Sequential compact set in $mathbb R^2$
$begingroup$
The problem : Show that the set of points in $mathbb R^2$ of the form $(x,sqrt{x})$ for
$x in [0, 2]$ is a sequentially compact subset of $mathbb R^2$.
My approach is to show that I can find a subsequence that converges to a point in the set.
Let $S$ = {$(x,sqrt{x}):x in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.
Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.
general-topology
asked Jan 14 at 20:53
AllorjaAllorja
739
$endgroup$
add a comment |
$begingroup$
The problem : Show that the set of points in $mathbb R^2$ of the form $(x,sqrt{x})$ for
$x in [0, 2]$ is a sequentially compact subset of $mathbb R^2$.
My approach is to show that I can find a subsequence that converges to a point in the set.
Let $S$ = {$(x,sqrt{x}):x in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.
Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.
general-topology
asked Jan 14 at 20:53
AllorjaAllorja
739
$endgroup$
$begingroup$
I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
$endgroup$
– Don Thousand
Jan 14 at 20:58
$begingroup$
What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
$endgroup$
– DanielWainfleet
Jan 14 at 22:00
add a comment |
$begingroup$
The problem : Show that the set of points in $mathbb R^2$ of the form $(x,sqrt{x})$ for
$x in [0, 2]$ is a sequentially compact subset of $mathbb R^2$.
My approach is to show that I can find a subsequence that converges to a point in the set.
Let $S$ = {$(x,sqrt{x}):x in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.
Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.
general-topology
asked Jan 14 at 20:53
AllorjaAllorja
739
$endgroup$
The problem : Show that the set of points in $mathbb R^2$ of the form $(x,sqrt{x})$ for
$x in [0, 2]$ is a sequentially compact subset of $mathbb R^2$.
My approach is to show that I can find a subsequence that converges to a point in the set.
Let $S$ = {$(x,sqrt{x}):x in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.
Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.
general-topology
general-topology
asked Jan 14 at 20:53
AllorjaAllorja
739
asked Jan 14 at 20:53
AllorjaAllorja
739
asked Jan 14 at 20:53
AllorjaAllorja
739
asked Jan 14 at 20:53
AllorjaAllorja
739
asked Jan 14 at 20:53
AllorjaAllorja
739
739
$begingroup$
I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
$endgroup$
– Don Thousand
Jan 14 at 20:58
$begingroup$
What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
$endgroup$
– DanielWainfleet
Jan 14 at 22:00
add a comment |
$begingroup$
I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
$endgroup$
– Don Thousand
Jan 14 at 20:58
$begingroup$
What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
$endgroup$
– DanielWainfleet
Jan 14 at 22:00
$begingroup$
I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
$endgroup$
– Don Thousand
Jan 14 at 20:58
$begingroup$
I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
$endgroup$
– Don Thousand
Jan 14 at 20:58
$begingroup$
What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
$endgroup$
– DanielWainfleet
Jan 14 at 22:00
$begingroup$
What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
$endgroup$
– DanielWainfleet
Jan 14 at 22:00
add a comment |
2 Answers
2
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oldest
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$begingroup$
Let’s try to be more general.
You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.
You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.
So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.
The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.
We’re done.
answered Jan 14 at 21:06
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.
answered Jan 14 at 21:04
GianniGianni
387
$endgroup$
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Let’s try to be more general.
You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.
You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.
So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.
The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.
We’re done.
answered Jan 14 at 21:06
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
Let’s try to be more general.
You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.
You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.
So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.
The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.
We’re done.
answered Jan 14 at 21:06
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
Let’s try to be more general.
You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.
You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.
So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.
The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.
We’re done.
answered Jan 14 at 21:06
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
Let’s try to be more general.
You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.
You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.
So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.
The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.
We’re done.
answered Jan 14 at 21:06
mathcounterexamples.netmathcounterexamples.net
27k22158
edited Jan 14 at 21:29
answered Jan 14 at 21:06
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 14 at 21:06
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 14 at 21:06
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
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In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.
answered Jan 14 at 21:04
GianniGianni
387
$endgroup$
add a comment |
$begingroup$
In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.
answered Jan 14 at 21:04
GianniGianni
387
$endgroup$
add a comment |
$begingroup$
In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.
answered Jan 14 at 21:04
GianniGianni
387
$endgroup$
In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.
answered Jan 14 at 21:04
GianniGianni
387
answered Jan 14 at 21:04
GianniGianni
387
answered Jan 14 at 21:04
GianniGianni
387
answered Jan 14 at 21:04
GianniGianni
387
387
add a comment |
add a comment |
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$begingroup$
I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
$endgroup$
– Don Thousand
Jan 14 at 20:58
$begingroup$
What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
$endgroup$
– DanielWainfleet
Jan 14 at 22:00