show that f is homomorphism?
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0
down vote
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under the usual Topology on $mathbb{R}^3$ ,the map $f : mathbb{R}^3 rightarrow mathbb{R}^3 $ defined by $f(x,y,z) =(x+1,y-1,z)$ is
choose the correct option
$a)$ open
$b)$ closed
$c)$ neither open nor closed
$d)$ both open and closed
I thinks option $d)$ will correct because $mathbb{R}$ is both open and closed
Is its True ??
Any hints/solution will be appreciated
thanks u
general-topology
asked Nov 23 at 12:29
Messi fifa
51611
add a comment |
up vote
0
down vote
favorite
under the usual Topology on $mathbb{R}^3$ ,the map $f : mathbb{R}^3 rightarrow mathbb{R}^3 $ defined by $f(x,y,z) =(x+1,y-1,z)$ is
choose the correct option
$a)$ open
$b)$ closed
$c)$ neither open nor closed
$d)$ both open and closed
I thinks option $d)$ will correct because $mathbb{R}$ is both open and closed
Is its True ??
Any hints/solution will be appreciated
thanks u
general-topology
asked Nov 23 at 12:29
Messi fifa
51611
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
under the usual Topology on $mathbb{R}^3$ ,the map $f : mathbb{R}^3 rightarrow mathbb{R}^3 $ defined by $f(x,y,z) =(x+1,y-1,z)$ is
choose the correct option
$a)$ open
$b)$ closed
$c)$ neither open nor closed
$d)$ both open and closed
I thinks option $d)$ will correct because $mathbb{R}$ is both open and closed
Is its True ??
Any hints/solution will be appreciated
thanks u
general-topology
asked Nov 23 at 12:29
Messi fifa
51611
under the usual Topology on $mathbb{R}^3$ ,the map $f : mathbb{R}^3 rightarrow mathbb{R}^3 $ defined by $f(x,y,z) =(x+1,y-1,z)$ is
choose the correct option
$a)$ open
$b)$ closed
$c)$ neither open nor closed
$d)$ both open and closed
I thinks option $d)$ will correct because $mathbb{R}$ is both open and closed
Is its True ??
Any hints/solution will be appreciated
thanks u
general-topology
general-topology
asked Nov 23 at 12:29
Messi fifa
51611
asked Nov 23 at 12:29
Messi fifa
51611
edited Nov 23 at 12:48
asked Nov 23 at 12:29
Messi fifa
51611
asked Nov 23 at 12:29
Messi fifa
51611
asked Nov 23 at 12:29
Messi fifa
51611
51611
add a comment |
add a comment |
1 Answer
1
active
oldest
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up vote
3
down vote
accepted
Prescribe $g:mathbb R^3tomathbb R^3$ by:$$(x,y,z)mapsto(x-1,y+1,z)$$
Then $fcirc g=gcirc f=mathsf{id}_{mathbb R^3}$.
Further both functions are continuous.
From this it follows that $f$ is a homeomorphism, hence is open and closed.
answered Nov 23 at 12:34
drhab
96.2k543126
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
up vote
3
down vote
accepted
Prescribe $g:mathbb R^3tomathbb R^3$ by:$$(x,y,z)mapsto(x-1,y+1,z)$$
Then $fcirc g=gcirc f=mathsf{id}_{mathbb R^3}$.
Further both functions are continuous.
From this it follows that $f$ is a homeomorphism, hence is open and closed.
answered Nov 23 at 12:34
drhab
96.2k543126
add a comment |
up vote
3
down vote
accepted
Prescribe $g:mathbb R^3tomathbb R^3$ by:$$(x,y,z)mapsto(x-1,y+1,z)$$
Then $fcirc g=gcirc f=mathsf{id}_{mathbb R^3}$.
Further both functions are continuous.
From this it follows that $f$ is a homeomorphism, hence is open and closed.
answered Nov 23 at 12:34
drhab
96.2k543126
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Prescribe $g:mathbb R^3tomathbb R^3$ by:$$(x,y,z)mapsto(x-1,y+1,z)$$
Then $fcirc g=gcirc f=mathsf{id}_{mathbb R^3}$.
Further both functions are continuous.
From this it follows that $f$ is a homeomorphism, hence is open and closed.
answered Nov 23 at 12:34
drhab
96.2k543126
Prescribe $g:mathbb R^3tomathbb R^3$ by:$$(x,y,z)mapsto(x-1,y+1,z)$$
Then $fcirc g=gcirc f=mathsf{id}_{mathbb R^3}$.
Further both functions are continuous.
From this it follows that $f$ is a homeomorphism, hence is open and closed.
answered Nov 23 at 12:34
drhab
96.2k543126
answered Nov 23 at 12:34
drhab
96.2k543126
answered Nov 23 at 12:34
drhab
96.2k543126
answered Nov 23 at 12:34
drhab
96.2k543126
96.2k543126
add a comment |
add a comment |
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