Mixing
If $A$ and $B$ are closed, then $A+B$ is Borel
$begingroup$
I am trying to prove this statement by proving first that $A+B = {a+b : a in A, b in B}$ is $F_sigma$.
I am lost, I saw that I had to prove first the case when $A$ and $B$ were compact then $A+B$ was compact but I can't see how this helps.
In any case, is this part of the proof also right then?
If $A$ and $B$ are compact, then for any sequences $(a_n)$ and $(b_n)$ (in $A$ and $B$ respectively) there is a convergent subsequence $(a_{n_k})$ and $(b_{n_k})$. Therefore for any sequence $(a_n + b_n)$, there is a convergent subsequence $(a_{n_k} + b_{n_k})$ in $A+B$.
real-analysis general-topology measure-theory compactness
asked Jan 25 at 11:06
The BoscoThe Bosco
613212
$endgroup$
add a comment |
$begingroup$
I am trying to prove this statement by proving first that $A+B = {a+b : a in A, b in B}$ is $F_sigma$.
I am lost, I saw that I had to prove first the case when $A$ and $B$ were compact then $A+B$ was compact but I can't see how this helps.
In any case, is this part of the proof also right then?
If $A$ and $B$ are compact, then for any sequences $(a_n)$ and $(b_n)$ (in $A$ and $B$ respectively) there is a convergent subsequence $(a_{n_k})$ and $(b_{n_k})$. Therefore for any sequence $(a_n + b_n)$, there is a convergent subsequence $(a_{n_k} + b_{n_k})$ in $A+B$.
real-analysis general-topology measure-theory compactness
asked Jan 25 at 11:06
The BoscoThe Bosco
613212
$endgroup$
add a comment |
$begingroup$
I am trying to prove this statement by proving first that $A+B = {a+b : a in A, b in B}$ is $F_sigma$.
I am lost, I saw that I had to prove first the case when $A$ and $B$ were compact then $A+B$ was compact but I can't see how this helps.
In any case, is this part of the proof also right then?
If $A$ and $B$ are compact, then for any sequences $(a_n)$ and $(b_n)$ (in $A$ and $B$ respectively) there is a convergent subsequence $(a_{n_k})$ and $(b_{n_k})$. Therefore for any sequence $(a_n + b_n)$, there is a convergent subsequence $(a_{n_k} + b_{n_k})$ in $A+B$.
real-analysis general-topology measure-theory compactness
asked Jan 25 at 11:06
The BoscoThe Bosco
613212
$endgroup$
I am trying to prove this statement by proving first that $A+B = {a+b : a in A, b in B}$ is $F_sigma$.
I am lost, I saw that I had to prove first the case when $A$ and $B$ were compact then $A+B$ was compact but I can't see how this helps.
In any case, is this part of the proof also right then?
If $A$ and $B$ are compact, then for any sequences $(a_n)$ and $(b_n)$ (in $A$ and $B$ respectively) there is a convergent subsequence $(a_{n_k})$ and $(b_{n_k})$. Therefore for any sequence $(a_n + b_n)$, there is a convergent subsequence $(a_{n_k} + b_{n_k})$ in $A+B$.
real-analysis general-topology measure-theory compactness
real-analysis general-topology measure-theory compactness
asked Jan 25 at 11:06
The BoscoThe Bosco
613212
asked Jan 25 at 11:06
The BoscoThe Bosco
613212
asked Jan 25 at 11:06
The BoscoThe Bosco
613212
asked Jan 25 at 11:06
The BoscoThe Bosco
613212
asked Jan 25 at 11:06
The BoscoThe Bosco
613212
613212
add a comment |
add a comment |
1 Answer
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This is true in the real line. $A$ and $B$ are both countable unions of compact sets. The sum of two compact sets is a compact set (your proof of this is OK). So $A+B$ is a countable union of compact sets. And a compact set is closed. Conclude that $A+B$ is an $F_sigma$.
answered Jan 25 at 11:11
GEdgarGEdgar
63.1k267171
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1 Answer
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1 Answer
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$begingroup$
This is true in the real line. $A$ and $B$ are both countable unions of compact sets. The sum of two compact sets is a compact set (your proof of this is OK). So $A+B$ is a countable union of compact sets. And a compact set is closed. Conclude that $A+B$ is an $F_sigma$.
answered Jan 25 at 11:11
GEdgarGEdgar
63.1k267171
$endgroup$
add a comment |
$begingroup$
This is true in the real line. $A$ and $B$ are both countable unions of compact sets. The sum of two compact sets is a compact set (your proof of this is OK). So $A+B$ is a countable union of compact sets. And a compact set is closed. Conclude that $A+B$ is an $F_sigma$.
answered Jan 25 at 11:11
GEdgarGEdgar
63.1k267171
$endgroup$
add a comment |
$begingroup$
This is true in the real line. $A$ and $B$ are both countable unions of compact sets. The sum of two compact sets is a compact set (your proof of this is OK). So $A+B$ is a countable union of compact sets. And a compact set is closed. Conclude that $A+B$ is an $F_sigma$.
answered Jan 25 at 11:11
GEdgarGEdgar
63.1k267171
$endgroup$
This is true in the real line. $A$ and $B$ are both countable unions of compact sets. The sum of two compact sets is a compact set (your proof of this is OK). So $A+B$ is a countable union of compact sets. And a compact set is closed. Conclude that $A+B$ is an $F_sigma$.
answered Jan 25 at 11:11
GEdgarGEdgar
63.1k267171
answered Jan 25 at 11:11
GEdgarGEdgar
63.1k267171
answered Jan 25 at 11:11
GEdgarGEdgar
63.1k267171
answered Jan 25 at 11:11
GEdgarGEdgar
63.1k267171
63.1k267171
add a comment |
add a comment |
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