={(x,y)|y=0}$ and $B={(x,y)|x>0text{ and}; y=1/x}$.
To show that $A$ is open I need to find an open set $N$ of $mathbb{R}^2$ such that $Xcap N=A$. I am not able to proceed further.
I welcome any hints.
Thanks for reading.
general-topology
asked Jan 6 at 10:59
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
add a comment |
$begingroup$
Consider the following subsets of the plane $mathbb{R}^2$:
$$X={(x,y)|y=0}cup {(x,y)|x>0text{ and}; y=1/x}$$
How to show that $A$ and $B$ are open in $X$ under subspace topology.
Efforts:
Let's define $A={(x,y)|y=0}$ and $B={(x,y)|x>0text{ and}; y=1/x}$.
To show that $A$ is open I need to find an open set $N$ of $mathbb{R}^2$ such that $Xcap N=A$. I am not able to proceed further.
I welcome any hints.
Thanks for reading.
general-topology
asked Jan 6 at 10:59
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
Consider the following subsets of the plane $mathbb{R}^2$:
$$X={(x,y)|y=0}cup {(x,y)|x>0text{ and}; y=1/x}$$
How to show that $A$ and $B$ are open in $X$ under subspace topology.
Efforts:
Let's define $A={(x,y)|y=0}$ and $B={(x,y)|x>0text{ and}; y=1/x}$.
To show that $A$ is open I need to find an open set $N$ of $mathbb{R}^2$ such that $Xcap N=A$. I am not able to proceed further.
I welcome any hints.
Thanks for reading.
general-topology
general-topology
asked Jan 6 at 10:59
StammeringMathematicianStammeringMathematician
2,7951324
asked Jan 6 at 10:59
StammeringMathematicianStammeringMathematician
2,7951324
asked Jan 6 at 10:59
StammeringMathematicianStammeringMathematician
2,7951324
asked Jan 6 at 10:59
StammeringMathematicianStammeringMathematician
2,7951324
asked Jan 6 at 10:59
StammeringMathematicianStammeringMathematician
2,7951324
2,7951324
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
Note that both $A$ and $B$ are closed in $mathbb R^2$ (why?), and so $M:=mathbb R^2backslash B$ and $N:=mathbb R^2backslash A$ are open in $mathbb R^2$. Consequently, since $X=Acup B$ and $Acap B=emptyset$, we have $A=Xcap M$ and $B=Xcap N$, showing that both $A$ and $B$ are also open in $X$.
Note that this implies that $A$ and $B$ are also closed in $X$, since $A=Xbackslash B$ and $B=Xbackslash A$.
answered Jan 6 at 11:45
sranthropsranthrop
7,1311925
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add a comment |
$begingroup$
Hint: It might be easier to show that both $A,B$ are closed in $X$, and then since $X = A cup B$, we immediately have that $A,B$ are both open in $X$.
answered Jan 6 at 11:39
Adam HigginsAdam Higgins
615113
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that both $A$ and $B$ are closed in $mathbb R^2$ (why?), and so $M:=mathbb R^2backslash B$ and $N:=mathbb R^2backslash A$ are open in $mathbb R^2$. Consequently, since $X=Acup B$ and $Acap B=emptyset$, we have $A=Xcap M$ and $B=Xcap N$, showing that both $A$ and $B$ are also open in $X$.
Note that this implies that $A$ and $B$ are also closed in $X$, since $A=Xbackslash B$ and $B=Xbackslash A$.
answered Jan 6 at 11:45
sranthropsranthrop
7,1311925
$endgroup$
add a comment |
$begingroup$
Note that both $A$ and $B$ are closed in $mathbb R^2$ (why?), and so $M:=mathbb R^2backslash B$ and $N:=mathbb R^2backslash A$ are open in $mathbb R^2$. Consequently, since $X=Acup B$ and $Acap B=emptyset$, we have $A=Xcap M$ and $B=Xcap N$, showing that both $A$ and $B$ are also open in $X$.
Note that this implies that $A$ and $B$ are also closed in $X$, since $A=Xbackslash B$ and $B=Xbackslash A$.
answered Jan 6 at 11:45
sranthropsranthrop
7,1311925
$endgroup$
add a comment |
$begingroup$
Note that both $A$ and $B$ are closed in $mathbb R^2$ (why?), and so $M:=mathbb R^2backslash B$ and $N:=mathbb R^2backslash A$ are open in $mathbb R^2$. Consequently, since $X=Acup B$ and $Acap B=emptyset$, we have $A=Xcap M$ and $B=Xcap N$, showing that both $A$ and $B$ are also open in $X$.
Note that this implies that $A$ and $B$ are also closed in $X$, since $A=Xbackslash B$ and $B=Xbackslash A$.
answered Jan 6 at 11:45
sranthropsranthrop
7,1311925
$endgroup$
Note that both $A$ and $B$ are closed in $mathbb R^2$ (why?), and so $M:=mathbb R^2backslash B$ and $N:=mathbb R^2backslash A$ are open in $mathbb R^2$. Consequently, since $X=Acup B$ and $Acap B=emptyset$, we have $A=Xcap M$ and $B=Xcap N$, showing that both $A$ and $B$ are also open in $X$.
Note that this implies that $A$ and $B$ are also closed in $X$, since $A=Xbackslash B$ and $B=Xbackslash A$.
answered Jan 6 at 11:45
sranthropsranthrop
7,1311925
answered Jan 6 at 11:45
sranthropsranthrop
7,1311925
answered Jan 6 at 11:45
sranthropsranthrop
7,1311925
answered Jan 6 at 11:45
sranthropsranthrop
7,1311925
7,1311925
add a comment |
add a comment |
$begingroup$
Hint: It might be easier to show that both $A,B$ are closed in $X$, and then since $X = A cup B$, we immediately have that $A,B$ are both open in $X$.
answered Jan 6 at 11:39
Adam HigginsAdam Higgins
615113
$endgroup$
add a comment |
$begingroup$
Hint: It might be easier to show that both $A,B$ are closed in $X$, and then since $X = A cup B$, we immediately have that $A,B$ are both open in $X$.
answered Jan 6 at 11:39
Adam HigginsAdam Higgins
615113
$endgroup$
add a comment |
$begingroup$
Hint: It might be easier to show that both $A,B$ are closed in $X$, and then since $X = A cup B$, we immediately have that $A,B$ are both open in $X$.
answered Jan 6 at 11:39
Adam HigginsAdam Higgins
615113
$endgroup$
Hint: It might be easier to show that both $A,B$ are closed in $X$, and then since $X = A cup B$, we immediately have that $A,B$ are both open in $X$.
answered Jan 6 at 11:39
Adam HigginsAdam Higgins
615113
answered Jan 6 at 11:39
Adam HigginsAdam Higgins
615113
answered Jan 6 at 11:39
Adam HigginsAdam Higgins
615113
answered Jan 6 at 11:39
Adam HigginsAdam Higgins
615113
615113
add a comment |
add a comment |
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