Topology in $Bbb {R}^n$ Proof Exercise [closed]
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Given set A$ subset Bbb {R}^n$ and set B$ subset Bbb {R}^n$, we define A+B={$text{ a+b such that a $in$ A and b $in$ B}}$.Prove: $$text{a) If A closed set, B compact set then A+B is a closed set}$$ $$text{b) If A,B are compact sets then A+B is a compact set}$$
general-topology compactness
asked Dec 17 '18 at 20:12
juan deutschjuan deutsch
156
$endgroup$
closed as off-topic by Carl Schildkraut, Cesareo, Davide Giraudo, Brahadeesh, Adrian Keister Dec 18 '18 at 13:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Schildkraut, Cesareo, Davide Giraudo, Brahadeesh, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Given set A$ subset Bbb {R}^n$ and set B$ subset Bbb {R}^n$, we define A+B={$text{ a+b such that a $in$ A and b $in$ B}}$.Prove: $$text{a) If A closed set, B compact set then A+B is a closed set}$$ $$text{b) If A,B are compact sets then A+B is a compact set}$$
general-topology compactness
asked Dec 17 '18 at 20:12
juan deutschjuan deutsch
156
$endgroup$
closed as off-topic by Carl Schildkraut, Cesareo, Davide Giraudo, Brahadeesh, Adrian Keister Dec 18 '18 at 13:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Schildkraut, Cesareo, Davide Giraudo, Brahadeesh, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What have you tried?
$endgroup$
– EDZ
Dec 17 '18 at 20:20
$begingroup$
Hint for b): write $A+B$ as a continuous image of $Atimes B subset mathbb{R}^{2n}$.
$endgroup$
– Slade
Dec 17 '18 at 20:22
add a comment |
$begingroup$
Given set A$ subset Bbb {R}^n$ and set B$ subset Bbb {R}^n$, we define A+B={$text{ a+b such that a $in$ A and b $in$ B}}$.Prove: $$text{a) If A closed set, B compact set then A+B is a closed set}$$ $$text{b) If A,B are compact sets then A+B is a compact set}$$
general-topology compactness
asked Dec 17 '18 at 20:12
juan deutschjuan deutsch
156
$endgroup$
Given set A$ subset Bbb {R}^n$ and set B$ subset Bbb {R}^n$, we define A+B={$text{ a+b such that a $in$ A and b $in$ B}}$.Prove: $$text{a) If A closed set, B compact set then A+B is a closed set}$$ $$text{b) If A,B are compact sets then A+B is a compact set}$$
general-topology compactness
general-topology compactness
asked Dec 17 '18 at 20:12
juan deutschjuan deutsch
156
asked Dec 17 '18 at 20:12
juan deutschjuan deutsch
156
asked Dec 17 '18 at 20:12
juan deutschjuan deutsch
156
asked Dec 17 '18 at 20:12
juan deutschjuan deutsch
156
asked Dec 17 '18 at 20:12
juan deutschjuan deutsch
156
156
closed as off-topic by Carl Schildkraut, Cesareo, Davide Giraudo, Brahadeesh, Adrian Keister Dec 18 '18 at 13:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Schildkraut, Cesareo, Davide Giraudo, Brahadeesh, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Carl Schildkraut, Cesareo, Davide Giraudo, Brahadeesh, Adrian Keister Dec 18 '18 at 13:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Schildkraut, Cesareo, Davide Giraudo, Brahadeesh, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What have you tried?
$endgroup$
– EDZ
Dec 17 '18 at 20:20
$begingroup$
Hint for b): write $A+B$ as a continuous image of $Atimes B subset mathbb{R}^{2n}$.
$endgroup$
– Slade
Dec 17 '18 at 20:22
add a comment |
$begingroup$
What have you tried?
$endgroup$
– EDZ
Dec 17 '18 at 20:20
$begingroup$
Hint for b): write $A+B$ as a continuous image of $Atimes B subset mathbb{R}^{2n}$.
$endgroup$
– Slade
Dec 17 '18 at 20:22
$begingroup$
What have you tried?
$endgroup$
– EDZ
Dec 17 '18 at 20:20
$begingroup$
What have you tried?
$endgroup$
– EDZ
Dec 17 '18 at 20:20
$begingroup$
Hint for b): write $A+B$ as a continuous image of $Atimes B subset mathbb{R}^{2n}$.
$endgroup$
– Slade
Dec 17 '18 at 20:22
$begingroup$
Hint for b): write $A+B$ as a continuous image of $Atimes B subset mathbb{R}^{2n}$.
$endgroup$
– Slade
Dec 17 '18 at 20:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given that $mathbb{R}^n$ is a metric space, we consider closeness and compactness from the point of view of sequences. Let ${y_n}$ be a sequence in $A+B$, than $y_n$ can be written as $a_n+b_n$ where $a_nin A$, $b_nin B$.
First we want to show that if $y_n$ converges to a certain $y$, than $y$ is an element of $A+B$. The hypothesis tells us that $a_n+b_n rightarrow y$. Now because $B$ is compact there exists a $bin B$ such that $b_nrightarrow b$. Now $a_n rightarrow y-b$ and for closedness of $A$ we deduce that $y-bin A$. Call $a=y-b$, then $y=a+b$ where $ain A$ and $bin B$. So $yin A+B$ implies that $A+B$ is closed.
Second we want to show that if ${y_n= a_n+b_n}$ is a sequence of $A+B$, then it admits a converging subsequence. From ${a_n}in A$ we can take a converging subsequence ${a_{n_j}}$, so $y_{n_j} = a_{n_j}+b_{n_j}$. From ${b_{n_j}}in B$ we can take a converging subsequence ${b_{n_{j_k}}}$, so $y_{n_{j_k}} = a_{n_{j_k}}+b_{n_{j_k}}rightarrow y$, so $A+B$ is compact.
edited Dec 18 '18 at 5:17
Henno Brandsma
110k347117
answered Dec 17 '18 at 20:38
PiccoloPaoloPiccoloPaolo
413
$endgroup$
$begingroup$
Thank you very much! Super clear.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:46
$begingroup$
Nitpick: for the fisrt proof, there is a subsequence of the $b_n$ that converges to some $b in B$. You then have to work in that subsequence.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:25
$begingroup$
As an alternative to the last one: $f(x,y) = x+y$ is continuous and $A times B$ is compact, hence so is $f[A times B] = A + B$.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:26
add a comment |
$begingroup$
Hint for a)
Take any convergent sequence ${a_k+b_k}_{k=1}^{infty}$ where $a_kin A$ and $b_kin B$.
We can pick out a convergent subsequence of ${b_k}_{k=1}^{infty}$ denoted ${b_{k_j}}_{j=1}^{infty}$ (why?). Say $b_{k_j}rightarrow b$. What can you now say about ${a_{k_j}}_{j=1}^{infty}$?
answered Dec 17 '18 at 20:22
Olof RubinOlof Rubin
1,156316
$endgroup$
$begingroup$
How can you know that a convergent sequence ${a_k+b_k}_{k=1}^{infty}$ exists with the conditions you specified? Answering the why ${b_{k_j}}_{j=1}^{infty}$ exists as a convergent sequence, would be because B is a bounded set, therefore ${b_k}_{k=1}^{infty}$ would be bounded and would therefore have a convergent subsequence.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:35
$begingroup$
Any sequence in $A+B$ will be a sequence ${x_k}_{k=1}^{infty}$ where $x_k = a_k+b_k$ where $a_kin A$ and $b_kin B$. In general any element $xin A+B$ is of the form $a+b$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:37
$begingroup$
Right, but how do you know a convergent one exists?
$endgroup$
– juan deutsch
Dec 17 '18 at 20:41
$begingroup$
A set is closed if and only if every convergent sequence has its limit in the set. Therefore to prove that a $A+$ is closed we consider an ARBITRARY sequence and prove that its limit lies in $A+B$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:42
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given that $mathbb{R}^n$ is a metric space, we consider closeness and compactness from the point of view of sequences. Let ${y_n}$ be a sequence in $A+B$, than $y_n$ can be written as $a_n+b_n$ where $a_nin A$, $b_nin B$.
First we want to show that if $y_n$ converges to a certain $y$, than $y$ is an element of $A+B$. The hypothesis tells us that $a_n+b_n rightarrow y$. Now because $B$ is compact there exists a $bin B$ such that $b_nrightarrow b$. Now $a_n rightarrow y-b$ and for closedness of $A$ we deduce that $y-bin A$. Call $a=y-b$, then $y=a+b$ where $ain A$ and $bin B$. So $yin A+B$ implies that $A+B$ is closed.
Second we want to show that if ${y_n= a_n+b_n}$ is a sequence of $A+B$, then it admits a converging subsequence. From ${a_n}in A$ we can take a converging subsequence ${a_{n_j}}$, so $y_{n_j} = a_{n_j}+b_{n_j}$. From ${b_{n_j}}in B$ we can take a converging subsequence ${b_{n_{j_k}}}$, so $y_{n_{j_k}} = a_{n_{j_k}}+b_{n_{j_k}}rightarrow y$, so $A+B$ is compact.
edited Dec 18 '18 at 5:17
Henno Brandsma
110k347117
answered Dec 17 '18 at 20:38
PiccoloPaoloPiccoloPaolo
413
$endgroup$
$begingroup$
Thank you very much! Super clear.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:46
$begingroup$
Nitpick: for the fisrt proof, there is a subsequence of the $b_n$ that converges to some $b in B$. You then have to work in that subsequence.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:25
$begingroup$
As an alternative to the last one: $f(x,y) = x+y$ is continuous and $A times B$ is compact, hence so is $f[A times B] = A + B$.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:26
add a comment |
$begingroup$
Given that $mathbb{R}^n$ is a metric space, we consider closeness and compactness from the point of view of sequences. Let ${y_n}$ be a sequence in $A+B$, than $y_n$ can be written as $a_n+b_n$ where $a_nin A$, $b_nin B$.
First we want to show that if $y_n$ converges to a certain $y$, than $y$ is an element of $A+B$. The hypothesis tells us that $a_n+b_n rightarrow y$. Now because $B$ is compact there exists a $bin B$ such that $b_nrightarrow b$. Now $a_n rightarrow y-b$ and for closedness of $A$ we deduce that $y-bin A$. Call $a=y-b$, then $y=a+b$ where $ain A$ and $bin B$. So $yin A+B$ implies that $A+B$ is closed.
Second we want to show that if ${y_n= a_n+b_n}$ is a sequence of $A+B$, then it admits a converging subsequence. From ${a_n}in A$ we can take a converging subsequence ${a_{n_j}}$, so $y_{n_j} = a_{n_j}+b_{n_j}$. From ${b_{n_j}}in B$ we can take a converging subsequence ${b_{n_{j_k}}}$, so $y_{n_{j_k}} = a_{n_{j_k}}+b_{n_{j_k}}rightarrow y$, so $A+B$ is compact.
edited Dec 18 '18 at 5:17
Henno Brandsma
110k347117
answered Dec 17 '18 at 20:38
PiccoloPaoloPiccoloPaolo
413
$endgroup$
$begingroup$
Thank you very much! Super clear.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:46
$begingroup$
Nitpick: for the fisrt proof, there is a subsequence of the $b_n$ that converges to some $b in B$. You then have to work in that subsequence.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:25
$begingroup$
As an alternative to the last one: $f(x,y) = x+y$ is continuous and $A times B$ is compact, hence so is $f[A times B] = A + B$.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:26
add a comment |
$begingroup$
Given that $mathbb{R}^n$ is a metric space, we consider closeness and compactness from the point of view of sequences. Let ${y_n}$ be a sequence in $A+B$, than $y_n$ can be written as $a_n+b_n$ where $a_nin A$, $b_nin B$.
First we want to show that if $y_n$ converges to a certain $y$, than $y$ is an element of $A+B$. The hypothesis tells us that $a_n+b_n rightarrow y$. Now because $B$ is compact there exists a $bin B$ such that $b_nrightarrow b$. Now $a_n rightarrow y-b$ and for closedness of $A$ we deduce that $y-bin A$. Call $a=y-b$, then $y=a+b$ where $ain A$ and $bin B$. So $yin A+B$ implies that $A+B$ is closed.
Second we want to show that if ${y_n= a_n+b_n}$ is a sequence of $A+B$, then it admits a converging subsequence. From ${a_n}in A$ we can take a converging subsequence ${a_{n_j}}$, so $y_{n_j} = a_{n_j}+b_{n_j}$. From ${b_{n_j}}in B$ we can take a converging subsequence ${b_{n_{j_k}}}$, so $y_{n_{j_k}} = a_{n_{j_k}}+b_{n_{j_k}}rightarrow y$, so $A+B$ is compact.
edited Dec 18 '18 at 5:17
Henno Brandsma
110k347117
answered Dec 17 '18 at 20:38
PiccoloPaoloPiccoloPaolo
413
$endgroup$
Given that $mathbb{R}^n$ is a metric space, we consider closeness and compactness from the point of view of sequences. Let ${y_n}$ be a sequence in $A+B$, than $y_n$ can be written as $a_n+b_n$ where $a_nin A$, $b_nin B$.
First we want to show that if $y_n$ converges to a certain $y$, than $y$ is an element of $A+B$. The hypothesis tells us that $a_n+b_n rightarrow y$. Now because $B$ is compact there exists a $bin B$ such that $b_nrightarrow b$. Now $a_n rightarrow y-b$ and for closedness of $A$ we deduce that $y-bin A$. Call $a=y-b$, then $y=a+b$ where $ain A$ and $bin B$. So $yin A+B$ implies that $A+B$ is closed.
Second we want to show that if ${y_n= a_n+b_n}$ is a sequence of $A+B$, then it admits a converging subsequence. From ${a_n}in A$ we can take a converging subsequence ${a_{n_j}}$, so $y_{n_j} = a_{n_j}+b_{n_j}$. From ${b_{n_j}}in B$ we can take a converging subsequence ${b_{n_{j_k}}}$, so $y_{n_{j_k}} = a_{n_{j_k}}+b_{n_{j_k}}rightarrow y$, so $A+B$ is compact.
edited Dec 18 '18 at 5:17
Henno Brandsma
110k347117
answered Dec 17 '18 at 20:38
PiccoloPaoloPiccoloPaolo
413
edited Dec 18 '18 at 5:17
Henno Brandsma
110k347117
edited Dec 18 '18 at 5:17
Henno Brandsma
110k347117
edited Dec 18 '18 at 5:17
Henno Brandsma
110k347117
110k347117
answered Dec 17 '18 at 20:38
PiccoloPaoloPiccoloPaolo
413
answered Dec 17 '18 at 20:38
PiccoloPaoloPiccoloPaolo
413
answered Dec 17 '18 at 20:38
PiccoloPaoloPiccoloPaolo
413
413
$begingroup$
Thank you very much! Super clear.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:46
$begingroup$
Nitpick: for the fisrt proof, there is a subsequence of the $b_n$ that converges to some $b in B$. You then have to work in that subsequence.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:25
$begingroup$
As an alternative to the last one: $f(x,y) = x+y$ is continuous and $A times B$ is compact, hence so is $f[A times B] = A + B$.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:26
add a comment |
$begingroup$
Thank you very much! Super clear.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:46
$begingroup$
Nitpick: for the fisrt proof, there is a subsequence of the $b_n$ that converges to some $b in B$. You then have to work in that subsequence.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:25
$begingroup$
As an alternative to the last one: $f(x,y) = x+y$ is continuous and $A times B$ is compact, hence so is $f[A times B] = A + B$.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:26
$begingroup$
Thank you very much! Super clear.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:46
$begingroup$
Thank you very much! Super clear.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:46
$begingroup$
Nitpick: for the fisrt proof, there is a subsequence of the $b_n$ that converges to some $b in B$. You then have to work in that subsequence.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:25
$begingroup$
Nitpick: for the fisrt proof, there is a subsequence of the $b_n$ that converges to some $b in B$. You then have to work in that subsequence.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:25
$begingroup$
As an alternative to the last one: $f(x,y) = x+y$ is continuous and $A times B$ is compact, hence so is $f[A times B] = A + B$.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:26
$begingroup$
As an alternative to the last one: $f(x,y) = x+y$ is continuous and $A times B$ is compact, hence so is $f[A times B] = A + B$.
$endgroup$
– Henno Brandsma
Dec 17 '18 at 22:26
add a comment |
$begingroup$
Hint for a)
Take any convergent sequence ${a_k+b_k}_{k=1}^{infty}$ where $a_kin A$ and $b_kin B$.
We can pick out a convergent subsequence of ${b_k}_{k=1}^{infty}$ denoted ${b_{k_j}}_{j=1}^{infty}$ (why?). Say $b_{k_j}rightarrow b$. What can you now say about ${a_{k_j}}_{j=1}^{infty}$?
answered Dec 17 '18 at 20:22
Olof RubinOlof Rubin
1,156316
$endgroup$
$begingroup$
How can you know that a convergent sequence ${a_k+b_k}_{k=1}^{infty}$ exists with the conditions you specified? Answering the why ${b_{k_j}}_{j=1}^{infty}$ exists as a convergent sequence, would be because B is a bounded set, therefore ${b_k}_{k=1}^{infty}$ would be bounded and would therefore have a convergent subsequence.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:35
$begingroup$
Any sequence in $A+B$ will be a sequence ${x_k}_{k=1}^{infty}$ where $x_k = a_k+b_k$ where $a_kin A$ and $b_kin B$. In general any element $xin A+B$ is of the form $a+b$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:37
$begingroup$
Right, but how do you know a convergent one exists?
$endgroup$
– juan deutsch
Dec 17 '18 at 20:41
$begingroup$
A set is closed if and only if every convergent sequence has its limit in the set. Therefore to prove that a $A+$ is closed we consider an ARBITRARY sequence and prove that its limit lies in $A+B$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:42
add a comment |
$begingroup$
Hint for a)
Take any convergent sequence ${a_k+b_k}_{k=1}^{infty}$ where $a_kin A$ and $b_kin B$.
We can pick out a convergent subsequence of ${b_k}_{k=1}^{infty}$ denoted ${b_{k_j}}_{j=1}^{infty}$ (why?). Say $b_{k_j}rightarrow b$. What can you now say about ${a_{k_j}}_{j=1}^{infty}$?
answered Dec 17 '18 at 20:22
Olof RubinOlof Rubin
1,156316
$endgroup$
$begingroup$
How can you know that a convergent sequence ${a_k+b_k}_{k=1}^{infty}$ exists with the conditions you specified? Answering the why ${b_{k_j}}_{j=1}^{infty}$ exists as a convergent sequence, would be because B is a bounded set, therefore ${b_k}_{k=1}^{infty}$ would be bounded and would therefore have a convergent subsequence.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:35
$begingroup$
Any sequence in $A+B$ will be a sequence ${x_k}_{k=1}^{infty}$ where $x_k = a_k+b_k$ where $a_kin A$ and $b_kin B$. In general any element $xin A+B$ is of the form $a+b$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:37
$begingroup$
Right, but how do you know a convergent one exists?
$endgroup$
– juan deutsch
Dec 17 '18 at 20:41
$begingroup$
A set is closed if and only if every convergent sequence has its limit in the set. Therefore to prove that a $A+$ is closed we consider an ARBITRARY sequence and prove that its limit lies in $A+B$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:42
add a comment |
$begingroup$
Hint for a)
Take any convergent sequence ${a_k+b_k}_{k=1}^{infty}$ where $a_kin A$ and $b_kin B$.
We can pick out a convergent subsequence of ${b_k}_{k=1}^{infty}$ denoted ${b_{k_j}}_{j=1}^{infty}$ (why?). Say $b_{k_j}rightarrow b$. What can you now say about ${a_{k_j}}_{j=1}^{infty}$?
answered Dec 17 '18 at 20:22
Olof RubinOlof Rubin
1,156316
$endgroup$
Hint for a)
Take any convergent sequence ${a_k+b_k}_{k=1}^{infty}$ where $a_kin A$ and $b_kin B$.
We can pick out a convergent subsequence of ${b_k}_{k=1}^{infty}$ denoted ${b_{k_j}}_{j=1}^{infty}$ (why?). Say $b_{k_j}rightarrow b$. What can you now say about ${a_{k_j}}_{j=1}^{infty}$?
answered Dec 17 '18 at 20:22
Olof RubinOlof Rubin
1,156316
answered Dec 17 '18 at 20:22
Olof RubinOlof Rubin
1,156316
answered Dec 17 '18 at 20:22
Olof RubinOlof Rubin
1,156316
answered Dec 17 '18 at 20:22
Olof RubinOlof Rubin
1,156316
1,156316
$begingroup$
How can you know that a convergent sequence ${a_k+b_k}_{k=1}^{infty}$ exists with the conditions you specified? Answering the why ${b_{k_j}}_{j=1}^{infty}$ exists as a convergent sequence, would be because B is a bounded set, therefore ${b_k}_{k=1}^{infty}$ would be bounded and would therefore have a convergent subsequence.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:35
$begingroup$
Any sequence in $A+B$ will be a sequence ${x_k}_{k=1}^{infty}$ where $x_k = a_k+b_k$ where $a_kin A$ and $b_kin B$. In general any element $xin A+B$ is of the form $a+b$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:37
$begingroup$
Right, but how do you know a convergent one exists?
$endgroup$
– juan deutsch
Dec 17 '18 at 20:41
$begingroup$
A set is closed if and only if every convergent sequence has its limit in the set. Therefore to prove that a $A+$ is closed we consider an ARBITRARY sequence and prove that its limit lies in $A+B$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:42
add a comment |
$begingroup$
How can you know that a convergent sequence ${a_k+b_k}_{k=1}^{infty}$ exists with the conditions you specified? Answering the why ${b_{k_j}}_{j=1}^{infty}$ exists as a convergent sequence, would be because B is a bounded set, therefore ${b_k}_{k=1}^{infty}$ would be bounded and would therefore have a convergent subsequence.
$endgroup$
– juan deutsch
Dec 17 '18 at 20:35
$begingroup$
Any sequence in $A+B$ will be a sequence ${x_k}_{k=1}^{infty}$ where $x_k = a_k+b_k$ where $a_kin A$ and $b_kin B$. In general any element $xin A+B$ is of the form $a+b$.
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:37
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Right, but how do you know a convergent one exists?
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– juan deutsch
Dec 17 '18 at 20:41
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A set is closed if and only if every convergent sequence has its limit in the set. Therefore to prove that a $A+$ is closed we consider an ARBITRARY sequence and prove that its limit lies in $A+B$.
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– Olof Rubin
Dec 17 '18 at 20:42
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How can you know that a convergent sequence ${a_k+b_k}_{k=1}^{infty}$ exists with the conditions you specified? Answering the why ${b_{k_j}}_{j=1}^{infty}$ exists as a convergent sequence, would be because B is a bounded set, therefore ${b_k}_{k=1}^{infty}$ would be bounded and would therefore have a convergent subsequence.
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– juan deutsch
Dec 17 '18 at 20:35
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How can you know that a convergent sequence ${a_k+b_k}_{k=1}^{infty}$ exists with the conditions you specified? Answering the why ${b_{k_j}}_{j=1}^{infty}$ exists as a convergent sequence, would be because B is a bounded set, therefore ${b_k}_{k=1}^{infty}$ would be bounded and would therefore have a convergent subsequence.
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– juan deutsch
Dec 17 '18 at 20:35
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Any sequence in $A+B$ will be a sequence ${x_k}_{k=1}^{infty}$ where $x_k = a_k+b_k$ where $a_kin A$ and $b_kin B$. In general any element $xin A+B$ is of the form $a+b$.
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– Olof Rubin
Dec 17 '18 at 20:37
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Any sequence in $A+B$ will be a sequence ${x_k}_{k=1}^{infty}$ where $x_k = a_k+b_k$ where $a_kin A$ and $b_kin B$. In general any element $xin A+B$ is of the form $a+b$.
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– Olof Rubin
Dec 17 '18 at 20:37
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Right, but how do you know a convergent one exists?
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– juan deutsch
Dec 17 '18 at 20:41
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Right, but how do you know a convergent one exists?
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– juan deutsch
Dec 17 '18 at 20:41
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A set is closed if and only if every convergent sequence has its limit in the set. Therefore to prove that a $A+$ is closed we consider an ARBITRARY sequence and prove that its limit lies in $A+B$.
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– Olof Rubin
Dec 17 '18 at 20:42
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A set is closed if and only if every convergent sequence has its limit in the set. Therefore to prove that a $A+$ is closed we consider an ARBITRARY sequence and prove that its limit lies in $A+B$.
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– Olof Rubin
Dec 17 '18 at 20:42
add a comment |
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What have you tried?
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– EDZ
Dec 17 '18 at 20:20
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Hint for b): write $A+B$ as a continuous image of $Atimes B subset mathbb{R}^{2n}$.
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– Slade
Dec 17 '18 at 20:22