Is this set meager? $A = {xin mathbb{R}: exists c>0, |x-j2^{-k}|geq c2^{-k}, forall jin mathbb{Z}, kgeq 0...
$begingroup$
We define the subset $Asubset mathbb{R}$ as follows: $xin A Longleftrightarrow$ if $exists c>0$ so that
$$ |x-j2^{-k}|geq c2^{-k} $$
holds $forall jin mathbb{Z}$ and integers $kgeq 0$. Show that $A$ is meager and dense.
I am completely lost on this one. It looks like it is saying that $xin A$ if the difference between $x$ and any dyadic rational can be made greater than $c2^{-k}$. But I have no intuition for this at all.
Anyone have a hint (not a solution) for how to approach this problem?
Edit:
So I tried to see if for each $k$, say $0,1,2,...$, I could create a set $A_k$ which is nowhere dense. But $A_0 = mathbb{R}/mathbb{Z}$ is not nowhere dense. So that approach isn't working unless I am confused.
real-analysis general-topology functional-analysis complete-spaces
asked Dec 1 '18 at 5:03
Joe Man AnalysisJoe Man Analysis
33419
$endgroup$
add a comment |
$begingroup$
We define the subset $Asubset mathbb{R}$ as follows: $xin A Longleftrightarrow$ if $exists c>0$ so that
$$ |x-j2^{-k}|geq c2^{-k} $$
holds $forall jin mathbb{Z}$ and integers $kgeq 0$. Show that $A$ is meager and dense.
I am completely lost on this one. It looks like it is saying that $xin A$ if the difference between $x$ and any dyadic rational can be made greater than $c2^{-k}$. But I have no intuition for this at all.
Anyone have a hint (not a solution) for how to approach this problem?
Edit:
So I tried to see if for each $k$, say $0,1,2,...$, I could create a set $A_k$ which is nowhere dense. But $A_0 = mathbb{R}/mathbb{Z}$ is not nowhere dense. So that approach isn't working unless I am confused.
real-analysis general-topology functional-analysis complete-spaces
asked Dec 1 '18 at 5:03
Joe Man AnalysisJoe Man Analysis
33419
$endgroup$
add a comment |
$begingroup$
We define the subset $Asubset mathbb{R}$ as follows: $xin A Longleftrightarrow$ if $exists c>0$ so that
$$ |x-j2^{-k}|geq c2^{-k} $$
holds $forall jin mathbb{Z}$ and integers $kgeq 0$. Show that $A$ is meager and dense.
I am completely lost on this one. It looks like it is saying that $xin A$ if the difference between $x$ and any dyadic rational can be made greater than $c2^{-k}$. But I have no intuition for this at all.
Anyone have a hint (not a solution) for how to approach this problem?
Edit:
So I tried to see if for each $k$, say $0,1,2,...$, I could create a set $A_k$ which is nowhere dense. But $A_0 = mathbb{R}/mathbb{Z}$ is not nowhere dense. So that approach isn't working unless I am confused.
real-analysis general-topology functional-analysis complete-spaces
asked Dec 1 '18 at 5:03
Joe Man AnalysisJoe Man Analysis
33419
$endgroup$
We define the subset $Asubset mathbb{R}$ as follows: $xin A Longleftrightarrow$ if $exists c>0$ so that
$$ |x-j2^{-k}|geq c2^{-k} $$
holds $forall jin mathbb{Z}$ and integers $kgeq 0$. Show that $A$ is meager and dense.
I am completely lost on this one. It looks like it is saying that $xin A$ if the difference between $x$ and any dyadic rational can be made greater than $c2^{-k}$. But I have no intuition for this at all.
Anyone have a hint (not a solution) for how to approach this problem?
Edit:
So I tried to see if for each $k$, say $0,1,2,...$, I could create a set $A_k$ which is nowhere dense. But $A_0 = mathbb{R}/mathbb{Z}$ is not nowhere dense. So that approach isn't working unless I am confused.
real-analysis general-topology functional-analysis complete-spaces
real-analysis general-topology functional-analysis complete-spaces
asked Dec 1 '18 at 5:03
Joe Man AnalysisJoe Man Analysis
33419
asked Dec 1 '18 at 5:03
Joe Man AnalysisJoe Man Analysis
33419
edited Dec 1 '18 at 9:11
Joe Man Analysis
asked Dec 1 '18 at 5:03
Joe Man AnalysisJoe Man Analysis
33419
asked Dec 1 '18 at 5:03
Joe Man AnalysisJoe Man Analysis
33419
asked Dec 1 '18 at 5:03
Joe Man AnalysisJoe Man Analysis
33419
33419
add a comment |
add a comment |
1 Answer
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$begingroup$
We have $A=bigcup A_n$, where $$A_n={xin mathbb{R}: |x-j2^{-k}|geq 2^{-n-k}, forall jin mathbb{Z}, kgeq 0 }.$$
Spoiler:
So it suffices to show that each $A_n$ is meager. Since $$A_n=bigcap_{jinBbb Z, >! kge 0} Bbb Rsetminus (j2^{-k}-2^{-n-k}, j2^{-k}+2^{-n-k}),$$
it is closed. Since $A_n$ is disjoint with the set of diadic rationals, the set $A_n$ is nowhere dense.
answered Dec 2 '18 at 6:50
Alex RavskyAlex Ravsky
39.5k32181
$endgroup$
add a comment |
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$begingroup$
We have $A=bigcup A_n$, where $$A_n={xin mathbb{R}: |x-j2^{-k}|geq 2^{-n-k}, forall jin mathbb{Z}, kgeq 0 }.$$
Spoiler:
So it suffices to show that each $A_n$ is meager. Since $$A_n=bigcap_{jinBbb Z, >! kge 0} Bbb Rsetminus (j2^{-k}-2^{-n-k}, j2^{-k}+2^{-n-k}),$$
it is closed. Since $A_n$ is disjoint with the set of diadic rationals, the set $A_n$ is nowhere dense.
answered Dec 2 '18 at 6:50
Alex RavskyAlex Ravsky
39.5k32181
$endgroup$
add a comment |
$begingroup$
We have $A=bigcup A_n$, where $$A_n={xin mathbb{R}: |x-j2^{-k}|geq 2^{-n-k}, forall jin mathbb{Z}, kgeq 0 }.$$
Spoiler:
So it suffices to show that each $A_n$ is meager. Since $$A_n=bigcap_{jinBbb Z, >! kge 0} Bbb Rsetminus (j2^{-k}-2^{-n-k}, j2^{-k}+2^{-n-k}),$$
it is closed. Since $A_n$ is disjoint with the set of diadic rationals, the set $A_n$ is nowhere dense.
answered Dec 2 '18 at 6:50
Alex RavskyAlex Ravsky
39.5k32181
$endgroup$
add a comment |
$begingroup$
We have $A=bigcup A_n$, where $$A_n={xin mathbb{R}: |x-j2^{-k}|geq 2^{-n-k}, forall jin mathbb{Z}, kgeq 0 }.$$
Spoiler:
So it suffices to show that each $A_n$ is meager. Since $$A_n=bigcap_{jinBbb Z, >! kge 0} Bbb Rsetminus (j2^{-k}-2^{-n-k}, j2^{-k}+2^{-n-k}),$$
it is closed. Since $A_n$ is disjoint with the set of diadic rationals, the set $A_n$ is nowhere dense.
answered Dec 2 '18 at 6:50
Alex RavskyAlex Ravsky
39.5k32181
$endgroup$
We have $A=bigcup A_n$, where $$A_n={xin mathbb{R}: |x-j2^{-k}|geq 2^{-n-k}, forall jin mathbb{Z}, kgeq 0 }.$$
Spoiler:
So it suffices to show that each $A_n$ is meager. Since $$A_n=bigcap_{jinBbb Z, >! kge 0} Bbb Rsetminus (j2^{-k}-2^{-n-k}, j2^{-k}+2^{-n-k}),$$
it is closed. Since $A_n$ is disjoint with the set of diadic rationals, the set $A_n$ is nowhere dense.
answered Dec 2 '18 at 6:50
Alex RavskyAlex Ravsky
39.5k32181
edited Dec 2 '18 at 6:56
answered Dec 2 '18 at 6:50
Alex RavskyAlex Ravsky
39.5k32181
answered Dec 2 '18 at 6:50
Alex RavskyAlex Ravsky
39.5k32181
answered Dec 2 '18 at 6:50
Alex RavskyAlex Ravsky
39.5k32181
39.5k32181
add a comment |
add a comment |
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