Mixing
Does $partial B(x_0, r) subseteq {x in X : d(x_0,x) = r }$ hold in an arbitrary metric space?
$begingroup$
Let $X$ be a metric space, and let $B(x_0, r)$ denote the open ball of radius $r$ centred at $x_0 in X$.
Does the statement $partial B(x_0, r) subseteq {x in X : d(x_0,x) = r }$ hold true always?
A similar question posted in Math StackExchange
As answered in the link above, we know that the boundary is not the set ${x in X : d(x_0,x) = r }$. however, it seems that the statement $partial B subseteq {x in X : d(x_0,x) = r }$ is true. Considering the example in the link above, the boundary of any set in the discrete metric is the empty set which is a subset of ${x in X : d(x_0,x) = r }$. But I am not too certain. Anyone mind showing me some counterexample? Thank you.
analysis metric-spaces
edited Jan 22 at 21:58
Kenny Wong
19.1k21441
asked Jan 22 at 21:54
RicoRico
558
$endgroup$
add a comment |
$begingroup$
Let $X$ be a metric space, and let $B(x_0, r)$ denote the open ball of radius $r$ centred at $x_0 in X$.
Does the statement $partial B(x_0, r) subseteq {x in X : d(x_0,x) = r }$ hold true always?
A similar question posted in Math StackExchange
As answered in the link above, we know that the boundary is not the set ${x in X : d(x_0,x) = r }$. however, it seems that the statement $partial B subseteq {x in X : d(x_0,x) = r }$ is true. Considering the example in the link above, the boundary of any set in the discrete metric is the empty set which is a subset of ${x in X : d(x_0,x) = r }$. But I am not too certain. Anyone mind showing me some counterexample? Thank you.
analysis metric-spaces
edited Jan 22 at 21:58
Kenny Wong
19.1k21441
asked Jan 22 at 21:54
RicoRico
558
$endgroup$
$begingroup$
I guess, $B={x:d(x_0,x)<r }$ here.
$endgroup$
– Berci
Jan 22 at 22:00
add a comment |
$begingroup$
Let $X$ be a metric space, and let $B(x_0, r)$ denote the open ball of radius $r$ centred at $x_0 in X$.
Does the statement $partial B(x_0, r) subseteq {x in X : d(x_0,x) = r }$ hold true always?
A similar question posted in Math StackExchange
As answered in the link above, we know that the boundary is not the set ${x in X : d(x_0,x) = r }$. however, it seems that the statement $partial B subseteq {x in X : d(x_0,x) = r }$ is true. Considering the example in the link above, the boundary of any set in the discrete metric is the empty set which is a subset of ${x in X : d(x_0,x) = r }$. But I am not too certain. Anyone mind showing me some counterexample? Thank you.
analysis metric-spaces
edited Jan 22 at 21:58
Kenny Wong
19.1k21441
asked Jan 22 at 21:54
RicoRico
558
$endgroup$
Let $X$ be a metric space, and let $B(x_0, r)$ denote the open ball of radius $r$ centred at $x_0 in X$.
Does the statement $partial B(x_0, r) subseteq {x in X : d(x_0,x) = r }$ hold true always?
A similar question posted in Math StackExchange
As answered in the link above, we know that the boundary is not the set ${x in X : d(x_0,x) = r }$. however, it seems that the statement $partial B subseteq {x in X : d(x_0,x) = r }$ is true. Considering the example in the link above, the boundary of any set in the discrete metric is the empty set which is a subset of ${x in X : d(x_0,x) = r }$. But I am not too certain. Anyone mind showing me some counterexample? Thank you.
analysis metric-spaces
analysis metric-spaces
edited Jan 22 at 21:58
Kenny Wong
19.1k21441
asked Jan 22 at 21:54
RicoRico
558
edited Jan 22 at 21:58
Kenny Wong
19.1k21441
asked Jan 22 at 21:54
RicoRico
558
edited Jan 22 at 21:58
Kenny Wong
19.1k21441
edited Jan 22 at 21:58
Kenny Wong
19.1k21441
edited Jan 22 at 21:58
Kenny Wong
19.1k21441
19.1k21441
asked Jan 22 at 21:54
RicoRico
558
asked Jan 22 at 21:54
RicoRico
558
asked Jan 22 at 21:54
RicoRico
558
558
$begingroup$
I guess, $B={x:d(x_0,x)<r }$ here.
$endgroup$
– Berci
Jan 22 at 22:00
add a comment |
$begingroup$
I guess, $B={x:d(x_0,x)<r }$ here.
$endgroup$
– Berci
Jan 22 at 22:00
$begingroup$
I guess, $B={x:d(x_0,x)<r }$ here.
$endgroup$
– Berci
Jan 22 at 22:00
$begingroup$
I guess, $B={x:d(x_0,x)<r }$ here.
$endgroup$
– Berci
Jan 22 at 22:00
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Take $yinpartial B(x_0,r)$. Can you have $d(y,x_0)neq r$? If so, there are two possibilities:
$d(y,x_0)<r$. But then $Bbigl(y,r-d(y,x_0)bigr)subset B(x_0,r)$ and so $yinoperatorname{int}bigl(B(x_0,r)bigr)$, which implies that $ynotinpartial B(x_0,r)$.
$d(y,x_0)>r$- But then $Bbigl(y,d(y,x_0)-rbigr)cap B(x_0,r)=emptyset$. So, $ynotinoverline{B(x_0,r)}$, which implies again that $ynotinpartial B(x_0,r)$.
answered Jan 22 at 22:03
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
This inclusion is true for $B$ the open ball of radius $r$ centered at $x_0$, for the following reasons :
1) $overline{B} subset {xin Xmid d(x,x_0)leq r}$ : this is because this set is closed.
2) If $mathrm{Int}(B) = B = {xin Xmid d(x,x_0)<r}$.
So if $xin partial B$, $xin overline{B}$ so $d(x,x_0)leq r$, but $xnotin B$ so $d(x,x_0)geq r$; so $d(x,x_0)=r$.
answered Jan 22 at 22:00
MaxMax
15.2k11143
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be a metric space, and let $B(x,r)subseteq X$ be an open ball. Since $text{Int}left(B(x,r)right)=B(x,r)$ (an open ball is an open set) and $text{Cl}left(B(x,r)right)subseteq{y in X :d(y,x)leq r}$ (a closed ball is a closed set), it follows that we have
begin{aligned}partial B(x,r)=text{Cl}left(B(x,r)right) setminus B(x,r) &subseteq {y in X :d(y,x)leq r} setminus B(x,r) \&={y in X :d(y,x)= r}.end{aligned}
Therefore, $partial B(x,r) subseteq {y in X :d(y,x) = r}$ holds for any metric space.
answered Jan 22 at 22:29
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
Take $yinpartial B(x_0,r)$. Can you have $d(y,x_0)neq r$? If so, there are two possibilities:
$d(y,x_0)<r$. But then $Bbigl(y,r-d(y,x_0)bigr)subset B(x_0,r)$ and so $yinoperatorname{int}bigl(B(x_0,r)bigr)$, which implies that $ynotinpartial B(x_0,r)$.
$d(y,x_0)>r$- But then $Bbigl(y,d(y,x_0)-rbigr)cap B(x_0,r)=emptyset$. So, $ynotinoverline{B(x_0,r)}$, which implies again that $ynotinpartial B(x_0,r)$.
answered Jan 22 at 22:03
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
Take $yinpartial B(x_0,r)$. Can you have $d(y,x_0)neq r$? If so, there are two possibilities:
$d(y,x_0)<r$. But then $Bbigl(y,r-d(y,x_0)bigr)subset B(x_0,r)$ and so $yinoperatorname{int}bigl(B(x_0,r)bigr)$, which implies that $ynotinpartial B(x_0,r)$.
$d(y,x_0)>r$- But then $Bbigl(y,d(y,x_0)-rbigr)cap B(x_0,r)=emptyset$. So, $ynotinoverline{B(x_0,r)}$, which implies again that $ynotinpartial B(x_0,r)$.
answered Jan 22 at 22:03
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
Take $yinpartial B(x_0,r)$. Can you have $d(y,x_0)neq r$? If so, there are two possibilities:
$d(y,x_0)<r$. But then $Bbigl(y,r-d(y,x_0)bigr)subset B(x_0,r)$ and so $yinoperatorname{int}bigl(B(x_0,r)bigr)$, which implies that $ynotinpartial B(x_0,r)$.
$d(y,x_0)>r$- But then $Bbigl(y,d(y,x_0)-rbigr)cap B(x_0,r)=emptyset$. So, $ynotinoverline{B(x_0,r)}$, which implies again that $ynotinpartial B(x_0,r)$.
answered Jan 22 at 22:03
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
Take $yinpartial B(x_0,r)$. Can you have $d(y,x_0)neq r$? If so, there are two possibilities:
$d(y,x_0)<r$. But then $Bbigl(y,r-d(y,x_0)bigr)subset B(x_0,r)$ and so $yinoperatorname{int}bigl(B(x_0,r)bigr)$, which implies that $ynotinpartial B(x_0,r)$.
$d(y,x_0)>r$- But then $Bbigl(y,d(y,x_0)-rbigr)cap B(x_0,r)=emptyset$. So, $ynotinoverline{B(x_0,r)}$, which implies again that $ynotinpartial B(x_0,r)$.
answered Jan 22 at 22:03
José Carlos SantosJosé Carlos Santos
166k22132235
edited Jan 22 at 22:14
answered Jan 22 at 22:03
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 22:03
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 22:03
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
add a comment |
add a comment |
$begingroup$
This inclusion is true for $B$ the open ball of radius $r$ centered at $x_0$, for the following reasons :
1) $overline{B} subset {xin Xmid d(x,x_0)leq r}$ : this is because this set is closed.
2) If $mathrm{Int}(B) = B = {xin Xmid d(x,x_0)<r}$.
So if $xin partial B$, $xin overline{B}$ so $d(x,x_0)leq r$, but $xnotin B$ so $d(x,x_0)geq r$; so $d(x,x_0)=r$.
answered Jan 22 at 22:00
MaxMax
15.2k11143
$endgroup$
add a comment |
$begingroup$
This inclusion is true for $B$ the open ball of radius $r$ centered at $x_0$, for the following reasons :
1) $overline{B} subset {xin Xmid d(x,x_0)leq r}$ : this is because this set is closed.
2) If $mathrm{Int}(B) = B = {xin Xmid d(x,x_0)<r}$.
So if $xin partial B$, $xin overline{B}$ so $d(x,x_0)leq r$, but $xnotin B$ so $d(x,x_0)geq r$; so $d(x,x_0)=r$.
answered Jan 22 at 22:00
MaxMax
15.2k11143
$endgroup$
add a comment |
$begingroup$
This inclusion is true for $B$ the open ball of radius $r$ centered at $x_0$, for the following reasons :
1) $overline{B} subset {xin Xmid d(x,x_0)leq r}$ : this is because this set is closed.
2) If $mathrm{Int}(B) = B = {xin Xmid d(x,x_0)<r}$.
So if $xin partial B$, $xin overline{B}$ so $d(x,x_0)leq r$, but $xnotin B$ so $d(x,x_0)geq r$; so $d(x,x_0)=r$.
answered Jan 22 at 22:00
MaxMax
15.2k11143
$endgroup$
This inclusion is true for $B$ the open ball of radius $r$ centered at $x_0$, for the following reasons :
1) $overline{B} subset {xin Xmid d(x,x_0)leq r}$ : this is because this set is closed.
2) If $mathrm{Int}(B) = B = {xin Xmid d(x,x_0)<r}$.
So if $xin partial B$, $xin overline{B}$ so $d(x,x_0)leq r$, but $xnotin B$ so $d(x,x_0)geq r$; so $d(x,x_0)=r$.
answered Jan 22 at 22:00
MaxMax
15.2k11143
answered Jan 22 at 22:00
MaxMax
15.2k11143
answered Jan 22 at 22:00
MaxMax
15.2k11143
answered Jan 22 at 22:00
MaxMax
15.2k11143
15.2k11143
add a comment |
add a comment |
$begingroup$
Let $(X,d)$ be a metric space, and let $B(x,r)subseteq X$ be an open ball. Since $text{Int}left(B(x,r)right)=B(x,r)$ (an open ball is an open set) and $text{Cl}left(B(x,r)right)subseteq{y in X :d(y,x)leq r}$ (a closed ball is a closed set), it follows that we have
begin{aligned}partial B(x,r)=text{Cl}left(B(x,r)right) setminus B(x,r) &subseteq {y in X :d(y,x)leq r} setminus B(x,r) \&={y in X :d(y,x)= r}.end{aligned}
Therefore, $partial B(x,r) subseteq {y in X :d(y,x) = r}$ holds for any metric space.
answered Jan 22 at 22:29
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be a metric space, and let $B(x,r)subseteq X$ be an open ball. Since $text{Int}left(B(x,r)right)=B(x,r)$ (an open ball is an open set) and $text{Cl}left(B(x,r)right)subseteq{y in X :d(y,x)leq r}$ (a closed ball is a closed set), it follows that we have
begin{aligned}partial B(x,r)=text{Cl}left(B(x,r)right) setminus B(x,r) &subseteq {y in X :d(y,x)leq r} setminus B(x,r) \&={y in X :d(y,x)= r}.end{aligned}
Therefore, $partial B(x,r) subseteq {y in X :d(y,x) = r}$ holds for any metric space.
answered Jan 22 at 22:29
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be a metric space, and let $B(x,r)subseteq X$ be an open ball. Since $text{Int}left(B(x,r)right)=B(x,r)$ (an open ball is an open set) and $text{Cl}left(B(x,r)right)subseteq{y in X :d(y,x)leq r}$ (a closed ball is a closed set), it follows that we have
begin{aligned}partial B(x,r)=text{Cl}left(B(x,r)right) setminus B(x,r) &subseteq {y in X :d(y,x)leq r} setminus B(x,r) \&={y in X :d(y,x)= r}.end{aligned}
Therefore, $partial B(x,r) subseteq {y in X :d(y,x) = r}$ holds for any metric space.
answered Jan 22 at 22:29
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
Let $(X,d)$ be a metric space, and let $B(x,r)subseteq X$ be an open ball. Since $text{Int}left(B(x,r)right)=B(x,r)$ (an open ball is an open set) and $text{Cl}left(B(x,r)right)subseteq{y in X :d(y,x)leq r}$ (a closed ball is a closed set), it follows that we have
begin{aligned}partial B(x,r)=text{Cl}left(B(x,r)right) setminus B(x,r) &subseteq {y in X :d(y,x)leq r} setminus B(x,r) \&={y in X :d(y,x)= r}.end{aligned}
Therefore, $partial B(x,r) subseteq {y in X :d(y,x) = r}$ holds for any metric space.
answered Jan 22 at 22:29
Matt A PeltoMatt A Pelto
2,602621
edited Jan 22 at 22:36
answered Jan 22 at 22:29
Matt A PeltoMatt A Pelto
2,602621
answered Jan 22 at 22:29
Matt A PeltoMatt A Pelto
2,602621
answered Jan 22 at 22:29
Matt A PeltoMatt A Pelto
2,602621
2,602621
add a comment |
add a comment |
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$begingroup$
I guess, $B={x:d(x_0,x)<r }$ here.
$endgroup$
– Berci
Jan 22 at 22:00