Is $[1,2]times Bbb R$ closed?
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Is $[1,2]times Bbb R$ closed in $Bbb R^2$? The answer is no. But I suggest it is yes, since it contains all its boundary points right? Am I right or wrong?
real-analysis
asked Nov 22 at 4:48
Shara
524
|
show 3 more comments
up vote
1
down vote
favorite
Is $[1,2]times Bbb R$ closed in $Bbb R^2$? The answer is no. But I suggest it is yes, since it contains all its boundary points right? Am I right or wrong?
real-analysis
asked Nov 22 at 4:48
Shara
524
1
Can you prove its complement is open i.e.${(-infty,1)timesmathbb{R}}cup{(2,infty)timesmathbb{R}}$ is open?
– Yadati Kiran
Nov 22 at 4:51
1
You are right. answer is wrong. Of course it isn't compact in $mathbb{R}^2.$
– coffeemath
Nov 22 at 4:52
yes, $[1,2]timesBbb R$ is closed in the euclidean topology
– Masacroso
Nov 22 at 4:53
@coffeemath: Compactness is not relevant here I guess.
– Yadati Kiran
Nov 22 at 4:56
@YadatiKiran You're right-- just threw that in in case OP wondered about that, or in case the wording of the question was diferent. [seems odd that the "answer" would be wrong about such a simple question. wonder what book said that..]
– coffeemath
Nov 22 at 4:59
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is $[1,2]times Bbb R$ closed in $Bbb R^2$? The answer is no. But I suggest it is yes, since it contains all its boundary points right? Am I right or wrong?
real-analysis
asked Nov 22 at 4:48
Shara
524
Is $[1,2]times Bbb R$ closed in $Bbb R^2$? The answer is no. But I suggest it is yes, since it contains all its boundary points right? Am I right or wrong?
real-analysis
real-analysis
asked Nov 22 at 4:48
Shara
524
asked Nov 22 at 4:48
Shara
524
asked Nov 22 at 4:48
Shara
524
asked Nov 22 at 4:48
Shara
524
asked Nov 22 at 4:48
Shara
524
524
1
Can you prove its complement is open i.e.${(-infty,1)timesmathbb{R}}cup{(2,infty)timesmathbb{R}}$ is open?
– Yadati Kiran
Nov 22 at 4:51
1
You are right. answer is wrong. Of course it isn't compact in $mathbb{R}^2.$
– coffeemath
Nov 22 at 4:52
yes, $[1,2]timesBbb R$ is closed in the euclidean topology
– Masacroso
Nov 22 at 4:53
@coffeemath: Compactness is not relevant here I guess.
– Yadati Kiran
Nov 22 at 4:56
@YadatiKiran You're right-- just threw that in in case OP wondered about that, or in case the wording of the question was diferent. [seems odd that the "answer" would be wrong about such a simple question. wonder what book said that..]
– coffeemath
Nov 22 at 4:59
|
show 3 more comments
1
Can you prove its complement is open i.e.${(-infty,1)timesmathbb{R}}cup{(2,infty)timesmathbb{R}}$ is open?
– Yadati Kiran
Nov 22 at 4:51
1
You are right. answer is wrong. Of course it isn't compact in $mathbb{R}^2.$
– coffeemath
Nov 22 at 4:52
yes, $[1,2]timesBbb R$ is closed in the euclidean topology
– Masacroso
Nov 22 at 4:53
@coffeemath: Compactness is not relevant here I guess.
– Yadati Kiran
Nov 22 at 4:56
@YadatiKiran You're right-- just threw that in in case OP wondered about that, or in case the wording of the question was diferent. [seems odd that the "answer" would be wrong about such a simple question. wonder what book said that..]
– coffeemath
Nov 22 at 4:59
1
1
Can you prove its complement is open i.e.${(-infty,1)timesmathbb{R}}cup{(2,infty)timesmathbb{R}}$ is open?
– Yadati Kiran
Nov 22 at 4:51
Can you prove its complement is open i.e.${(-infty,1)timesmathbb{R}}cup{(2,infty)timesmathbb{R}}$ is open?
– Yadati Kiran
Nov 22 at 4:51
1
1
You are right. answer is wrong. Of course it isn't compact in $mathbb{R}^2.$
– coffeemath
Nov 22 at 4:52
You are right. answer is wrong. Of course it isn't compact in $mathbb{R}^2.$
– coffeemath
Nov 22 at 4:52
yes, $[1,2]timesBbb R$ is closed in the euclidean topology
– Masacroso
Nov 22 at 4:53
yes, $[1,2]timesBbb R$ is closed in the euclidean topology
– Masacroso
Nov 22 at 4:53
@coffeemath: Compactness is not relevant here I guess.
– Yadati Kiran
Nov 22 at 4:56
@coffeemath: Compactness is not relevant here I guess.
– Yadati Kiran
Nov 22 at 4:56
@YadatiKiran You're right-- just threw that in in case OP wondered about that, or in case the wording of the question was diferent. [seems odd that the "answer" would be wrong about such a simple question. wonder what book said that..]
– coffeemath
Nov 22 at 4:59
@YadatiKiran You're right-- just threw that in in case OP wondered about that, or in case the wording of the question was diferent. [seems odd that the "answer" would be wrong about such a simple question. wonder what book said that..]
– coffeemath
Nov 22 at 4:59
|
show 3 more comments
1 Answer
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Take a sequence $(x_n)$ of points in $[1,2]times mathbb{R}$. If we write $x_n=(y_n,z_n)$ in the coordinates on the product space then we see that if the sequence $x_n$ converges, then so do the sequences $(y_n)$ and $(z_n)$ in $[1,2]$ and $mathbb{R}$, respectively. Now, $[1,2]$ and $mathbb{R}$ are closed so that $y_nto yin [1,2]$ and $z_nto zin mathbb{R}$. So, $x_nto x=(y,z)in [1,2]times mathbb{R}$. Indeed, $[1,2]times mathbb{R}$ is closed.
Remark: As mentioned in the comments: $[1,2]times mathbb{R}$ is not compact because it is not bounded. The Heine-Borel Theorem says that in the topology on $mathbb{R}^2$, $Omegasubseteq mathbb{R}^2$ is compact if and only if it is closed and bounded.
answered Nov 22 at 4:56
Antonios-Alexandros Robotis
9,11541640
Antonius.Nice answer +.
– Peter Szilas
Nov 22 at 8:57
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
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up vote
3
down vote
accepted
Take a sequence $(x_n)$ of points in $[1,2]times mathbb{R}$. If we write $x_n=(y_n,z_n)$ in the coordinates on the product space then we see that if the sequence $x_n$ converges, then so do the sequences $(y_n)$ and $(z_n)$ in $[1,2]$ and $mathbb{R}$, respectively. Now, $[1,2]$ and $mathbb{R}$ are closed so that $y_nto yin [1,2]$ and $z_nto zin mathbb{R}$. So, $x_nto x=(y,z)in [1,2]times mathbb{R}$. Indeed, $[1,2]times mathbb{R}$ is closed.
Remark: As mentioned in the comments: $[1,2]times mathbb{R}$ is not compact because it is not bounded. The Heine-Borel Theorem says that in the topology on $mathbb{R}^2$, $Omegasubseteq mathbb{R}^2$ is compact if and only if it is closed and bounded.
answered Nov 22 at 4:56
Antonios-Alexandros Robotis
9,11541640
Antonius.Nice answer +.
– Peter Szilas
Nov 22 at 8:57
add a comment |
up vote
3
down vote
accepted
Take a sequence $(x_n)$ of points in $[1,2]times mathbb{R}$. If we write $x_n=(y_n,z_n)$ in the coordinates on the product space then we see that if the sequence $x_n$ converges, then so do the sequences $(y_n)$ and $(z_n)$ in $[1,2]$ and $mathbb{R}$, respectively. Now, $[1,2]$ and $mathbb{R}$ are closed so that $y_nto yin [1,2]$ and $z_nto zin mathbb{R}$. So, $x_nto x=(y,z)in [1,2]times mathbb{R}$. Indeed, $[1,2]times mathbb{R}$ is closed.
Remark: As mentioned in the comments: $[1,2]times mathbb{R}$ is not compact because it is not bounded. The Heine-Borel Theorem says that in the topology on $mathbb{R}^2$, $Omegasubseteq mathbb{R}^2$ is compact if and only if it is closed and bounded.
answered Nov 22 at 4:56
Antonios-Alexandros Robotis
9,11541640
Antonius.Nice answer +.
– Peter Szilas
Nov 22 at 8:57
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Take a sequence $(x_n)$ of points in $[1,2]times mathbb{R}$. If we write $x_n=(y_n,z_n)$ in the coordinates on the product space then we see that if the sequence $x_n$ converges, then so do the sequences $(y_n)$ and $(z_n)$ in $[1,2]$ and $mathbb{R}$, respectively. Now, $[1,2]$ and $mathbb{R}$ are closed so that $y_nto yin [1,2]$ and $z_nto zin mathbb{R}$. So, $x_nto x=(y,z)in [1,2]times mathbb{R}$. Indeed, $[1,2]times mathbb{R}$ is closed.
Remark: As mentioned in the comments: $[1,2]times mathbb{R}$ is not compact because it is not bounded. The Heine-Borel Theorem says that in the topology on $mathbb{R}^2$, $Omegasubseteq mathbb{R}^2$ is compact if and only if it is closed and bounded.
answered Nov 22 at 4:56
Antonios-Alexandros Robotis
9,11541640
Take a sequence $(x_n)$ of points in $[1,2]times mathbb{R}$. If we write $x_n=(y_n,z_n)$ in the coordinates on the product space then we see that if the sequence $x_n$ converges, then so do the sequences $(y_n)$ and $(z_n)$ in $[1,2]$ and $mathbb{R}$, respectively. Now, $[1,2]$ and $mathbb{R}$ are closed so that $y_nto yin [1,2]$ and $z_nto zin mathbb{R}$. So, $x_nto x=(y,z)in [1,2]times mathbb{R}$. Indeed, $[1,2]times mathbb{R}$ is closed.
Remark: As mentioned in the comments: $[1,2]times mathbb{R}$ is not compact because it is not bounded. The Heine-Borel Theorem says that in the topology on $mathbb{R}^2$, $Omegasubseteq mathbb{R}^2$ is compact if and only if it is closed and bounded.
answered Nov 22 at 4:56
Antonios-Alexandros Robotis
9,11541640
answered Nov 22 at 4:56
Antonios-Alexandros Robotis
9,11541640
answered Nov 22 at 4:56
Antonios-Alexandros Robotis
9,11541640
answered Nov 22 at 4:56
Antonios-Alexandros Robotis
9,11541640
9,11541640
Antonius.Nice answer +.
– Peter Szilas
Nov 22 at 8:57
add a comment |
Antonius.Nice answer +.
– Peter Szilas
Nov 22 at 8:57
Antonius.Nice answer +.
– Peter Szilas
Nov 22 at 8:57
Antonius.Nice answer +.
– Peter Szilas
Nov 22 at 8:57
add a comment |
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1
Can you prove its complement is open i.e.${(-infty,1)timesmathbb{R}}cup{(2,infty)timesmathbb{R}}$ is open?
– Yadati Kiran
Nov 22 at 4:51
1
You are right. answer is wrong. Of course it isn't compact in $mathbb{R}^2.$
– coffeemath
Nov 22 at 4:52
yes, $[1,2]timesBbb R$ is closed in the euclidean topology
– Masacroso
Nov 22 at 4:53
@coffeemath: Compactness is not relevant here I guess.
– Yadati Kiran
Nov 22 at 4:56
@YadatiKiran You're right-- just threw that in in case OP wondered about that, or in case the wording of the question was diferent. [seems odd that the "answer" would be wrong about such a simple question. wonder what book said that..]
– coffeemath
Nov 22 at 4:59