Mixing
The map $D$ is choose the correct satements
$begingroup$
Consider the closed interval $[0, 1]$ in the real line $mathbb{R}$ and the product space $([0, 1]^{mathbb{N}}, τ ),$
where $τ$ is a topology on $[0, 1]^mathbb{N} $. Let $ D : [0, 1] rightarrow [0, 1]^mathbb{N} $ be the map defined by
$D(x) := (x, x, · · · , x, · · ·)$ for $x in [0, 1].$
. The map $D$ is
choose the correct satements
$(a)$ not continuous if $τ$ is the box topology and also not continuous if $τ$ is the product
topology.
$(b)$ continuous if $τ$ is the product topology and also continuous if $τ$ is the box
topology.
$(c)$ continuous if $τ$ is the box topology and not continuous if $τ$ is the product
topology.
$(d)$ continuous if $τ$ is the product topology and not continuous if $τ$ is the box
topology
My attempt : i thinks option $c)$ is true because box topology is finer then product
Is its true ?
Any hints/solution
general-topology
asked Jan 6 at 11:33
jasminejasmine
1,687417
$endgroup$
add a comment |
$begingroup$
Consider the closed interval $[0, 1]$ in the real line $mathbb{R}$ and the product space $([0, 1]^{mathbb{N}}, τ ),$
where $τ$ is a topology on $[0, 1]^mathbb{N} $. Let $ D : [0, 1] rightarrow [0, 1]^mathbb{N} $ be the map defined by
$D(x) := (x, x, · · · , x, · · ·)$ for $x in [0, 1].$
. The map $D$ is
choose the correct satements
$(a)$ not continuous if $τ$ is the box topology and also not continuous if $τ$ is the product
topology.
$(b)$ continuous if $τ$ is the product topology and also continuous if $τ$ is the box
topology.
$(c)$ continuous if $τ$ is the box topology and not continuous if $τ$ is the product
topology.
$(d)$ continuous if $τ$ is the product topology and not continuous if $τ$ is the box
topology
My attempt : i thinks option $c)$ is true because box topology is finer then product
Is its true ?
Any hints/solution
general-topology
asked Jan 6 at 11:33
jasminejasmine
1,687417
$endgroup$
add a comment |
$begingroup$
Consider the closed interval $[0, 1]$ in the real line $mathbb{R}$ and the product space $([0, 1]^{mathbb{N}}, τ ),$
where $τ$ is a topology on $[0, 1]^mathbb{N} $. Let $ D : [0, 1] rightarrow [0, 1]^mathbb{N} $ be the map defined by
$D(x) := (x, x, · · · , x, · · ·)$ for $x in [0, 1].$
. The map $D$ is
choose the correct satements
$(a)$ not continuous if $τ$ is the box topology and also not continuous if $τ$ is the product
topology.
$(b)$ continuous if $τ$ is the product topology and also continuous if $τ$ is the box
topology.
$(c)$ continuous if $τ$ is the box topology and not continuous if $τ$ is the product
topology.
$(d)$ continuous if $τ$ is the product topology and not continuous if $τ$ is the box
topology
My attempt : i thinks option $c)$ is true because box topology is finer then product
Is its true ?
Any hints/solution
general-topology
asked Jan 6 at 11:33
jasminejasmine
1,687417
$endgroup$
Consider the closed interval $[0, 1]$ in the real line $mathbb{R}$ and the product space $([0, 1]^{mathbb{N}}, τ ),$
where $τ$ is a topology on $[0, 1]^mathbb{N} $. Let $ D : [0, 1] rightarrow [0, 1]^mathbb{N} $ be the map defined by
$D(x) := (x, x, · · · , x, · · ·)$ for $x in [0, 1].$
. The map $D$ is
choose the correct satements
$(a)$ not continuous if $τ$ is the box topology and also not continuous if $τ$ is the product
topology.
$(b)$ continuous if $τ$ is the product topology and also continuous if $τ$ is the box
topology.
$(c)$ continuous if $τ$ is the box topology and not continuous if $τ$ is the product
topology.
$(d)$ continuous if $τ$ is the product topology and not continuous if $τ$ is the box
topology
My attempt : i thinks option $c)$ is true because box topology is finer then product
Is its true ?
Any hints/solution
general-topology
general-topology
asked Jan 6 at 11:33
jasminejasmine
1,687417
asked Jan 6 at 11:33
jasminejasmine
1,687417
asked Jan 6 at 11:33
jasminejasmine
1,687417
asked Jan 6 at 11:33
jasminejasmine
1,687417
asked Jan 6 at 11:33
jasminejasmine
1,687417
1,687417
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.
answered Jan 6 at 12:15
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
$begingroup$
thanks u kavi sir
$endgroup$
– jasmine
Jan 6 at 12:20
add a comment |
$begingroup$
Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?
answered Jan 6 at 11:37
MindlackMindlack
3,15217
$endgroup$
$begingroup$
for box i thinks product ??
$endgroup$
– jasmine
Jan 6 at 11:40
$begingroup$
I do not understand your comment.
$endgroup$
– Mindlack
Jan 6 at 11:41
$begingroup$
@Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
$endgroup$
– jasmine
Jan 6 at 11:50
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.
answered Jan 6 at 12:15
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
$begingroup$
thanks u kavi sir
$endgroup$
– jasmine
Jan 6 at 12:20
add a comment |
$begingroup$
The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.
answered Jan 6 at 12:15
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
$begingroup$
thanks u kavi sir
$endgroup$
– jasmine
Jan 6 at 12:20
add a comment |
$begingroup$
The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.
answered Jan 6 at 12:15
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.
answered Jan 6 at 12:15
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Jan 6 at 12:15
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Jan 6 at 12:15
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Jan 6 at 12:15
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
55.7k42158
$begingroup$
thanks u kavi sir
$endgroup$
– jasmine
Jan 6 at 12:20
add a comment |
$begingroup$
thanks u kavi sir
$endgroup$
– jasmine
Jan 6 at 12:20
$begingroup$
thanks u kavi sir
$endgroup$
– jasmine
Jan 6 at 12:20
$begingroup$
thanks u kavi sir
$endgroup$
– jasmine
Jan 6 at 12:20
add a comment |
$begingroup$
Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?
answered Jan 6 at 11:37
MindlackMindlack
3,15217
$endgroup$
$begingroup$
for box i thinks product ??
$endgroup$
– jasmine
Jan 6 at 11:40
$begingroup$
I do not understand your comment.
$endgroup$
– Mindlack
Jan 6 at 11:41
$begingroup$
@Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
$endgroup$
– jasmine
Jan 6 at 11:50
add a comment |
$begingroup$
Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?
answered Jan 6 at 11:37
MindlackMindlack
3,15217
$endgroup$
$begingroup$
for box i thinks product ??
$endgroup$
– jasmine
Jan 6 at 11:40
$begingroup$
I do not understand your comment.
$endgroup$
– Mindlack
Jan 6 at 11:41
$begingroup$
@Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
$endgroup$
– jasmine
Jan 6 at 11:50
add a comment |
$begingroup$
Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?
answered Jan 6 at 11:37
MindlackMindlack
3,15217
$endgroup$
Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?
answered Jan 6 at 11:37
MindlackMindlack
3,15217
answered Jan 6 at 11:37
MindlackMindlack
3,15217
answered Jan 6 at 11:37
MindlackMindlack
3,15217
answered Jan 6 at 11:37
MindlackMindlack
3,15217
3,15217
$begingroup$
for box i thinks product ??
$endgroup$
– jasmine
Jan 6 at 11:40
$begingroup$
I do not understand your comment.
$endgroup$
– Mindlack
Jan 6 at 11:41
$begingroup$
@Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
$endgroup$
– jasmine
Jan 6 at 11:50
add a comment |
$begingroup$
for box i thinks product ??
$endgroup$
– jasmine
Jan 6 at 11:40
$begingroup$
I do not understand your comment.
$endgroup$
– Mindlack
Jan 6 at 11:41
$begingroup$
@Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
$endgroup$
– jasmine
Jan 6 at 11:50
$begingroup$
for box i thinks product ??
$endgroup$
– jasmine
Jan 6 at 11:40
$begingroup$
for box i thinks product ??
$endgroup$
– jasmine
Jan 6 at 11:40
$begingroup$
I do not understand your comment.
$endgroup$
– Mindlack
Jan 6 at 11:41
$begingroup$
I do not understand your comment.
$endgroup$
– Mindlack
Jan 6 at 11:41
$begingroup$
@Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
$endgroup$
– jasmine
Jan 6 at 11:50
$begingroup$
@Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
$endgroup$
– jasmine
Jan 6 at 11:50
add a comment |
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