Intersection of balls in $l^{infty}$
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Must it be the case that the intersection of any two closed balls in $l^{infty}$ either is empty, a single point, or contains a non-empty open ball?
real-analysis general-topology functional-analysis
asked Dec 24 '18 at 0:04
Andrew LizarragaAndrew Lizarraga
4215
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add a comment |
$begingroup$
Must it be the case that the intersection of any two closed balls in $l^{infty}$ either is empty, a single point, or contains a non-empty open ball?
real-analysis general-topology functional-analysis
asked Dec 24 '18 at 0:04
Andrew LizarragaAndrew Lizarraga
4215
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5
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Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
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– Nate Eldredge
Dec 24 '18 at 1:07
add a comment |
$begingroup$
Must it be the case that the intersection of any two closed balls in $l^{infty}$ either is empty, a single point, or contains a non-empty open ball?
real-analysis general-topology functional-analysis
asked Dec 24 '18 at 0:04
Andrew LizarragaAndrew Lizarraga
4215
$endgroup$
Must it be the case that the intersection of any two closed balls in $l^{infty}$ either is empty, a single point, or contains a non-empty open ball?
real-analysis general-topology functional-analysis
real-analysis general-topology functional-analysis
asked Dec 24 '18 at 0:04
Andrew LizarragaAndrew Lizarraga
4215
asked Dec 24 '18 at 0:04
Andrew LizarragaAndrew Lizarraga
4215
asked Dec 24 '18 at 0:04
Andrew LizarragaAndrew Lizarraga
4215
asked Dec 24 '18 at 0:04
Andrew LizarragaAndrew Lizarraga
4215
asked Dec 24 '18 at 0:04
Andrew LizarragaAndrew Lizarraga
4215
4215
5
$begingroup$
Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
$endgroup$
– Nate Eldredge
Dec 24 '18 at 1:07
add a comment |
5
$begingroup$
Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
$endgroup$
– Nate Eldredge
Dec 24 '18 at 1:07
5
5
$begingroup$
Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
$endgroup$
– Nate Eldredge
Dec 24 '18 at 1:07
$begingroup$
Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
$endgroup$
– Nate Eldredge
Dec 24 '18 at 1:07
add a comment |
1 Answer
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To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
$$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
So: the intersection is
- clearly non-empty
- contains infinitely many points
- does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.
answered Dec 24 '18 at 15:37
postmortespostmortes
2,04121119
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1 Answer
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$begingroup$
To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
$$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
So: the intersection is
- clearly non-empty
- contains infinitely many points
- does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.
answered Dec 24 '18 at 15:37
postmortespostmortes
2,04121119
$endgroup$
add a comment |
$begingroup$
To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
$$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
So: the intersection is
- clearly non-empty
- contains infinitely many points
- does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.
answered Dec 24 '18 at 15:37
postmortespostmortes
2,04121119
$endgroup$
add a comment |
$begingroup$
To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
$$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
So: the intersection is
- clearly non-empty
- contains infinitely many points
- does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.
answered Dec 24 '18 at 15:37
postmortespostmortes
2,04121119
$endgroup$
To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
$$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
So: the intersection is
- clearly non-empty
- contains infinitely many points
- does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.
answered Dec 24 '18 at 15:37
postmortespostmortes
2,04121119
answered Dec 24 '18 at 15:37
postmortespostmortes
2,04121119
answered Dec 24 '18 at 15:37
postmortespostmortes
2,04121119
answered Dec 24 '18 at 15:37
postmortespostmortes
2,04121119
2,04121119
add a comment |
add a comment |
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$begingroup$
Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
$endgroup$
– Nate Eldredge
Dec 24 '18 at 1:07