choose the correct option regarding complete metric space
$begingroup$
let $X= mathbb{N}$,the set of postive integer consider the metrics $d_1,d_2 $ on $X$ given by
$d_1(m,n) =|m-n|$ , $m , n in X $
$d_2(m,n) =|frac{1}{m} - frac{1}{n}|$ , $m ,n in X $
let $X_1,X_2$ denotes the metric space $(X,d_1),(X,d_2)$ respectively .Then
choose the coorect option
$1.$ $X_1$ is complete
$2.$ $X_2$ is complete
$3.$ $X_1$ is totally bounded
$4.$ $X_2$ is totally bounded
My attempt : I got option $1)$ and option $3)$ beacuse the sequence $x_n = n in mathbb{N}$ is a Cauchy sequence for $d$. And $(x_n)$ don't converge to a limit.
is its true
Any hints/solution will be appreciated
thanks u
general-topology
asked Dec 16 '18 at 19:28
jasminejasmine
1,884418
$endgroup$
add a comment |
$begingroup$
let $X= mathbb{N}$,the set of postive integer consider the metrics $d_1,d_2 $ on $X$ given by
$d_1(m,n) =|m-n|$ , $m , n in X $
$d_2(m,n) =|frac{1}{m} - frac{1}{n}|$ , $m ,n in X $
let $X_1,X_2$ denotes the metric space $(X,d_1),(X,d_2)$ respectively .Then
choose the coorect option
$1.$ $X_1$ is complete
$2.$ $X_2$ is complete
$3.$ $X_1$ is totally bounded
$4.$ $X_2$ is totally bounded
My attempt : I got option $1)$ and option $3)$ beacuse the sequence $x_n = n in mathbb{N}$ is a Cauchy sequence for $d$. And $(x_n)$ don't converge to a limit.
is its true
Any hints/solution will be appreciated
thanks u
general-topology
asked Dec 16 '18 at 19:28
jasminejasmine
1,884418
$endgroup$
add a comment |
$begingroup$
let $X= mathbb{N}$,the set of postive integer consider the metrics $d_1,d_2 $ on $X$ given by
$d_1(m,n) =|m-n|$ , $m , n in X $
$d_2(m,n) =|frac{1}{m} - frac{1}{n}|$ , $m ,n in X $
let $X_1,X_2$ denotes the metric space $(X,d_1),(X,d_2)$ respectively .Then
choose the coorect option
$1.$ $X_1$ is complete
$2.$ $X_2$ is complete
$3.$ $X_1$ is totally bounded
$4.$ $X_2$ is totally bounded
My attempt : I got option $1)$ and option $3)$ beacuse the sequence $x_n = n in mathbb{N}$ is a Cauchy sequence for $d$. And $(x_n)$ don't converge to a limit.
is its true
Any hints/solution will be appreciated
thanks u
general-topology
asked Dec 16 '18 at 19:28
jasminejasmine
1,884418
$endgroup$
let $X= mathbb{N}$,the set of postive integer consider the metrics $d_1,d_2 $ on $X$ given by
$d_1(m,n) =|m-n|$ , $m , n in X $
$d_2(m,n) =|frac{1}{m} - frac{1}{n}|$ , $m ,n in X $
let $X_1,X_2$ denotes the metric space $(X,d_1),(X,d_2)$ respectively .Then
choose the coorect option
$1.$ $X_1$ is complete
$2.$ $X_2$ is complete
$3.$ $X_1$ is totally bounded
$4.$ $X_2$ is totally bounded
My attempt : I got option $1)$ and option $3)$ beacuse the sequence $x_n = n in mathbb{N}$ is a Cauchy sequence for $d$. And $(x_n)$ don't converge to a limit.
is its true
Any hints/solution will be appreciated
thanks u
general-topology
general-topology
asked Dec 16 '18 at 19:28
jasminejasmine
1,884418
asked Dec 16 '18 at 19:28
jasminejasmine
1,884418
edited Dec 17 '18 at 9:28
jasmine
asked Dec 16 '18 at 19:28
jasminejasmine
1,884418
asked Dec 16 '18 at 19:28
jasminejasmine
1,884418
asked Dec 16 '18 at 19:28
jasminejasmine
1,884418
1,884418
add a comment |
add a comment |
2 Answers
2
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$begingroup$
is OK, as Cauchy sequences are eventually constant.
is false, indeed, as $x_n= n$ is a non-convergent Cauchy sequence.
is false, as we cannot cover $X$ by finitely many balls of radius $1$.
is true, as we can compactify it by adding a single point, e.g.
answered Dec 16 '18 at 22:16
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
For point $4.$, to explicitly see that $X_2$ is totally bounded, fix $epsilon > 0$. Choose $N > 3/epsilon.$ Then for any $M > N$, we see $$d_2(N,M) = left lvert frac 1 N - frac 1 M right rvert le frac 1 N + frac 1 M le 2epsilon/3 < epsilon.$$ Thus the set ${ m in mathbb N , : , m ge N}$ is contained in $B_epsilon(N)$. Thus the collection ${B_epsilon(n)}^N_{n=1}$ forms a finite cover of $mathbb N$ by balls of radius $epsilon$.
The rest of the points can be elucidated with simple examples, as in the answer by @HennoBrandsma.
answered Dec 16 '18 at 22:43
User8128User8128
10.8k1622
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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votes
$begingroup$
is OK, as Cauchy sequences are eventually constant.
is false, indeed, as $x_n= n$ is a non-convergent Cauchy sequence.
is false, as we cannot cover $X$ by finitely many balls of radius $1$.
is true, as we can compactify it by adding a single point, e.g.
answered Dec 16 '18 at 22:16
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
is OK, as Cauchy sequences are eventually constant.
is false, indeed, as $x_n= n$ is a non-convergent Cauchy sequence.
is false, as we cannot cover $X$ by finitely many balls of radius $1$.
is true, as we can compactify it by adding a single point, e.g.
answered Dec 16 '18 at 22:16
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
is OK, as Cauchy sequences are eventually constant.
is false, indeed, as $x_n= n$ is a non-convergent Cauchy sequence.
is false, as we cannot cover $X$ by finitely many balls of radius $1$.
is true, as we can compactify it by adding a single point, e.g.
answered Dec 16 '18 at 22:16
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
is OK, as Cauchy sequences are eventually constant.
is false, indeed, as $x_n= n$ is a non-convergent Cauchy sequence.
is false, as we cannot cover $X$ by finitely many balls of radius $1$.
is true, as we can compactify it by adding a single point, e.g.
answered Dec 16 '18 at 22:16
Henno BrandsmaHenno Brandsma
112k348121
answered Dec 16 '18 at 22:16
Henno BrandsmaHenno Brandsma
112k348121
answered Dec 16 '18 at 22:16
Henno BrandsmaHenno Brandsma
112k348121
answered Dec 16 '18 at 22:16
Henno BrandsmaHenno Brandsma
112k348121
112k348121
add a comment |
add a comment |
$begingroup$
For point $4.$, to explicitly see that $X_2$ is totally bounded, fix $epsilon > 0$. Choose $N > 3/epsilon.$ Then for any $M > N$, we see $$d_2(N,M) = left lvert frac 1 N - frac 1 M right rvert le frac 1 N + frac 1 M le 2epsilon/3 < epsilon.$$ Thus the set ${ m in mathbb N , : , m ge N}$ is contained in $B_epsilon(N)$. Thus the collection ${B_epsilon(n)}^N_{n=1}$ forms a finite cover of $mathbb N$ by balls of radius $epsilon$.
The rest of the points can be elucidated with simple examples, as in the answer by @HennoBrandsma.
answered Dec 16 '18 at 22:43
User8128User8128
10.8k1622
$endgroup$
add a comment |
$begingroup$
For point $4.$, to explicitly see that $X_2$ is totally bounded, fix $epsilon > 0$. Choose $N > 3/epsilon.$ Then for any $M > N$, we see $$d_2(N,M) = left lvert frac 1 N - frac 1 M right rvert le frac 1 N + frac 1 M le 2epsilon/3 < epsilon.$$ Thus the set ${ m in mathbb N , : , m ge N}$ is contained in $B_epsilon(N)$. Thus the collection ${B_epsilon(n)}^N_{n=1}$ forms a finite cover of $mathbb N$ by balls of radius $epsilon$.
The rest of the points can be elucidated with simple examples, as in the answer by @HennoBrandsma.
answered Dec 16 '18 at 22:43
User8128User8128
10.8k1622
$endgroup$
add a comment |
$begingroup$
For point $4.$, to explicitly see that $X_2$ is totally bounded, fix $epsilon > 0$. Choose $N > 3/epsilon.$ Then for any $M > N$, we see $$d_2(N,M) = left lvert frac 1 N - frac 1 M right rvert le frac 1 N + frac 1 M le 2epsilon/3 < epsilon.$$ Thus the set ${ m in mathbb N , : , m ge N}$ is contained in $B_epsilon(N)$. Thus the collection ${B_epsilon(n)}^N_{n=1}$ forms a finite cover of $mathbb N$ by balls of radius $epsilon$.
The rest of the points can be elucidated with simple examples, as in the answer by @HennoBrandsma.
answered Dec 16 '18 at 22:43
User8128User8128
10.8k1622
$endgroup$
For point $4.$, to explicitly see that $X_2$ is totally bounded, fix $epsilon > 0$. Choose $N > 3/epsilon.$ Then for any $M > N$, we see $$d_2(N,M) = left lvert frac 1 N - frac 1 M right rvert le frac 1 N + frac 1 M le 2epsilon/3 < epsilon.$$ Thus the set ${ m in mathbb N , : , m ge N}$ is contained in $B_epsilon(N)$. Thus the collection ${B_epsilon(n)}^N_{n=1}$ forms a finite cover of $mathbb N$ by balls of radius $epsilon$.
The rest of the points can be elucidated with simple examples, as in the answer by @HennoBrandsma.
answered Dec 16 '18 at 22:43
User8128User8128
10.8k1622
answered Dec 16 '18 at 22:43
User8128User8128
10.8k1622
answered Dec 16 '18 at 22:43
User8128User8128
10.8k1622
answered Dec 16 '18 at 22:43
User8128User8128
10.8k1622
10.8k1622
add a comment |
add a comment |
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