Mixing
It is true that $overline{(0,epsilon)cup(mathbb Qcap(epsilon,1))}=[0,1]$ for $epsilonin(0,1)$?
$begingroup$
Let $epsilonin(0,1)$ and $E=(0,epsilon)cup(mathbb Qcap(epsilon,1))$. It is true that $overline{E}=[0,1]$?
I know that $overline{E}subset[0,1]$, but how to show that $overline{E}supset[0,1]$?
I've tried to show that if I have an open interval $(a,b)$ then $overline{mathbb Qcap(a,b)}=[a,b]$. By the way, is this true?
real-analysis general-topology
asked Jan 22 at 11:27
Ali KhanAli Khan
17312
$endgroup$
add a comment |
$begingroup$
Let $epsilonin(0,1)$ and $E=(0,epsilon)cup(mathbb Qcap(epsilon,1))$. It is true that $overline{E}=[0,1]$?
I know that $overline{E}subset[0,1]$, but how to show that $overline{E}supset[0,1]$?
I've tried to show that if I have an open interval $(a,b)$ then $overline{mathbb Qcap(a,b)}=[a,b]$. By the way, is this true?
real-analysis general-topology
asked Jan 22 at 11:27
Ali KhanAli Khan
17312
$endgroup$
1
$begingroup$
Yes it's true. Just note that $mathbb{Q}cap(0,1)setminus{epsilon}subset Esubset [0,1]$ and take the closure.
$endgroup$
– Ben W
Jan 22 at 11:30
$begingroup$
$mathbb{Q}cap(0,1)setminus{epsilon}=mathbb{Q}cap((0,epsilon)cup(epsilon,1))$, then $overline{mathbb{Q}cap(0,1)setminus{epsilon}}=overline{mathbb{Q}cap((0,epsilon)cup(epsilon,1))}subsetmathbb{R}cap[0,1]=[0,1]$. But how do I conclude that $[0,1]subset overline{mathbb{Q}cap(0,1)setminus{epsilon}}$
$endgroup$
– Ali Khan
Jan 22 at 13:42
add a comment |
$begingroup$
Let $epsilonin(0,1)$ and $E=(0,epsilon)cup(mathbb Qcap(epsilon,1))$. It is true that $overline{E}=[0,1]$?
I know that $overline{E}subset[0,1]$, but how to show that $overline{E}supset[0,1]$?
I've tried to show that if I have an open interval $(a,b)$ then $overline{mathbb Qcap(a,b)}=[a,b]$. By the way, is this true?
real-analysis general-topology
asked Jan 22 at 11:27
Ali KhanAli Khan
17312
$endgroup$
Let $epsilonin(0,1)$ and $E=(0,epsilon)cup(mathbb Qcap(epsilon,1))$. It is true that $overline{E}=[0,1]$?
I know that $overline{E}subset[0,1]$, but how to show that $overline{E}supset[0,1]$?
I've tried to show that if I have an open interval $(a,b)$ then $overline{mathbb Qcap(a,b)}=[a,b]$. By the way, is this true?
real-analysis general-topology
real-analysis general-topology
asked Jan 22 at 11:27
Ali KhanAli Khan
17312
asked Jan 22 at 11:27
Ali KhanAli Khan
17312
edited Jan 22 at 13:36
Ali Khan
asked Jan 22 at 11:27
Ali KhanAli Khan
17312
asked Jan 22 at 11:27
Ali KhanAli Khan
17312
asked Jan 22 at 11:27
Ali KhanAli Khan
17312
17312
1
$begingroup$
Yes it's true. Just note that $mathbb{Q}cap(0,1)setminus{epsilon}subset Esubset [0,1]$ and take the closure.
$endgroup$
– Ben W
Jan 22 at 11:30
$begingroup$
$mathbb{Q}cap(0,1)setminus{epsilon}=mathbb{Q}cap((0,epsilon)cup(epsilon,1))$, then $overline{mathbb{Q}cap(0,1)setminus{epsilon}}=overline{mathbb{Q}cap((0,epsilon)cup(epsilon,1))}subsetmathbb{R}cap[0,1]=[0,1]$. But how do I conclude that $[0,1]subset overline{mathbb{Q}cap(0,1)setminus{epsilon}}$
$endgroup$
– Ali Khan
Jan 22 at 13:42
add a comment |
1
$begingroup$
Yes it's true. Just note that $mathbb{Q}cap(0,1)setminus{epsilon}subset Esubset [0,1]$ and take the closure.
$endgroup$
– Ben W
Jan 22 at 11:30
$begingroup$
$mathbb{Q}cap(0,1)setminus{epsilon}=mathbb{Q}cap((0,epsilon)cup(epsilon,1))$, then $overline{mathbb{Q}cap(0,1)setminus{epsilon}}=overline{mathbb{Q}cap((0,epsilon)cup(epsilon,1))}subsetmathbb{R}cap[0,1]=[0,1]$. But how do I conclude that $[0,1]subset overline{mathbb{Q}cap(0,1)setminus{epsilon}}$
$endgroup$
– Ali Khan
Jan 22 at 13:42
1
1
$begingroup$
Yes it's true. Just note that $mathbb{Q}cap(0,1)setminus{epsilon}subset Esubset [0,1]$ and take the closure.
$endgroup$
– Ben W
Jan 22 at 11:30
$begingroup$
Yes it's true. Just note that $mathbb{Q}cap(0,1)setminus{epsilon}subset Esubset [0,1]$ and take the closure.
$endgroup$
– Ben W
Jan 22 at 11:30
$begingroup$
$mathbb{Q}cap(0,1)setminus{epsilon}=mathbb{Q}cap((0,epsilon)cup(epsilon,1))$, then $overline{mathbb{Q}cap(0,1)setminus{epsilon}}=overline{mathbb{Q}cap((0,epsilon)cup(epsilon,1))}subsetmathbb{R}cap[0,1]=[0,1]$. But how do I conclude that $[0,1]subset overline{mathbb{Q}cap(0,1)setminus{epsilon}}$
$endgroup$
– Ali Khan
Jan 22 at 13:42
$begingroup$
$mathbb{Q}cap(0,1)setminus{epsilon}=mathbb{Q}cap((0,epsilon)cup(epsilon,1))$, then $overline{mathbb{Q}cap(0,1)setminus{epsilon}}=overline{mathbb{Q}cap((0,epsilon)cup(epsilon,1))}subsetmathbb{R}cap[0,1]=[0,1]$. But how do I conclude that $[0,1]subset overline{mathbb{Q}cap(0,1)setminus{epsilon}}$
$endgroup$
– Ali Khan
Jan 22 at 13:42
add a comment |
1 Answer
1
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oldest
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$begingroup$
The result is indeed true. Here is a way to show it.
Since $E$ is just a finite union of sets, we have
$$
overline{E} = overline{(0,varepsilon) cup (mathbb{Q}cap(varepsilon,1))}=overline{(0,varepsilon)}cupoverline{mathbb{Q}cap(varepsilon,1)} = [0,varepsilon] cup overline{mathbb{Q}cap(varepsilon,1)}.
$$
If we manage to show that for an interval $(a,b)$, the equality $overline{mathbb{Q}cap(a,b)}=[a,b]$ is true, as you have asked, then we are done. To show this, first notice that from standard properties of the closure we already have the inclusion $overline{mathbb{Q}cap(a,b)} subseteq overline{mathbb{Q}}capoverline{(a,b)}=[a,b]$. Now, to show the reverse inclusion, we must show that for an arbitrary point $x in [a,b]$, any neighborhood of $x$ intersects the set $mathbb{Q}cap(a,b)$. Since any neighborhood of $x$ contains open balls of arbitrarily small radii centered at $x$, it is sufficient to prove that any sufficiently small ball centered at $x$ intercepts $mathbb{Q}cap(a,b)$.
If $x in (a,b)$, since this set is open, for any sufficiently small radius $r>0$, the interval $(x-r,x+r)$ is contained in $(a,b)$, but the density of $mathbb{Q}$ then implies the existence of a rational number $q in (x-r,x+r)$, and by construction this rational number is in $mathbb{Q}cap(a,b)$.
If $x=a$, even though in this case we cannot obtain $r>0$ such that $(x-r,x+r) subset (a,b)$, we can guarantee that the right-side $(x,x+r)$ of the interval will be contained in $(a,b)$. Density again implies that there is a rational number $q in (x,x+r)$, therefore in $mathbb{Q}cap(a,b)$, so $(x-r,x+r)$ intersects $mathbb{Q}cap(a,b)$.
The last case where $x=b$ is analogous to the previous one, just use the left-side $(x-r,x)$ of the interval and argue in the same way.
answered Jan 22 at 14:14
Edmundo MartinsEdmundo Martins
29228
$endgroup$
add a comment |
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$begingroup$
The result is indeed true. Here is a way to show it.
Since $E$ is just a finite union of sets, we have
$$
overline{E} = overline{(0,varepsilon) cup (mathbb{Q}cap(varepsilon,1))}=overline{(0,varepsilon)}cupoverline{mathbb{Q}cap(varepsilon,1)} = [0,varepsilon] cup overline{mathbb{Q}cap(varepsilon,1)}.
$$
If we manage to show that for an interval $(a,b)$, the equality $overline{mathbb{Q}cap(a,b)}=[a,b]$ is true, as you have asked, then we are done. To show this, first notice that from standard properties of the closure we already have the inclusion $overline{mathbb{Q}cap(a,b)} subseteq overline{mathbb{Q}}capoverline{(a,b)}=[a,b]$. Now, to show the reverse inclusion, we must show that for an arbitrary point $x in [a,b]$, any neighborhood of $x$ intersects the set $mathbb{Q}cap(a,b)$. Since any neighborhood of $x$ contains open balls of arbitrarily small radii centered at $x$, it is sufficient to prove that any sufficiently small ball centered at $x$ intercepts $mathbb{Q}cap(a,b)$.
If $x in (a,b)$, since this set is open, for any sufficiently small radius $r>0$, the interval $(x-r,x+r)$ is contained in $(a,b)$, but the density of $mathbb{Q}$ then implies the existence of a rational number $q in (x-r,x+r)$, and by construction this rational number is in $mathbb{Q}cap(a,b)$.
If $x=a$, even though in this case we cannot obtain $r>0$ such that $(x-r,x+r) subset (a,b)$, we can guarantee that the right-side $(x,x+r)$ of the interval will be contained in $(a,b)$. Density again implies that there is a rational number $q in (x,x+r)$, therefore in $mathbb{Q}cap(a,b)$, so $(x-r,x+r)$ intersects $mathbb{Q}cap(a,b)$.
The last case where $x=b$ is analogous to the previous one, just use the left-side $(x-r,x)$ of the interval and argue in the same way.
answered Jan 22 at 14:14
Edmundo MartinsEdmundo Martins
29228
$endgroup$
add a comment |
$begingroup$
The result is indeed true. Here is a way to show it.
Since $E$ is just a finite union of sets, we have
$$
overline{E} = overline{(0,varepsilon) cup (mathbb{Q}cap(varepsilon,1))}=overline{(0,varepsilon)}cupoverline{mathbb{Q}cap(varepsilon,1)} = [0,varepsilon] cup overline{mathbb{Q}cap(varepsilon,1)}.
$$
If we manage to show that for an interval $(a,b)$, the equality $overline{mathbb{Q}cap(a,b)}=[a,b]$ is true, as you have asked, then we are done. To show this, first notice that from standard properties of the closure we already have the inclusion $overline{mathbb{Q}cap(a,b)} subseteq overline{mathbb{Q}}capoverline{(a,b)}=[a,b]$. Now, to show the reverse inclusion, we must show that for an arbitrary point $x in [a,b]$, any neighborhood of $x$ intersects the set $mathbb{Q}cap(a,b)$. Since any neighborhood of $x$ contains open balls of arbitrarily small radii centered at $x$, it is sufficient to prove that any sufficiently small ball centered at $x$ intercepts $mathbb{Q}cap(a,b)$.
If $x in (a,b)$, since this set is open, for any sufficiently small radius $r>0$, the interval $(x-r,x+r)$ is contained in $(a,b)$, but the density of $mathbb{Q}$ then implies the existence of a rational number $q in (x-r,x+r)$, and by construction this rational number is in $mathbb{Q}cap(a,b)$.
If $x=a$, even though in this case we cannot obtain $r>0$ such that $(x-r,x+r) subset (a,b)$, we can guarantee that the right-side $(x,x+r)$ of the interval will be contained in $(a,b)$. Density again implies that there is a rational number $q in (x,x+r)$, therefore in $mathbb{Q}cap(a,b)$, so $(x-r,x+r)$ intersects $mathbb{Q}cap(a,b)$.
The last case where $x=b$ is analogous to the previous one, just use the left-side $(x-r,x)$ of the interval and argue in the same way.
answered Jan 22 at 14:14
Edmundo MartinsEdmundo Martins
29228
$endgroup$
add a comment |
$begingroup$
The result is indeed true. Here is a way to show it.
Since $E$ is just a finite union of sets, we have
$$
overline{E} = overline{(0,varepsilon) cup (mathbb{Q}cap(varepsilon,1))}=overline{(0,varepsilon)}cupoverline{mathbb{Q}cap(varepsilon,1)} = [0,varepsilon] cup overline{mathbb{Q}cap(varepsilon,1)}.
$$
If we manage to show that for an interval $(a,b)$, the equality $overline{mathbb{Q}cap(a,b)}=[a,b]$ is true, as you have asked, then we are done. To show this, first notice that from standard properties of the closure we already have the inclusion $overline{mathbb{Q}cap(a,b)} subseteq overline{mathbb{Q}}capoverline{(a,b)}=[a,b]$. Now, to show the reverse inclusion, we must show that for an arbitrary point $x in [a,b]$, any neighborhood of $x$ intersects the set $mathbb{Q}cap(a,b)$. Since any neighborhood of $x$ contains open balls of arbitrarily small radii centered at $x$, it is sufficient to prove that any sufficiently small ball centered at $x$ intercepts $mathbb{Q}cap(a,b)$.
If $x in (a,b)$, since this set is open, for any sufficiently small radius $r>0$, the interval $(x-r,x+r)$ is contained in $(a,b)$, but the density of $mathbb{Q}$ then implies the existence of a rational number $q in (x-r,x+r)$, and by construction this rational number is in $mathbb{Q}cap(a,b)$.
If $x=a$, even though in this case we cannot obtain $r>0$ such that $(x-r,x+r) subset (a,b)$, we can guarantee that the right-side $(x,x+r)$ of the interval will be contained in $(a,b)$. Density again implies that there is a rational number $q in (x,x+r)$, therefore in $mathbb{Q}cap(a,b)$, so $(x-r,x+r)$ intersects $mathbb{Q}cap(a,b)$.
The last case where $x=b$ is analogous to the previous one, just use the left-side $(x-r,x)$ of the interval and argue in the same way.
answered Jan 22 at 14:14
Edmundo MartinsEdmundo Martins
29228
$endgroup$
The result is indeed true. Here is a way to show it.
Since $E$ is just a finite union of sets, we have
$$
overline{E} = overline{(0,varepsilon) cup (mathbb{Q}cap(varepsilon,1))}=overline{(0,varepsilon)}cupoverline{mathbb{Q}cap(varepsilon,1)} = [0,varepsilon] cup overline{mathbb{Q}cap(varepsilon,1)}.
$$
If we manage to show that for an interval $(a,b)$, the equality $overline{mathbb{Q}cap(a,b)}=[a,b]$ is true, as you have asked, then we are done. To show this, first notice that from standard properties of the closure we already have the inclusion $overline{mathbb{Q}cap(a,b)} subseteq overline{mathbb{Q}}capoverline{(a,b)}=[a,b]$. Now, to show the reverse inclusion, we must show that for an arbitrary point $x in [a,b]$, any neighborhood of $x$ intersects the set $mathbb{Q}cap(a,b)$. Since any neighborhood of $x$ contains open balls of arbitrarily small radii centered at $x$, it is sufficient to prove that any sufficiently small ball centered at $x$ intercepts $mathbb{Q}cap(a,b)$.
If $x in (a,b)$, since this set is open, for any sufficiently small radius $r>0$, the interval $(x-r,x+r)$ is contained in $(a,b)$, but the density of $mathbb{Q}$ then implies the existence of a rational number $q in (x-r,x+r)$, and by construction this rational number is in $mathbb{Q}cap(a,b)$.
If $x=a$, even though in this case we cannot obtain $r>0$ such that $(x-r,x+r) subset (a,b)$, we can guarantee that the right-side $(x,x+r)$ of the interval will be contained in $(a,b)$. Density again implies that there is a rational number $q in (x,x+r)$, therefore in $mathbb{Q}cap(a,b)$, so $(x-r,x+r)$ intersects $mathbb{Q}cap(a,b)$.
The last case where $x=b$ is analogous to the previous one, just use the left-side $(x-r,x)$ of the interval and argue in the same way.
answered Jan 22 at 14:14
Edmundo MartinsEdmundo Martins
29228
answered Jan 22 at 14:14
Edmundo MartinsEdmundo Martins
29228
answered Jan 22 at 14:14
Edmundo MartinsEdmundo Martins
29228
answered Jan 22 at 14:14
Edmundo MartinsEdmundo Martins
29228
29228
add a comment |
add a comment |
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$begingroup$
Yes it's true. Just note that $mathbb{Q}cap(0,1)setminus{epsilon}subset Esubset [0,1]$ and take the closure.
$endgroup$
– Ben W
Jan 22 at 11:30
$begingroup$
$mathbb{Q}cap(0,1)setminus{epsilon}=mathbb{Q}cap((0,epsilon)cup(epsilon,1))$, then $overline{mathbb{Q}cap(0,1)setminus{epsilon}}=overline{mathbb{Q}cap((0,epsilon)cup(epsilon,1))}subsetmathbb{R}cap[0,1]=[0,1]$. But how do I conclude that $[0,1]subset overline{mathbb{Q}cap(0,1)setminus{epsilon}}$
$endgroup$
– Ali Khan
Jan 22 at 13:42