State and prove the Bolzano-Weierstrass Theorem for $mathbb{R}^3$ with metric?
I'm given a metric on $(mathbb{R}^3, d_{mathbb{R^3}})$, where
$$
d_{mathbb{R}^3}((x_1, x_2, x_3),(y_1, y_2, y_3))
= |x_1 - y_1| + |x_2 - y_2| + |x_3 - y_3|
$$
And then I'm asked to state and prove the Bolzano-Weierstrass Theorem for ℝ3 with this metric. So, I'm confused about this task. Is it right that I have to prove the following:
Every bounded sequence in R3 has a convergent subsequence?
I have found information for this theorem in Rn, but now I can not understand how to apply given metric.
Thank you in advance!
real-analysis metric-spaces
edited Nov 26 at 19:22
Viktor Glombik
571425
asked Nov 26 at 19:17
Sergey Malinov
154
add a comment |
I'm given a metric on $(mathbb{R}^3, d_{mathbb{R^3}})$, where
$$
d_{mathbb{R}^3}((x_1, x_2, x_3),(y_1, y_2, y_3))
= |x_1 - y_1| + |x_2 - y_2| + |x_3 - y_3|
$$
And then I'm asked to state and prove the Bolzano-Weierstrass Theorem for ℝ3 with this metric. So, I'm confused about this task. Is it right that I have to prove the following:
Every bounded sequence in R3 has a convergent subsequence?
I have found information for this theorem in Rn, but now I can not understand how to apply given metric.
Thank you in advance!
real-analysis metric-spaces
edited Nov 26 at 19:22
Viktor Glombik
571425
asked Nov 26 at 19:17
Sergey Malinov
154
add a comment |
I'm given a metric on $(mathbb{R}^3, d_{mathbb{R^3}})$, where
$$
d_{mathbb{R}^3}((x_1, x_2, x_3),(y_1, y_2, y_3))
= |x_1 - y_1| + |x_2 - y_2| + |x_3 - y_3|
$$
And then I'm asked to state and prove the Bolzano-Weierstrass Theorem for ℝ3 with this metric. So, I'm confused about this task. Is it right that I have to prove the following:
Every bounded sequence in R3 has a convergent subsequence?
I have found information for this theorem in Rn, but now I can not understand how to apply given metric.
Thank you in advance!
real-analysis metric-spaces
edited Nov 26 at 19:22
Viktor Glombik
571425
asked Nov 26 at 19:17
Sergey Malinov
154
I'm given a metric on $(mathbb{R}^3, d_{mathbb{R^3}})$, where
$$
d_{mathbb{R}^3}((x_1, x_2, x_3),(y_1, y_2, y_3))
= |x_1 - y_1| + |x_2 - y_2| + |x_3 - y_3|
$$
And then I'm asked to state and prove the Bolzano-Weierstrass Theorem for ℝ3 with this metric. So, I'm confused about this task. Is it right that I have to prove the following:
Every bounded sequence in R3 has a convergent subsequence?
I have found information for this theorem in Rn, but now I can not understand how to apply given metric.
Thank you in advance!
real-analysis metric-spaces
real-analysis metric-spaces
edited Nov 26 at 19:22
Viktor Glombik
571425
asked Nov 26 at 19:17
Sergey Malinov
154
edited Nov 26 at 19:22
Viktor Glombik
571425
asked Nov 26 at 19:17
Sergey Malinov
154
edited Nov 26 at 19:22
Viktor Glombik
571425
edited Nov 26 at 19:22
Viktor Glombik
571425
edited Nov 26 at 19:22
Viktor Glombik
571425
571425
asked Nov 26 at 19:17
Sergey Malinov
154
asked Nov 26 at 19:17
Sergey Malinov
154
asked Nov 26 at 19:17
Sergey Malinov
154
154
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
I shall call your metric simply $d$ in the sequel. As you say, we have to prove that any $d$-bounded sequence $bigl({bf r}_nbigr)_{ngeq1}$ in ${mathbb R}^3=:X$ has a convergent subsequence. Now "$d$-bounded" means that there is a point ${bf p}in X$ and an $M>0$ such that $d({bf p},{bf r}_n)leq M$ for all $ngeq1$. We may assume WLG that ${bf p}={bf 0}$, i.e., that $d({bf 0},{bf r}_n)=|x_n|+|y_n|+|z_n|leq M$ for all $ngeq1$.
The key idea in the proof is sieving. As $|x_n|leq M$ for all $ngeq1$, by Bolzano's theorem for ${mathbb R}$ the first coordinates $x_n$ of the points ${bf r}_n$ have a convergent subsequence $jmapsto x_{n_j}=:x'_j$ with $x'_jtoxi$ $(jtoinfty)$. Put $y_{n_j}=:y'_j$, $z_{n_j}=:z'_j$. As $|y'_j|leq M$ for all $jgeq1$, by Bolzano's theorem for ${mathbb R}$ the second coordinates $y'_j$ of the points ${bf r'}_j$ have a convergent subsequence $kmapsto y'_{j_k}=:y''_k$ with $y''_ktoeta$ $(ktoinfty)$. Put $x'_{j_k}=:x''_k$ and $z'_{j_k}=:z''_k$. And then the same thing a third time. In the end we have a sub-sub-sub sequence ${bf r}'''_l=(x'''_l,y'''_l,z'''_l)$ $(lgeq 1)$ with $x'''_ltoxi$, $>y_l'''toeta$, $>z'''_ltozeta$ when $ltoinfty$.
It is then easy to see that $$lim_{ltoinfty}dbigl({bf r}'''_l, (xi,eta,zeta)bigr)=0 .$$
answered Nov 27 at 20:03
Christian Blatter
172k7112325
add a comment |
Answering this is easier if one has some basic understanding of functional analysis. First note that $mathbb{R} ^3$ is a linear space. The norm $||x||_1:=|x_1|+|x_2|+|x_3|$ induces the given metric. Now in any finite dimensional normed linear space any two norms are equivalent i. e there exists constants $M_1,M_2>0$ s.t $||x||_1 leq M_1||x||_2$ and $||x||_2 leq M_2||x||_1$, where $||cdot||_2$ is our well known euclidean norm. So if $y_n$ is a sequence bounded w.r.t $||cdot||_1$ then its also bounded w.r.t $||cdot||_2$ and so it has a convergent subseqence w.r.t $||cdot||_2$. This subseqence is now convergent w.r.t $||cdot||_1$ by equivalence of the norms.
answered Nov 27 at 18:09
Arpan Das
613
add a comment |
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2 Answers
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2 Answers
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I shall call your metric simply $d$ in the sequel. As you say, we have to prove that any $d$-bounded sequence $bigl({bf r}_nbigr)_{ngeq1}$ in ${mathbb R}^3=:X$ has a convergent subsequence. Now "$d$-bounded" means that there is a point ${bf p}in X$ and an $M>0$ such that $d({bf p},{bf r}_n)leq M$ for all $ngeq1$. We may assume WLG that ${bf p}={bf 0}$, i.e., that $d({bf 0},{bf r}_n)=|x_n|+|y_n|+|z_n|leq M$ for all $ngeq1$.
The key idea in the proof is sieving. As $|x_n|leq M$ for all $ngeq1$, by Bolzano's theorem for ${mathbb R}$ the first coordinates $x_n$ of the points ${bf r}_n$ have a convergent subsequence $jmapsto x_{n_j}=:x'_j$ with $x'_jtoxi$ $(jtoinfty)$. Put $y_{n_j}=:y'_j$, $z_{n_j}=:z'_j$. As $|y'_j|leq M$ for all $jgeq1$, by Bolzano's theorem for ${mathbb R}$ the second coordinates $y'_j$ of the points ${bf r'}_j$ have a convergent subsequence $kmapsto y'_{j_k}=:y''_k$ with $y''_ktoeta$ $(ktoinfty)$. Put $x'_{j_k}=:x''_k$ and $z'_{j_k}=:z''_k$. And then the same thing a third time. In the end we have a sub-sub-sub sequence ${bf r}'''_l=(x'''_l,y'''_l,z'''_l)$ $(lgeq 1)$ with $x'''_ltoxi$, $>y_l'''toeta$, $>z'''_ltozeta$ when $ltoinfty$.
It is then easy to see that $$lim_{ltoinfty}dbigl({bf r}'''_l, (xi,eta,zeta)bigr)=0 .$$
answered Nov 27 at 20:03
Christian Blatter
172k7112325
add a comment |
I shall call your metric simply $d$ in the sequel. As you say, we have to prove that any $d$-bounded sequence $bigl({bf r}_nbigr)_{ngeq1}$ in ${mathbb R}^3=:X$ has a convergent subsequence. Now "$d$-bounded" means that there is a point ${bf p}in X$ and an $M>0$ such that $d({bf p},{bf r}_n)leq M$ for all $ngeq1$. We may assume WLG that ${bf p}={bf 0}$, i.e., that $d({bf 0},{bf r}_n)=|x_n|+|y_n|+|z_n|leq M$ for all $ngeq1$.
The key idea in the proof is sieving. As $|x_n|leq M$ for all $ngeq1$, by Bolzano's theorem for ${mathbb R}$ the first coordinates $x_n$ of the points ${bf r}_n$ have a convergent subsequence $jmapsto x_{n_j}=:x'_j$ with $x'_jtoxi$ $(jtoinfty)$. Put $y_{n_j}=:y'_j$, $z_{n_j}=:z'_j$. As $|y'_j|leq M$ for all $jgeq1$, by Bolzano's theorem for ${mathbb R}$ the second coordinates $y'_j$ of the points ${bf r'}_j$ have a convergent subsequence $kmapsto y'_{j_k}=:y''_k$ with $y''_ktoeta$ $(ktoinfty)$. Put $x'_{j_k}=:x''_k$ and $z'_{j_k}=:z''_k$. And then the same thing a third time. In the end we have a sub-sub-sub sequence ${bf r}'''_l=(x'''_l,y'''_l,z'''_l)$ $(lgeq 1)$ with $x'''_ltoxi$, $>y_l'''toeta$, $>z'''_ltozeta$ when $ltoinfty$.
It is then easy to see that $$lim_{ltoinfty}dbigl({bf r}'''_l, (xi,eta,zeta)bigr)=0 .$$
answered Nov 27 at 20:03
Christian Blatter
172k7112325
add a comment |
I shall call your metric simply $d$ in the sequel. As you say, we have to prove that any $d$-bounded sequence $bigl({bf r}_nbigr)_{ngeq1}$ in ${mathbb R}^3=:X$ has a convergent subsequence. Now "$d$-bounded" means that there is a point ${bf p}in X$ and an $M>0$ such that $d({bf p},{bf r}_n)leq M$ for all $ngeq1$. We may assume WLG that ${bf p}={bf 0}$, i.e., that $d({bf 0},{bf r}_n)=|x_n|+|y_n|+|z_n|leq M$ for all $ngeq1$.
The key idea in the proof is sieving. As $|x_n|leq M$ for all $ngeq1$, by Bolzano's theorem for ${mathbb R}$ the first coordinates $x_n$ of the points ${bf r}_n$ have a convergent subsequence $jmapsto x_{n_j}=:x'_j$ with $x'_jtoxi$ $(jtoinfty)$. Put $y_{n_j}=:y'_j$, $z_{n_j}=:z'_j$. As $|y'_j|leq M$ for all $jgeq1$, by Bolzano's theorem for ${mathbb R}$ the second coordinates $y'_j$ of the points ${bf r'}_j$ have a convergent subsequence $kmapsto y'_{j_k}=:y''_k$ with $y''_ktoeta$ $(ktoinfty)$. Put $x'_{j_k}=:x''_k$ and $z'_{j_k}=:z''_k$. And then the same thing a third time. In the end we have a sub-sub-sub sequence ${bf r}'''_l=(x'''_l,y'''_l,z'''_l)$ $(lgeq 1)$ with $x'''_ltoxi$, $>y_l'''toeta$, $>z'''_ltozeta$ when $ltoinfty$.
It is then easy to see that $$lim_{ltoinfty}dbigl({bf r}'''_l, (xi,eta,zeta)bigr)=0 .$$
answered Nov 27 at 20:03
Christian Blatter
172k7112325
I shall call your metric simply $d$ in the sequel. As you say, we have to prove that any $d$-bounded sequence $bigl({bf r}_nbigr)_{ngeq1}$ in ${mathbb R}^3=:X$ has a convergent subsequence. Now "$d$-bounded" means that there is a point ${bf p}in X$ and an $M>0$ such that $d({bf p},{bf r}_n)leq M$ for all $ngeq1$. We may assume WLG that ${bf p}={bf 0}$, i.e., that $d({bf 0},{bf r}_n)=|x_n|+|y_n|+|z_n|leq M$ for all $ngeq1$.
The key idea in the proof is sieving. As $|x_n|leq M$ for all $ngeq1$, by Bolzano's theorem for ${mathbb R}$ the first coordinates $x_n$ of the points ${bf r}_n$ have a convergent subsequence $jmapsto x_{n_j}=:x'_j$ with $x'_jtoxi$ $(jtoinfty)$. Put $y_{n_j}=:y'_j$, $z_{n_j}=:z'_j$. As $|y'_j|leq M$ for all $jgeq1$, by Bolzano's theorem for ${mathbb R}$ the second coordinates $y'_j$ of the points ${bf r'}_j$ have a convergent subsequence $kmapsto y'_{j_k}=:y''_k$ with $y''_ktoeta$ $(ktoinfty)$. Put $x'_{j_k}=:x''_k$ and $z'_{j_k}=:z''_k$. And then the same thing a third time. In the end we have a sub-sub-sub sequence ${bf r}'''_l=(x'''_l,y'''_l,z'''_l)$ $(lgeq 1)$ with $x'''_ltoxi$, $>y_l'''toeta$, $>z'''_ltozeta$ when $ltoinfty$.
It is then easy to see that $$lim_{ltoinfty}dbigl({bf r}'''_l, (xi,eta,zeta)bigr)=0 .$$
answered Nov 27 at 20:03
Christian Blatter
172k7112325
answered Nov 27 at 20:03
Christian Blatter
172k7112325
answered Nov 27 at 20:03
Christian Blatter
172k7112325
answered Nov 27 at 20:03
Christian Blatter
172k7112325
172k7112325
add a comment |
add a comment |
Answering this is easier if one has some basic understanding of functional analysis. First note that $mathbb{R} ^3$ is a linear space. The norm $||x||_1:=|x_1|+|x_2|+|x_3|$ induces the given metric. Now in any finite dimensional normed linear space any two norms are equivalent i. e there exists constants $M_1,M_2>0$ s.t $||x||_1 leq M_1||x||_2$ and $||x||_2 leq M_2||x||_1$, where $||cdot||_2$ is our well known euclidean norm. So if $y_n$ is a sequence bounded w.r.t $||cdot||_1$ then its also bounded w.r.t $||cdot||_2$ and so it has a convergent subseqence w.r.t $||cdot||_2$. This subseqence is now convergent w.r.t $||cdot||_1$ by equivalence of the norms.
answered Nov 27 at 18:09
Arpan Das
613
add a comment |
Answering this is easier if one has some basic understanding of functional analysis. First note that $mathbb{R} ^3$ is a linear space. The norm $||x||_1:=|x_1|+|x_2|+|x_3|$ induces the given metric. Now in any finite dimensional normed linear space any two norms are equivalent i. e there exists constants $M_1,M_2>0$ s.t $||x||_1 leq M_1||x||_2$ and $||x||_2 leq M_2||x||_1$, where $||cdot||_2$ is our well known euclidean norm. So if $y_n$ is a sequence bounded w.r.t $||cdot||_1$ then its also bounded w.r.t $||cdot||_2$ and so it has a convergent subseqence w.r.t $||cdot||_2$. This subseqence is now convergent w.r.t $||cdot||_1$ by equivalence of the norms.
answered Nov 27 at 18:09
Arpan Das
613
add a comment |
Answering this is easier if one has some basic understanding of functional analysis. First note that $mathbb{R} ^3$ is a linear space. The norm $||x||_1:=|x_1|+|x_2|+|x_3|$ induces the given metric. Now in any finite dimensional normed linear space any two norms are equivalent i. e there exists constants $M_1,M_2>0$ s.t $||x||_1 leq M_1||x||_2$ and $||x||_2 leq M_2||x||_1$, where $||cdot||_2$ is our well known euclidean norm. So if $y_n$ is a sequence bounded w.r.t $||cdot||_1$ then its also bounded w.r.t $||cdot||_2$ and so it has a convergent subseqence w.r.t $||cdot||_2$. This subseqence is now convergent w.r.t $||cdot||_1$ by equivalence of the norms.
answered Nov 27 at 18:09
Arpan Das
613
Answering this is easier if one has some basic understanding of functional analysis. First note that $mathbb{R} ^3$ is a linear space. The norm $||x||_1:=|x_1|+|x_2|+|x_3|$ induces the given metric. Now in any finite dimensional normed linear space any two norms are equivalent i. e there exists constants $M_1,M_2>0$ s.t $||x||_1 leq M_1||x||_2$ and $||x||_2 leq M_2||x||_1$, where $||cdot||_2$ is our well known euclidean norm. So if $y_n$ is a sequence bounded w.r.t $||cdot||_1$ then its also bounded w.r.t $||cdot||_2$ and so it has a convergent subseqence w.r.t $||cdot||_2$. This subseqence is now convergent w.r.t $||cdot||_1$ by equivalence of the norms.
answered Nov 27 at 18:09
Arpan Das
613
answered Nov 27 at 18:09
Arpan Das
613
answered Nov 27 at 18:09
Arpan Das
613
answered Nov 27 at 18:09
Arpan Das
613
613
add a comment |
add a comment |
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