Set of all sequences converging to $0$ forms a Banach Space.
Let $c_0 ={(x_n)in mathbb{R}:x_{n}to 0},$ then show that $c_0subset l^{infty}$ and that $c_0$ forms a Banach Space.
Since convergent sequences are bounded clearly $c_0subset l^{infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{infty}to 0$ as $nto infty.$ Then we want to show that $u_{k}in c_{0}.$ Clearly for a given $epsilon>0$ there exists $N$ such that for all $ngeq N$ we have that
$$||u_{k}^{n}-u_{k}||<epsilon$$
for all $kin mathbb{N}.$ Therefore for each $kin mathbb{N}$ we have
$$|u_{k}^{n}-u_{k}|leq ||u_{k}^{n}-u_{k}||<epsilon.$$ If we send $kto infty$ we get that
$$|u_k|leq epsilon$$
and therefore $lim_{kto infty}u_k=0.$ This show that $c_0$ is closed and hence complete.
Is this argument correct?
real-analysis proof-verification banach-spaces
edited Nov 25 at 15:15
José Carlos Santos
148k22117218
asked Nov 25 at 15:01
Hello_World
3,83521630
add a comment |
Let $c_0 ={(x_n)in mathbb{R}:x_{n}to 0},$ then show that $c_0subset l^{infty}$ and that $c_0$ forms a Banach Space.
Since convergent sequences are bounded clearly $c_0subset l^{infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{infty}to 0$ as $nto infty.$ Then we want to show that $u_{k}in c_{0}.$ Clearly for a given $epsilon>0$ there exists $N$ such that for all $ngeq N$ we have that
$$||u_{k}^{n}-u_{k}||<epsilon$$
for all $kin mathbb{N}.$ Therefore for each $kin mathbb{N}$ we have
$$|u_{k}^{n}-u_{k}|leq ||u_{k}^{n}-u_{k}||<epsilon.$$ If we send $kto infty$ we get that
$$|u_k|leq epsilon$$
and therefore $lim_{kto infty}u_k=0.$ This show that $c_0$ is closed and hence complete.
Is this argument correct?
real-analysis proof-verification banach-spaces
edited Nov 25 at 15:15
José Carlos Santos
148k22117218
asked Nov 25 at 15:01
Hello_World
3,83521630
add a comment |
Let $c_0 ={(x_n)in mathbb{R}:x_{n}to 0},$ then show that $c_0subset l^{infty}$ and that $c_0$ forms a Banach Space.
Since convergent sequences are bounded clearly $c_0subset l^{infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{infty}to 0$ as $nto infty.$ Then we want to show that $u_{k}in c_{0}.$ Clearly for a given $epsilon>0$ there exists $N$ such that for all $ngeq N$ we have that
$$||u_{k}^{n}-u_{k}||<epsilon$$
for all $kin mathbb{N}.$ Therefore for each $kin mathbb{N}$ we have
$$|u_{k}^{n}-u_{k}|leq ||u_{k}^{n}-u_{k}||<epsilon.$$ If we send $kto infty$ we get that
$$|u_k|leq epsilon$$
and therefore $lim_{kto infty}u_k=0.$ This show that $c_0$ is closed and hence complete.
Is this argument correct?
real-analysis proof-verification banach-spaces
edited Nov 25 at 15:15
José Carlos Santos
148k22117218
asked Nov 25 at 15:01
Hello_World
3,83521630
Let $c_0 ={(x_n)in mathbb{R}:x_{n}to 0},$ then show that $c_0subset l^{infty}$ and that $c_0$ forms a Banach Space.
Since convergent sequences are bounded clearly $c_0subset l^{infty}.$ Next, my strategy is to show that $c_0$ is closed which would imply that it is complete since $l^{infty}$ with the infinite norm is complete. Therefore let $u_{k}^{n}$ be a convergent sequence in $c_0$ such that $||u_{k}^{n}-u_{k}||_{infty}to 0$ as $nto infty.$ Then we want to show that $u_{k}in c_{0}.$ Clearly for a given $epsilon>0$ there exists $N$ such that for all $ngeq N$ we have that
$$||u_{k}^{n}-u_{k}||<epsilon$$
for all $kin mathbb{N}.$ Therefore for each $kin mathbb{N}$ we have
$$|u_{k}^{n}-u_{k}|leq ||u_{k}^{n}-u_{k}||<epsilon.$$ If we send $kto infty$ we get that
$$|u_k|leq epsilon$$
and therefore $lim_{kto infty}u_k=0.$ This show that $c_0$ is closed and hence complete.
Is this argument correct?
real-analysis proof-verification banach-spaces
real-analysis proof-verification banach-spaces
edited Nov 25 at 15:15
José Carlos Santos
148k22117218
asked Nov 25 at 15:01
Hello_World
3,83521630
edited Nov 25 at 15:15
José Carlos Santos
148k22117218
asked Nov 25 at 15:01
Hello_World
3,83521630
edited Nov 25 at 15:15
José Carlos Santos
148k22117218
edited Nov 25 at 15:15
José Carlos Santos
148k22117218
edited Nov 25 at 15:15
José Carlos Santos
148k22117218
148k22117218
asked Nov 25 at 15:01
Hello_World
3,83521630
asked Nov 25 at 15:01
Hello_World
3,83521630
asked Nov 25 at 15:01
Hello_World
3,83521630
3,83521630
add a comment |
add a comment |
1 Answer
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Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:
$(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.
$(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.
answered Nov 25 at 15:12
José Carlos Santos
148k22117218
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
votes
Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:
$(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.
$(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.
answered Nov 25 at 15:12
José Carlos Santos
148k22117218
add a comment |
Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:
$(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.
$(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.
answered Nov 25 at 15:12
José Carlos Santos
148k22117218
add a comment |
Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:
$(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.
$(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.
answered Nov 25 at 15:12
José Carlos Santos
148k22117218
Your argument is indeed correct, but it is simpler to prove that ${c_0}^complement$ is open in $ell^infty$. This can be done by noting that if $(u_n)_{ninmathbb N}in{c_0}^complement$, then there are two possibilites:
$(u_n)_{ninmathbb N}$ converges to $lneq0$. Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $lvert lrvert$ is contained in ${c_0}^complement$.
$(u_n)_{ninmathbb N}$ doesn't converge. Then it is not a Cauchy sequence and therefore there is some $varepsilon>0$ such that$$(forall Ninmathbb{N})(exists m,ninmathbb{N}):m,ngeqslant Nwedgelvert u_m-u_nrvertgeqslantvarepsilon.$$Then the open ball centered at $(u_n)_{ninmathbb N}$ with radius $fracvarepsilon3$ is contained in ${c_0}^complement$.
answered Nov 25 at 15:12
José Carlos Santos
148k22117218
answered Nov 25 at 15:12
José Carlos Santos
148k22117218
answered Nov 25 at 15:12
José Carlos Santos
148k22117218
answered Nov 25 at 15:12
José Carlos Santos
148k22117218
148k22117218
add a comment |
add a comment |
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