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?






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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















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







share|cite|improve this question


















  • 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













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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










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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.






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  • Antonius.Nice answer +.
    – Peter Szilas
    Nov 22 at 8:57











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up vote
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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.






share|cite|improve this answer





















  • Antonius.Nice answer +.
    – Peter Szilas
    Nov 22 at 8:57















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.






share|cite|improve this answer





















  • Antonius.Nice answer +.
    – Peter Szilas
    Nov 22 at 8:57













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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 4:56









Antonios-Alexandros Robotis

9,11541640




9,11541640












  • 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




Antonius.Nice answer +.
– Peter Szilas
Nov 22 at 8:57


















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