Mixing
The exterior measure, outer measure.
Clash Royale CLAN TAG#URR8PPP
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I am struggling with this proof:
If for a fixed $epsilon$ we choose for each j a cube $Q_j$$subset$$S_j$ and such that $$left| S_jright| leq (1+e) left| Q_jright|$$
Then I have a struggle with the following:
My assumption is(please let me know if I am wrong!)
If $Q_j$$subset$$S_j$, then $left| Q_jright|<=left|S_jright|$
Now if that assumption is correct:
then choose $a=left|S_jright|$
and $b=left|Q_jright|$
Now if
$$a leq (1+e) b$$ for all e>0
then $$a leq b$$
proof: assume the opposite: $$b <a$$
then a-b>0, so we can choose $epsilon=a-b$
it the follows that $$a leq b$$ and we end up with a contradiction.
In other words:
$$left|S_jright|<=left| Q_jright|$$
Therefore,
$$left| S_jright| leq (1+e) left| Q_jright|$$
holds for a particular e and not for all if $Q_j$$subset$$S_j$
But this is then used later on in the proof, where they conclude with since e is arbitrary, but I have just shown that the equation does not hold for all e, and hence e is not arbitrary.
What am I doing wrong?
real-analysis measure-theory proof-explanation
asked Aug 1 at 23:11
ALEXANDER
854818
This question has an open bounty worth +50
reputation from ALEXANDER ending ending at 2018-08-11 01:49:01Z">in 3 days.
The current answers do not contain enough detail.
add a comment |Â
up vote
0
down vote
favorite
I am struggling with this proof:
If for a fixed $epsilon$ we choose for each j a cube $Q_j$$subset$$S_j$ and such that $$left| S_jright| leq (1+e) left| Q_jright|$$
Then I have a struggle with the following:
My assumption is(please let me know if I am wrong!)
If $Q_j$$subset$$S_j$, then $left| Q_jright|<=left|S_jright|$
Now if that assumption is correct:
then choose $a=left|S_jright|$
and $b=left|Q_jright|$
Now if
$$a leq (1+e) b$$ for all e>0
then $$a leq b$$
proof: assume the opposite: $$b <a$$
then a-b>0, so we can choose $epsilon=a-b$
it the follows that $$a leq b$$ and we end up with a contradiction.
In other words:
$$left|S_jright|<=left| Q_jright|$$
Therefore,
$$left| S_jright| leq (1+e) left| Q_jright|$$
holds for a particular e and not for all if $Q_j$$subset$$S_j$
But this is then used later on in the proof, where they conclude with since e is arbitrary, but I have just shown that the equation does not hold for all e, and hence e is not arbitrary.
What am I doing wrong?
real-analysis measure-theory proof-explanation
asked Aug 1 at 23:11
ALEXANDER
854818
This question has an open bounty worth +50
reputation from ALEXANDER ending ending at 2018-08-11 01:49:01Z">in 3 days.
The current answers do not contain enough detail.
@ Theoretical Economist Please have a look.
– ALEXANDER
Aug 1 at 23:26
I was not notified of this comment because I had not previously interacted with this post. In any case, the answer below appears adequate. I’m happy to try to help if you’re still unsure.
– Theoretical Economist
Aug 4 at 3:53
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am struggling with this proof:
If for a fixed $epsilon$ we choose for each j a cube $Q_j$$subset$$S_j$ and such that $$left| S_jright| leq (1+e) left| Q_jright|$$
Then I have a struggle with the following:
My assumption is(please let me know if I am wrong!)
If $Q_j$$subset$$S_j$, then $left| Q_jright|<=left|S_jright|$
Now if that assumption is correct:
then choose $a=left|S_jright|$
and $b=left|Q_jright|$
Now if
$$a leq (1+e) b$$ for all e>0
then $$a leq b$$
proof: assume the opposite: $$b <a$$
then a-b>0, so we can choose $epsilon=a-b$
it the follows that $$a leq b$$ and we end up with a contradiction.
In other words:
$$left|S_jright|<=left| Q_jright|$$
Therefore,
$$left| S_jright| leq (1+e) left| Q_jright|$$
holds for a particular e and not for all if $Q_j$$subset$$S_j$
But this is then used later on in the proof, where they conclude with since e is arbitrary, but I have just shown that the equation does not hold for all e, and hence e is not arbitrary.
What am I doing wrong?
real-analysis measure-theory proof-explanation
asked Aug 1 at 23:11
ALEXANDER
854818
I am struggling with this proof:
If for a fixed $epsilon$ we choose for each j a cube $Q_j$$subset$$S_j$ and such that $$left| S_jright| leq (1+e) left| Q_jright|$$
Then I have a struggle with the following:
My assumption is(please let me know if I am wrong!)
If $Q_j$$subset$$S_j$, then $left| Q_jright|<=left|S_jright|$
Now if that assumption is correct:
then choose $a=left|S_jright|$
and $b=left|Q_jright|$
Now if
$$a leq (1+e) b$$ for all e>0
then $$a leq b$$
proof: assume the opposite: $$b <a$$
then a-b>0, so we can choose $epsilon=a-b$
it the follows that $$a leq b$$ and we end up with a contradiction.
In other words:
$$left|S_jright|<=left| Q_jright|$$
Therefore,
$$left| S_jright| leq (1+e) left| Q_jright|$$
holds for a particular e and not for all if $Q_j$$subset$$S_j$
But this is then used later on in the proof, where they conclude with since e is arbitrary, but I have just shown that the equation does not hold for all e, and hence e is not arbitrary.
What am I doing wrong?
real-analysis measure-theory proof-explanation
asked Aug 1 at 23:11
ALEXANDER
854818
asked Aug 1 at 23:11
ALEXANDER
854818
asked Aug 1 at 23:11
ALEXANDER
854818
asked Aug 1 at 23:11
ALEXANDER
854818
854818
This question has an open bounty worth +50
reputation from ALEXANDER ending ending at 2018-08-11 01:49:01Z">in 3 days.
The current answers do not contain enough detail.
This question has an open bounty worth +50
reputation from ALEXANDER ending ending at 2018-08-11 01:49:01Z">in 3 days.
The current answers do not contain enough detail.
@ Theoretical Economist Please have a look.
– ALEXANDER
Aug 1 at 23:26
I was not notified of this comment because I had not previously interacted with this post. In any case, the answer below appears adequate. I’m happy to try to help if you’re still unsure.
– Theoretical Economist
Aug 4 at 3:53
add a comment |Â
@ Theoretical Economist Please have a look.
– ALEXANDER
Aug 1 at 23:26
I was not notified of this comment because I had not previously interacted with this post. In any case, the answer below appears adequate. I’m happy to try to help if you’re still unsure.
– Theoretical Economist
Aug 4 at 3:53
@ Theoretical Economist Please have a look.
– ALEXANDER
Aug 1 at 23:26
@ Theoretical Economist Please have a look.
– ALEXANDER
Aug 1 at 23:26
I was not notified of this comment because I had not previously interacted with this post. In any case, the answer below appears adequate. I’m happy to try to help if you’re still unsure.
– Theoretical Economist
Aug 4 at 3:53
I was not notified of this comment because I had not previously interacted with this post. In any case, the answer below appears adequate. I’m happy to try to help if you’re still unsure.
– Theoretical Economist
Aug 4 at 3:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
The cube $S_j$ depends on $epsilon$: after fixing $epsilon>0$, you choose $S_j$ such that $Q_jsubseteq S_j$ and $|S_j|leq(1+epsilon)|Q_j|$. So we have a different $S_j$ for each choice of $epsilon$, and we do not conclude that $|S_j|=|Q_j|$ for any single one of these choices.
This does not change the fact that $epsilon$ is arbitrary, since we fix an arbitrary $epsilon$ first before we choose $S_j$. For any particular $epsilon$, it is possible to choose such an $S_j$.
answered Aug 1 at 23:16
Eric Wofsey
161k12188297
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The cube $S_j$ depends on $epsilon$: after fixing $epsilon>0$, you choose $S_j$ such that $Q_jsubseteq S_j$ and $|S_j|leq(1+epsilon)|Q_j|$. So we have a different $S_j$ for each choice of $epsilon$, and we do not conclude that $|S_j|=|Q_j|$ for any single one of these choices.
This does not change the fact that $epsilon$ is arbitrary, since we fix an arbitrary $epsilon$ first before we choose $S_j$. For any particular $epsilon$, it is possible to choose such an $S_j$.
answered Aug 1 at 23:16
Eric Wofsey
161k12188297
add a comment |Â
up vote
0
down vote
The cube $S_j$ depends on $epsilon$: after fixing $epsilon>0$, you choose $S_j$ such that $Q_jsubseteq S_j$ and $|S_j|leq(1+epsilon)|Q_j|$. So we have a different $S_j$ for each choice of $epsilon$, and we do not conclude that $|S_j|=|Q_j|$ for any single one of these choices.
This does not change the fact that $epsilon$ is arbitrary, since we fix an arbitrary $epsilon$ first before we choose $S_j$. For any particular $epsilon$, it is possible to choose such an $S_j$.
answered Aug 1 at 23:16
Eric Wofsey
161k12188297
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The cube $S_j$ depends on $epsilon$: after fixing $epsilon>0$, you choose $S_j$ such that $Q_jsubseteq S_j$ and $|S_j|leq(1+epsilon)|Q_j|$. So we have a different $S_j$ for each choice of $epsilon$, and we do not conclude that $|S_j|=|Q_j|$ for any single one of these choices.
This does not change the fact that $epsilon$ is arbitrary, since we fix an arbitrary $epsilon$ first before we choose $S_j$. For any particular $epsilon$, it is possible to choose such an $S_j$.
answered Aug 1 at 23:16
Eric Wofsey
161k12188297
The cube $S_j$ depends on $epsilon$: after fixing $epsilon>0$, you choose $S_j$ such that $Q_jsubseteq S_j$ and $|S_j|leq(1+epsilon)|Q_j|$. So we have a different $S_j$ for each choice of $epsilon$, and we do not conclude that $|S_j|=|Q_j|$ for any single one of these choices.
This does not change the fact that $epsilon$ is arbitrary, since we fix an arbitrary $epsilon$ first before we choose $S_j$. For any particular $epsilon$, it is possible to choose such an $S_j$.
answered Aug 1 at 23:16
Eric Wofsey
161k12188297
answered Aug 1 at 23:16
Eric Wofsey
161k12188297
answered Aug 1 at 23:16
Eric Wofsey
161k12188297
answered Aug 1 at 23:16
Eric Wofsey
161k12188297
161k12188297
add a comment |Â
add a comment |Â
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@ Theoretical Economist Please have a look.
– ALEXANDER
Aug 1 at 23:26
I was not notified of this comment because I had not previously interacted with this post. In any case, the answer below appears adequate. I’m happy to try to help if you’re still unsure.
– Theoretical Economist
Aug 4 at 3:53