Closed subset of $mathbb R^2$
$begingroup$
The exercise is as follows:
Show that the set of points in $mathbb R^2 $ of the form $(x, sqrt{x})$
(for $x$ a nonnegative real number) is a closed subset of $mathbb R^2$.
I am new to topology and I'm getting familiar with the definitions of opened and closed sets,open balls and such, but I just don't know how to get around this exercise.
Let's name the set $S$ to be defined as ${(x,sqrt{x}): x ge 0}$. Should I take the complement of $S$ (and if so, how can I?) and find an $varepsilon$-ball not in $mathbb R^2setminus S$ or is there another way around it.
Thank you in prior.
general-topology
edited Jan 21 at 13:28
Thomas Shelby
4,0342625
asked Jan 12 at 13:06
AllorjaAllorja
789
$endgroup$
add a comment |
$begingroup$
The exercise is as follows:
Show that the set of points in $mathbb R^2 $ of the form $(x, sqrt{x})$
(for $x$ a nonnegative real number) is a closed subset of $mathbb R^2$.
I am new to topology and I'm getting familiar with the definitions of opened and closed sets,open balls and such, but I just don't know how to get around this exercise.
Let's name the set $S$ to be defined as ${(x,sqrt{x}): x ge 0}$. Should I take the complement of $S$ (and if so, how can I?) and find an $varepsilon$-ball not in $mathbb R^2setminus S$ or is there another way around it.
Thank you in prior.
general-topology
edited Jan 21 at 13:28
Thomas Shelby
4,0342625
asked Jan 12 at 13:06
AllorjaAllorja
789
$endgroup$
3
$begingroup$
There are many different ways, however your approach sounds good. Another way is to use the fact that $mathbb{R}^2$ is metric and so it is enough to show that every convergence sequence $(x_n,sqrt{x_n})$ converges to an element in your set (this one sounds easier to me). Equivalently you can use this theorem en.wikipedia.org/wiki/Closed_graph_theorem
$endgroup$
– Yanko
Jan 12 at 13:09
add a comment |
$begingroup$
The exercise is as follows:
Show that the set of points in $mathbb R^2 $ of the form $(x, sqrt{x})$
(for $x$ a nonnegative real number) is a closed subset of $mathbb R^2$.
I am new to topology and I'm getting familiar with the definitions of opened and closed sets,open balls and such, but I just don't know how to get around this exercise.
Let's name the set $S$ to be defined as ${(x,sqrt{x}): x ge 0}$. Should I take the complement of $S$ (and if so, how can I?) and find an $varepsilon$-ball not in $mathbb R^2setminus S$ or is there another way around it.
Thank you in prior.
general-topology
edited Jan 21 at 13:28
Thomas Shelby
4,0342625
asked Jan 12 at 13:06
AllorjaAllorja
789
$endgroup$
The exercise is as follows:
Show that the set of points in $mathbb R^2 $ of the form $(x, sqrt{x})$
(for $x$ a nonnegative real number) is a closed subset of $mathbb R^2$.
I am new to topology and I'm getting familiar with the definitions of opened and closed sets,open balls and such, but I just don't know how to get around this exercise.
Let's name the set $S$ to be defined as ${(x,sqrt{x}): x ge 0}$. Should I take the complement of $S$ (and if so, how can I?) and find an $varepsilon$-ball not in $mathbb R^2setminus S$ or is there another way around it.
Thank you in prior.
general-topology
general-topology
edited Jan 21 at 13:28
Thomas Shelby
4,0342625
asked Jan 12 at 13:06
AllorjaAllorja
789
edited Jan 21 at 13:28
Thomas Shelby
4,0342625
asked Jan 12 at 13:06
AllorjaAllorja
789
edited Jan 21 at 13:28
Thomas Shelby
4,0342625
edited Jan 21 at 13:28
Thomas Shelby
4,0342625
edited Jan 21 at 13:28
Thomas Shelby
4,0342625
4,0342625
asked Jan 12 at 13:06
AllorjaAllorja
789
asked Jan 12 at 13:06
AllorjaAllorja
789
asked Jan 12 at 13:06
AllorjaAllorja
789
789
3
$begingroup$
There are many different ways, however your approach sounds good. Another way is to use the fact that $mathbb{R}^2$ is metric and so it is enough to show that every convergence sequence $(x_n,sqrt{x_n})$ converges to an element in your set (this one sounds easier to me). Equivalently you can use this theorem en.wikipedia.org/wiki/Closed_graph_theorem
$endgroup$
– Yanko
Jan 12 at 13:09
add a comment |
3
$begingroup$
There are many different ways, however your approach sounds good. Another way is to use the fact that $mathbb{R}^2$ is metric and so it is enough to show that every convergence sequence $(x_n,sqrt{x_n})$ converges to an element in your set (this one sounds easier to me). Equivalently you can use this theorem en.wikipedia.org/wiki/Closed_graph_theorem
$endgroup$
– Yanko
Jan 12 at 13:09
3
3
$begingroup$
There are many different ways, however your approach sounds good. Another way is to use the fact that $mathbb{R}^2$ is metric and so it is enough to show that every convergence sequence $(x_n,sqrt{x_n})$ converges to an element in your set (this one sounds easier to me). Equivalently you can use this theorem en.wikipedia.org/wiki/Closed_graph_theorem
$endgroup$
– Yanko
Jan 12 at 13:09
$begingroup$
There are many different ways, however your approach sounds good. Another way is to use the fact that $mathbb{R}^2$ is metric and so it is enough to show that every convergence sequence $(x_n,sqrt{x_n})$ converges to an element in your set (this one sounds easier to me). Equivalently you can use this theorem en.wikipedia.org/wiki/Closed_graph_theorem
$endgroup$
– Yanko
Jan 12 at 13:09
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
There are multiple ways to show that a subset is closed. In case of metric spaces there is a useful characterization:
A subset $A$ of a metric space $X$ is closed if and only if for any convergent sequence $(a_n)subseteq X$ if $a_nin A$ for all $n$ then $lim a_nin A$.
So let's give it a try. Let $(v_n)$ be a convergent sequence in $S$. So $v_n=(x_n, sqrt{x_n})$ by the definiton of $S$. Since $v_n$ is convergent then so is $x_n$. Now since $sqrt{cdot}$ is a continuous function then $sqrt{x_n}$ is convergent as well and $limsqrt{x_n}=sqrt{lim x_n}$. It follows that
$$lim v_n=lim (x_n,sqrt{x_n})=(lim x_n, limsqrt{x_n})=$$
$$=(lim x_n,sqrt{lim x_n})$$
so $lim v_n$ is again of the form $(a,sqrt{a})$ and so $lim v_nin S$ proving that $S$ is closed.
answered Jan 12 at 17:12
freakishfreakish
12.7k1631
$endgroup$
add a comment |
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$begingroup$
There are multiple ways to show that a subset is closed. In case of metric spaces there is a useful characterization:
A subset $A$ of a metric space $X$ is closed if and only if for any convergent sequence $(a_n)subseteq X$ if $a_nin A$ for all $n$ then $lim a_nin A$.
So let's give it a try. Let $(v_n)$ be a convergent sequence in $S$. So $v_n=(x_n, sqrt{x_n})$ by the definiton of $S$. Since $v_n$ is convergent then so is $x_n$. Now since $sqrt{cdot}$ is a continuous function then $sqrt{x_n}$ is convergent as well and $limsqrt{x_n}=sqrt{lim x_n}$. It follows that
$$lim v_n=lim (x_n,sqrt{x_n})=(lim x_n, limsqrt{x_n})=$$
$$=(lim x_n,sqrt{lim x_n})$$
so $lim v_n$ is again of the form $(a,sqrt{a})$ and so $lim v_nin S$ proving that $S$ is closed.
answered Jan 12 at 17:12
freakishfreakish
12.7k1631
$endgroup$
add a comment |
$begingroup$
There are multiple ways to show that a subset is closed. In case of metric spaces there is a useful characterization:
A subset $A$ of a metric space $X$ is closed if and only if for any convergent sequence $(a_n)subseteq X$ if $a_nin A$ for all $n$ then $lim a_nin A$.
So let's give it a try. Let $(v_n)$ be a convergent sequence in $S$. So $v_n=(x_n, sqrt{x_n})$ by the definiton of $S$. Since $v_n$ is convergent then so is $x_n$. Now since $sqrt{cdot}$ is a continuous function then $sqrt{x_n}$ is convergent as well and $limsqrt{x_n}=sqrt{lim x_n}$. It follows that
$$lim v_n=lim (x_n,sqrt{x_n})=(lim x_n, limsqrt{x_n})=$$
$$=(lim x_n,sqrt{lim x_n})$$
so $lim v_n$ is again of the form $(a,sqrt{a})$ and so $lim v_nin S$ proving that $S$ is closed.
answered Jan 12 at 17:12
freakishfreakish
12.7k1631
$endgroup$
add a comment |
$begingroup$
There are multiple ways to show that a subset is closed. In case of metric spaces there is a useful characterization:
A subset $A$ of a metric space $X$ is closed if and only if for any convergent sequence $(a_n)subseteq X$ if $a_nin A$ for all $n$ then $lim a_nin A$.
So let's give it a try. Let $(v_n)$ be a convergent sequence in $S$. So $v_n=(x_n, sqrt{x_n})$ by the definiton of $S$. Since $v_n$ is convergent then so is $x_n$. Now since $sqrt{cdot}$ is a continuous function then $sqrt{x_n}$ is convergent as well and $limsqrt{x_n}=sqrt{lim x_n}$. It follows that
$$lim v_n=lim (x_n,sqrt{x_n})=(lim x_n, limsqrt{x_n})=$$
$$=(lim x_n,sqrt{lim x_n})$$
so $lim v_n$ is again of the form $(a,sqrt{a})$ and so $lim v_nin S$ proving that $S$ is closed.
answered Jan 12 at 17:12
freakishfreakish
12.7k1631
$endgroup$
There are multiple ways to show that a subset is closed. In case of metric spaces there is a useful characterization:
A subset $A$ of a metric space $X$ is closed if and only if for any convergent sequence $(a_n)subseteq X$ if $a_nin A$ for all $n$ then $lim a_nin A$.
So let's give it a try. Let $(v_n)$ be a convergent sequence in $S$. So $v_n=(x_n, sqrt{x_n})$ by the definiton of $S$. Since $v_n$ is convergent then so is $x_n$. Now since $sqrt{cdot}$ is a continuous function then $sqrt{x_n}$ is convergent as well and $limsqrt{x_n}=sqrt{lim x_n}$. It follows that
$$lim v_n=lim (x_n,sqrt{x_n})=(lim x_n, limsqrt{x_n})=$$
$$=(lim x_n,sqrt{lim x_n})$$
so $lim v_n$ is again of the form $(a,sqrt{a})$ and so $lim v_nin S$ proving that $S$ is closed.
answered Jan 12 at 17:12
freakishfreakish
12.7k1631
answered Jan 12 at 17:12
freakishfreakish
12.7k1631
answered Jan 12 at 17:12
freakishfreakish
12.7k1631
answered Jan 12 at 17:12
freakishfreakish
12.7k1631
12.7k1631
add a comment |
add a comment |
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$begingroup$
There are many different ways, however your approach sounds good. Another way is to use the fact that $mathbb{R}^2$ is metric and so it is enough to show that every convergence sequence $(x_n,sqrt{x_n})$ converges to an element in your set (this one sounds easier to me). Equivalently you can use this theorem en.wikipedia.org/wiki/Closed_graph_theorem
$endgroup$
– Yanko
Jan 12 at 13:09