Mixing
Identification Space in the plane
Clash Royale CLAN TAG#URR8PPP
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Let $X=mathbb R^2$. Describe (visualize) the space $X/ $~ if ~ is the smallest equivalence relation satisfying the following conditions.
a) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$.
b) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$ and $y = y' - 1$
c) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$ or $y = y' - 1$.
I'm a little confused as to how I need to approach these problems. The book I'm using doesn't really explain how to deal with identification space problems regarding $mathbb R^2$. In otherwords my intuition regarding how to deal with $mathbb R^2$/~ identification spaces is a little weak.
Any advice on how to approach and guide me through these problems would be greatly appreciated.
general-topology quotient-spaces
edited Jul 15 at 16:20
Cameron Buie
83.5k771153
asked Jul 15 at 15:08
Alexander King
418311
add a comment |Â
up vote
1
down vote
favorite
Let $X=mathbb R^2$. Describe (visualize) the space $X/ $~ if ~ is the smallest equivalence relation satisfying the following conditions.
a) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$.
b) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$ and $y = y' - 1$
c) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$ or $y = y' - 1$.
I'm a little confused as to how I need to approach these problems. The book I'm using doesn't really explain how to deal with identification space problems regarding $mathbb R^2$. In otherwords my intuition regarding how to deal with $mathbb R^2$/~ identification spaces is a little weak.
Any advice on how to approach and guide me through these problems would be greatly appreciated.
general-topology quotient-spaces
edited Jul 15 at 16:20
Cameron Buie
83.5k771153
asked Jul 15 at 15:08
Alexander King
418311
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X=mathbb R^2$. Describe (visualize) the space $X/ $~ if ~ is the smallest equivalence relation satisfying the following conditions.
a) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$.
b) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$ and $y = y' - 1$
c) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$ or $y = y' - 1$.
I'm a little confused as to how I need to approach these problems. The book I'm using doesn't really explain how to deal with identification space problems regarding $mathbb R^2$. In otherwords my intuition regarding how to deal with $mathbb R^2$/~ identification spaces is a little weak.
Any advice on how to approach and guide me through these problems would be greatly appreciated.
general-topology quotient-spaces
edited Jul 15 at 16:20
Cameron Buie
83.5k771153
asked Jul 15 at 15:08
Alexander King
418311
Let $X=mathbb R^2$. Describe (visualize) the space $X/ $~ if ~ is the smallest equivalence relation satisfying the following conditions.
a) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$.
b) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$ and $y = y' - 1$
c) $(x,y)$ ~$(x',y')$ if and only if $x = x' - 1$ or $y = y' - 1$.
I'm a little confused as to how I need to approach these problems. The book I'm using doesn't really explain how to deal with identification space problems regarding $mathbb R^2$. In otherwords my intuition regarding how to deal with $mathbb R^2$/~ identification spaces is a little weak.
Any advice on how to approach and guide me through these problems would be greatly appreciated.
general-topology quotient-spaces
edited Jul 15 at 16:20
Cameron Buie
83.5k771153
asked Jul 15 at 15:08
Alexander King
418311
edited Jul 15 at 16:20
Cameron Buie
83.5k771153
edited Jul 15 at 16:20
Cameron Buie
83.5k771153
edited Jul 15 at 16:20
Cameron Buie
83.5k771153
83.5k771153
asked Jul 15 at 15:08
Alexander King
418311
asked Jul 15 at 15:08
Alexander King
418311
asked Jul 15 at 15:08
Alexander King
418311
418311
add a comment |Â
add a comment |Â
1 Answer
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The rough idea is to think of an identification as "gluing" points together.
In your first example, note that points whose $x$-values differ by $1$ will be glued together. Consequently (by properties of equivalence relations), points whose $x$-values differ by any integer will be glued together. Note also that this is independent of their $y$-value. For example, given any integer $k$ and any real $y,$ we will have that $(k,y)$ is glued to the origin. For another example, given any real $x,y,z,$ we have that $(x,y)$ and $(x,z)$ will be glued together. To be more formal, we have $(x,y)sim(x',y')$ iff there is an integer $k$ such that $x'=x+k.$ Consequently, every point in $Bbb R^2$ not in the interval $[0,1)$ on the $x$-axis will be glued to a unique point in that interval, and the right end of the interval will be glued to the left end, forming a circle.
Now, for your second example, we have that $(x,y)sim(x',y')$ iff there exist integers $m,n$ such that $(x',y')=(x,y)+(m,n).$ Consequently, every point of $Bbb R^2$ outside of the half-closed square $[0,1)times[0,1)$ will be glued to a unique point of the half-open square. Also, the left and right sides of the half-open square will be glued to each other to form a cylinder with an "open edge" on the top, and the top and bottom of the resulting cylinder will then be glued together to form a torus (the surface of a doughnut).
I'll leave the third one to you. Let me know if you have any more questions about those, if you get stuck working through the other one, or if you just want to bounce your reasoning off me.
answered Jul 15 at 16:09
Cameron Buie
83.5k771153
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The rough idea is to think of an identification as "gluing" points together.
In your first example, note that points whose $x$-values differ by $1$ will be glued together. Consequently (by properties of equivalence relations), points whose $x$-values differ by any integer will be glued together. Note also that this is independent of their $y$-value. For example, given any integer $k$ and any real $y,$ we will have that $(k,y)$ is glued to the origin. For another example, given any real $x,y,z,$ we have that $(x,y)$ and $(x,z)$ will be glued together. To be more formal, we have $(x,y)sim(x',y')$ iff there is an integer $k$ such that $x'=x+k.$ Consequently, every point in $Bbb R^2$ not in the interval $[0,1)$ on the $x$-axis will be glued to a unique point in that interval, and the right end of the interval will be glued to the left end, forming a circle.
Now, for your second example, we have that $(x,y)sim(x',y')$ iff there exist integers $m,n$ such that $(x',y')=(x,y)+(m,n).$ Consequently, every point of $Bbb R^2$ outside of the half-closed square $[0,1)times[0,1)$ will be glued to a unique point of the half-open square. Also, the left and right sides of the half-open square will be glued to each other to form a cylinder with an "open edge" on the top, and the top and bottom of the resulting cylinder will then be glued together to form a torus (the surface of a doughnut).
I'll leave the third one to you. Let me know if you have any more questions about those, if you get stuck working through the other one, or if you just want to bounce your reasoning off me.
answered Jul 15 at 16:09
Cameron Buie
83.5k771153
add a comment |Â
up vote
0
down vote
The rough idea is to think of an identification as "gluing" points together.
In your first example, note that points whose $x$-values differ by $1$ will be glued together. Consequently (by properties of equivalence relations), points whose $x$-values differ by any integer will be glued together. Note also that this is independent of their $y$-value. For example, given any integer $k$ and any real $y,$ we will have that $(k,y)$ is glued to the origin. For another example, given any real $x,y,z,$ we have that $(x,y)$ and $(x,z)$ will be glued together. To be more formal, we have $(x,y)sim(x',y')$ iff there is an integer $k$ such that $x'=x+k.$ Consequently, every point in $Bbb R^2$ not in the interval $[0,1)$ on the $x$-axis will be glued to a unique point in that interval, and the right end of the interval will be glued to the left end, forming a circle.
Now, for your second example, we have that $(x,y)sim(x',y')$ iff there exist integers $m,n$ such that $(x',y')=(x,y)+(m,n).$ Consequently, every point of $Bbb R^2$ outside of the half-closed square $[0,1)times[0,1)$ will be glued to a unique point of the half-open square. Also, the left and right sides of the half-open square will be glued to each other to form a cylinder with an "open edge" on the top, and the top and bottom of the resulting cylinder will then be glued together to form a torus (the surface of a doughnut).
I'll leave the third one to you. Let me know if you have any more questions about those, if you get stuck working through the other one, or if you just want to bounce your reasoning off me.
answered Jul 15 at 16:09
Cameron Buie
83.5k771153
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The rough idea is to think of an identification as "gluing" points together.
In your first example, note that points whose $x$-values differ by $1$ will be glued together. Consequently (by properties of equivalence relations), points whose $x$-values differ by any integer will be glued together. Note also that this is independent of their $y$-value. For example, given any integer $k$ and any real $y,$ we will have that $(k,y)$ is glued to the origin. For another example, given any real $x,y,z,$ we have that $(x,y)$ and $(x,z)$ will be glued together. To be more formal, we have $(x,y)sim(x',y')$ iff there is an integer $k$ such that $x'=x+k.$ Consequently, every point in $Bbb R^2$ not in the interval $[0,1)$ on the $x$-axis will be glued to a unique point in that interval, and the right end of the interval will be glued to the left end, forming a circle.
Now, for your second example, we have that $(x,y)sim(x',y')$ iff there exist integers $m,n$ such that $(x',y')=(x,y)+(m,n).$ Consequently, every point of $Bbb R^2$ outside of the half-closed square $[0,1)times[0,1)$ will be glued to a unique point of the half-open square. Also, the left and right sides of the half-open square will be glued to each other to form a cylinder with an "open edge" on the top, and the top and bottom of the resulting cylinder will then be glued together to form a torus (the surface of a doughnut).
I'll leave the third one to you. Let me know if you have any more questions about those, if you get stuck working through the other one, or if you just want to bounce your reasoning off me.
answered Jul 15 at 16:09
Cameron Buie
83.5k771153
The rough idea is to think of an identification as "gluing" points together.
In your first example, note that points whose $x$-values differ by $1$ will be glued together. Consequently (by properties of equivalence relations), points whose $x$-values differ by any integer will be glued together. Note also that this is independent of their $y$-value. For example, given any integer $k$ and any real $y,$ we will have that $(k,y)$ is glued to the origin. For another example, given any real $x,y,z,$ we have that $(x,y)$ and $(x,z)$ will be glued together. To be more formal, we have $(x,y)sim(x',y')$ iff there is an integer $k$ such that $x'=x+k.$ Consequently, every point in $Bbb R^2$ not in the interval $[0,1)$ on the $x$-axis will be glued to a unique point in that interval, and the right end of the interval will be glued to the left end, forming a circle.
Now, for your second example, we have that $(x,y)sim(x',y')$ iff there exist integers $m,n$ such that $(x',y')=(x,y)+(m,n).$ Consequently, every point of $Bbb R^2$ outside of the half-closed square $[0,1)times[0,1)$ will be glued to a unique point of the half-open square. Also, the left and right sides of the half-open square will be glued to each other to form a cylinder with an "open edge" on the top, and the top and bottom of the resulting cylinder will then be glued together to form a torus (the surface of a doughnut).
I'll leave the third one to you. Let me know if you have any more questions about those, if you get stuck working through the other one, or if you just want to bounce your reasoning off me.
answered Jul 15 at 16:09
Cameron Buie
83.5k771153
edited Jul 15 at 16:16
answered Jul 15 at 16:09
Cameron Buie
83.5k771153
answered Jul 15 at 16:09
Cameron Buie
83.5k771153
answered Jul 15 at 16:09
Cameron Buie
83.5k771153
83.5k771153
add a comment |Â
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