Are $left{ (x,y) in mathbb{R^2} | |y| > x^2 , |y| x^4 , |y| < 10 right} $ homeomorphic?
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Are spaces $left{ (x,y) in mathbb{R^2} | |y| > x^2 , |y| < 10 right} $ and $left{(x,y) in mathbb{R^2} | |y| > x^4 , |y| < 10 right} $ homeomorphic? I think they might be but can't construct a homeomorphism because one is a subset of the other.
general-topology continuity
asked Nov 18 at 14:31
user15269
1588
add a comment |
up vote
0
down vote
favorite
Are spaces $left{ (x,y) in mathbb{R^2} | |y| > x^2 , |y| < 10 right} $ and $left{(x,y) in mathbb{R^2} | |y| > x^4 , |y| < 10 right} $ homeomorphic? I think they might be but can't construct a homeomorphism because one is a subset of the other.
general-topology continuity
asked Nov 18 at 14:31
user15269
1588
Can you construct a homeomorphism between e.g. ${(x,y)mid x^2+y^2leq1}$ and ${(x,y)mid x^2+y^2leq4}$? Also here one is a subset of the other. I don't see why that should be an obstacle.
– drhab
Nov 18 at 14:46
What is (x,y)(x,y)?
– William Elliot
Nov 19 at 2:41
sorry I corrected it
– user15269
Nov 19 at 9:19
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Are spaces $left{ (x,y) in mathbb{R^2} | |y| > x^2 , |y| < 10 right} $ and $left{(x,y) in mathbb{R^2} | |y| > x^4 , |y| < 10 right} $ homeomorphic? I think they might be but can't construct a homeomorphism because one is a subset of the other.
general-topology continuity
asked Nov 18 at 14:31
user15269
1588
Are spaces $left{ (x,y) in mathbb{R^2} | |y| > x^2 , |y| < 10 right} $ and $left{(x,y) in mathbb{R^2} | |y| > x^4 , |y| < 10 right} $ homeomorphic? I think they might be but can't construct a homeomorphism because one is a subset of the other.
general-topology continuity
general-topology continuity
asked Nov 18 at 14:31
user15269
1588
asked Nov 18 at 14:31
user15269
1588
edited Nov 19 at 9:15
asked Nov 18 at 14:31
user15269
1588
asked Nov 18 at 14:31
user15269
1588
asked Nov 18 at 14:31
user15269
1588
1588
Can you construct a homeomorphism between e.g. ${(x,y)mid x^2+y^2leq1}$ and ${(x,y)mid x^2+y^2leq4}$? Also here one is a subset of the other. I don't see why that should be an obstacle.
– drhab
Nov 18 at 14:46
What is (x,y)(x,y)?
– William Elliot
Nov 19 at 2:41
sorry I corrected it
– user15269
Nov 19 at 9:19
add a comment |
Can you construct a homeomorphism between e.g. ${(x,y)mid x^2+y^2leq1}$ and ${(x,y)mid x^2+y^2leq4}$? Also here one is a subset of the other. I don't see why that should be an obstacle.
– drhab
Nov 18 at 14:46
What is (x,y)(x,y)?
– William Elliot
Nov 19 at 2:41
sorry I corrected it
– user15269
Nov 19 at 9:19
Can you construct a homeomorphism between e.g. ${(x,y)mid x^2+y^2leq1}$ and ${(x,y)mid x^2+y^2leq4}$? Also here one is a subset of the other. I don't see why that should be an obstacle.
– drhab
Nov 18 at 14:46
Can you construct a homeomorphism between e.g. ${(x,y)mid x^2+y^2leq1}$ and ${(x,y)mid x^2+y^2leq4}$? Also here one is a subset of the other. I don't see why that should be an obstacle.
– drhab
Nov 18 at 14:46
What is (x,y)(x,y)?
– William Elliot
Nov 19 at 2:41
What is (x,y)(x,y)?
– William Elliot
Nov 19 at 2:41
sorry I corrected it
– user15269
Nov 19 at 9:19
sorry I corrected it
– user15269
Nov 19 at 9:19
add a comment |
1 Answer
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Let's disect those sets. Both of them consist of two connected components: when $y<0$ and when $y>0$. It is enough that we show that those components are pairwise homeomorphic. And since each upper half component is a reflection of the lower half then it is enough to show that $A={(x,y) | y>x^2; y< 10}$ and $B={(x,y) | y>x^4; y< 10}$ are homeomorphic.
To do that you have to realize that both $f(x)=x^2$ and $f(x)=x^4$ are convex functions. A bit of work has to be done to make sure that these conditions together with $y<10$ imply that both $A$ and $B$ are open and convex subsets of $mathbb{R}^2$. And it is well known that every open and convex subset of $mathbb{R}^n$ is homeomorphic to a $n$-ball.
answered Nov 19 at 9:29
freakish
10.8k1527
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let's disect those sets. Both of them consist of two connected components: when $y<0$ and when $y>0$. It is enough that we show that those components are pairwise homeomorphic. And since each upper half component is a reflection of the lower half then it is enough to show that $A={(x,y) | y>x^2; y< 10}$ and $B={(x,y) | y>x^4; y< 10}$ are homeomorphic.
To do that you have to realize that both $f(x)=x^2$ and $f(x)=x^4$ are convex functions. A bit of work has to be done to make sure that these conditions together with $y<10$ imply that both $A$ and $B$ are open and convex subsets of $mathbb{R}^2$. And it is well known that every open and convex subset of $mathbb{R}^n$ is homeomorphic to a $n$-ball.
answered Nov 19 at 9:29
freakish
10.8k1527
add a comment |
up vote
1
down vote
Let's disect those sets. Both of them consist of two connected components: when $y<0$ and when $y>0$. It is enough that we show that those components are pairwise homeomorphic. And since each upper half component is a reflection of the lower half then it is enough to show that $A={(x,y) | y>x^2; y< 10}$ and $B={(x,y) | y>x^4; y< 10}$ are homeomorphic.
To do that you have to realize that both $f(x)=x^2$ and $f(x)=x^4$ are convex functions. A bit of work has to be done to make sure that these conditions together with $y<10$ imply that both $A$ and $B$ are open and convex subsets of $mathbb{R}^2$. And it is well known that every open and convex subset of $mathbb{R}^n$ is homeomorphic to a $n$-ball.
answered Nov 19 at 9:29
freakish
10.8k1527
add a comment |
up vote
1
down vote
up vote
1
down vote
Let's disect those sets. Both of them consist of two connected components: when $y<0$ and when $y>0$. It is enough that we show that those components are pairwise homeomorphic. And since each upper half component is a reflection of the lower half then it is enough to show that $A={(x,y) | y>x^2; y< 10}$ and $B={(x,y) | y>x^4; y< 10}$ are homeomorphic.
To do that you have to realize that both $f(x)=x^2$ and $f(x)=x^4$ are convex functions. A bit of work has to be done to make sure that these conditions together with $y<10$ imply that both $A$ and $B$ are open and convex subsets of $mathbb{R}^2$. And it is well known that every open and convex subset of $mathbb{R}^n$ is homeomorphic to a $n$-ball.
answered Nov 19 at 9:29
freakish
10.8k1527
Let's disect those sets. Both of them consist of two connected components: when $y<0$ and when $y>0$. It is enough that we show that those components are pairwise homeomorphic. And since each upper half component is a reflection of the lower half then it is enough to show that $A={(x,y) | y>x^2; y< 10}$ and $B={(x,y) | y>x^4; y< 10}$ are homeomorphic.
To do that you have to realize that both $f(x)=x^2$ and $f(x)=x^4$ are convex functions. A bit of work has to be done to make sure that these conditions together with $y<10$ imply that both $A$ and $B$ are open and convex subsets of $mathbb{R}^2$. And it is well known that every open and convex subset of $mathbb{R}^n$ is homeomorphic to a $n$-ball.
answered Nov 19 at 9:29
freakish
10.8k1527
answered Nov 19 at 9:29
freakish
10.8k1527
answered Nov 19 at 9:29
freakish
10.8k1527
answered Nov 19 at 9:29
freakish
10.8k1527
10.8k1527
add a comment |
add a comment |
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Can you construct a homeomorphism between e.g. ${(x,y)mid x^2+y^2leq1}$ and ${(x,y)mid x^2+y^2leq4}$? Also here one is a subset of the other. I don't see why that should be an obstacle.
– drhab
Nov 18 at 14:46
What is (x,y)(x,y)?
– William Elliot
Nov 19 at 2:41
sorry I corrected it
– user15269
Nov 19 at 9:19