Mixing
Is $f(x,y) = d(x,y)$ a continuous function from a Metric space $X times X$ to $mathbb R$?
$begingroup$
Is $f(x,y) = d(x,y)$ a continuous function from a Metric space $X times X$ to $mathbb R$? where $(X , d)$ is a metric space.
My Attempt :
I think it is.
If we take any open ball of radius $epsilon > 0 $ , $(a-epsilon , a+epsilon)$ in $mathbb R$.
, We will get an open preimage in $X times X$.
The preimage will be the union of all $A times B$ where $A$ and $B $ are open balls of radius $frac{epsilon}{2}$ in $X$ whose centers are $a$ units apart.
Have I gone wrong anywhere?
Can anyone please check If I am correct or wrong?
general-topology proof-verification continuity metric-spaces
edited Jan 6 at 15:25
José Carlos Santos
156k22126227
asked Jan 6 at 15:07
cmicmi
1,126212
$endgroup$
add a comment |
$begingroup$
Is $f(x,y) = d(x,y)$ a continuous function from a Metric space $X times X$ to $mathbb R$? where $(X , d)$ is a metric space.
My Attempt :
I think it is.
If we take any open ball of radius $epsilon > 0 $ , $(a-epsilon , a+epsilon)$ in $mathbb R$.
, We will get an open preimage in $X times X$.
The preimage will be the union of all $A times B$ where $A$ and $B $ are open balls of radius $frac{epsilon}{2}$ in $X$ whose centers are $a$ units apart.
Have I gone wrong anywhere?
Can anyone please check If I am correct or wrong?
general-topology proof-verification continuity metric-spaces
edited Jan 6 at 15:25
José Carlos Santos
156k22126227
asked Jan 6 at 15:07
cmicmi
1,126212
$endgroup$
$begingroup$
Possible duplicate of How to prove the continuity of the metric function?
$endgroup$
– Cardioid_Ass_22
Jan 6 at 15:24
add a comment |
$begingroup$
Is $f(x,y) = d(x,y)$ a continuous function from a Metric space $X times X$ to $mathbb R$? where $(X , d)$ is a metric space.
My Attempt :
I think it is.
If we take any open ball of radius $epsilon > 0 $ , $(a-epsilon , a+epsilon)$ in $mathbb R$.
, We will get an open preimage in $X times X$.
The preimage will be the union of all $A times B$ where $A$ and $B $ are open balls of radius $frac{epsilon}{2}$ in $X$ whose centers are $a$ units apart.
Have I gone wrong anywhere?
Can anyone please check If I am correct or wrong?
general-topology proof-verification continuity metric-spaces
edited Jan 6 at 15:25
José Carlos Santos
156k22126227
asked Jan 6 at 15:07
cmicmi
1,126212
$endgroup$
Is $f(x,y) = d(x,y)$ a continuous function from a Metric space $X times X$ to $mathbb R$? where $(X , d)$ is a metric space.
My Attempt :
I think it is.
If we take any open ball of radius $epsilon > 0 $ , $(a-epsilon , a+epsilon)$ in $mathbb R$.
, We will get an open preimage in $X times X$.
The preimage will be the union of all $A times B$ where $A$ and $B $ are open balls of radius $frac{epsilon}{2}$ in $X$ whose centers are $a$ units apart.
Have I gone wrong anywhere?
Can anyone please check If I am correct or wrong?
general-topology proof-verification continuity metric-spaces
general-topology proof-verification continuity metric-spaces
edited Jan 6 at 15:25
José Carlos Santos
156k22126227
asked Jan 6 at 15:07
cmicmi
1,126212
edited Jan 6 at 15:25
José Carlos Santos
156k22126227
asked Jan 6 at 15:07
cmicmi
1,126212
edited Jan 6 at 15:25
José Carlos Santos
156k22126227
edited Jan 6 at 15:25
José Carlos Santos
156k22126227
edited Jan 6 at 15:25
José Carlos Santos
156k22126227
156k22126227
asked Jan 6 at 15:07
cmicmi
1,126212
asked Jan 6 at 15:07
cmicmi
1,126212
asked Jan 6 at 15:07
cmicmi
1,126212
1,126212
$begingroup$
Possible duplicate of How to prove the continuity of the metric function?
$endgroup$
– Cardioid_Ass_22
Jan 6 at 15:24
add a comment |
$begingroup$
Possible duplicate of How to prove the continuity of the metric function?
$endgroup$
– Cardioid_Ass_22
Jan 6 at 15:24
$begingroup$
Possible duplicate of How to prove the continuity of the metric function?
$endgroup$
– Cardioid_Ass_22
Jan 6 at 15:24
$begingroup$
Possible duplicate of How to prove the continuity of the metric function?
$endgroup$
– Cardioid_Ass_22
Jan 6 at 15:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your attempt is interesting, but bot correct. Consider the set$$S={(x,y)in Xtimes X,|,d(x,y)=a}.$$You claimed that$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)=bigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$This would indeed prove that $f$ is continuous, but actually you only have$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$I sugest that you prove that $f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)$ is a neighborhood of all of its points.
answered Jan 6 at 15:20
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
$begingroup$
I can not understand why you are saying $$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$ ? Can you give an element which is only in the bigger set@Jose Carlos Santos
$endgroup$
– cmi
Jan 6 at 15:28
2
$begingroup$
@cmi: It might even be that $S$ is empty such that the entire right-hand-side is empty too. For example, this is the case if if $X$ is $mathbb Q$ with the usual distance and $a=sqrt2$. Then for every $varepsilon>0$, the left hand side will contain plenty of pairs.
$endgroup$
– Henning Makholm
Jan 6 at 15:29
1
$begingroup$
@HenningMakholm Thank you for saving my time. You are right, of course.
$endgroup$
– José Carlos Santos
Jan 6 at 15:34
$begingroup$
Is the function uniformly continuous on $X times X$?@HenningMakholm
$endgroup$
– cmi
Jan 6 at 16:26
1
$begingroup$
@cmi Yes, it is.
$endgroup$
– José Carlos Santos
Jan 6 at 16:34
add a comment |
$begingroup$
You want to start from $(x,y)in Xtimes X$ and prove that $d$ is continuous at $(x,y)$.
Let $varepsilon>0$; take $x'in B(x,varepsilon/2)$ and $y'in B(x,varepsilon/2)$; then
$$
d(x',y')le d(x',x)+d(x,y')le d(x',x)+d(x,y)+d(y,y')le d(x,y)+varepsilon
$$
and therefore
$$
d(x',y')-d(x,y)levarepsilon
$$
Can you finish?
Your idea is good as well, but it should be better justified.
answered Jan 6 at 15:15
egregegreg
180k1485202
$endgroup$
$begingroup$
Is my attempt wrong?@egreg
$endgroup$
– cmi
Jan 6 at 15:16
$begingroup$
Yea I can conclude by the sequential argument.@egreg
$endgroup$
– cmi
Jan 6 at 15:17
1
$begingroup$
@cmi The idea is good, but it should be more detailed.
$endgroup$
– egreg
Jan 6 at 15:17
1
$begingroup$
@cmi You can do similarly to show that $d(x',y')-d(x,y)ge-varepsilon$, no sequences.
$endgroup$
– egreg
Jan 6 at 15:18
1
$begingroup$
@cmi Certainly so.
$endgroup$
– egreg
Jan 6 at 16:27
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your attempt is interesting, but bot correct. Consider the set$$S={(x,y)in Xtimes X,|,d(x,y)=a}.$$You claimed that$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)=bigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$This would indeed prove that $f$ is continuous, but actually you only have$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$I sugest that you prove that $f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)$ is a neighborhood of all of its points.
answered Jan 6 at 15:20
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
$begingroup$
I can not understand why you are saying $$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$ ? Can you give an element which is only in the bigger set@Jose Carlos Santos
$endgroup$
– cmi
Jan 6 at 15:28
2
$begingroup$
@cmi: It might even be that $S$ is empty such that the entire right-hand-side is empty too. For example, this is the case if if $X$ is $mathbb Q$ with the usual distance and $a=sqrt2$. Then for every $varepsilon>0$, the left hand side will contain plenty of pairs.
$endgroup$
– Henning Makholm
Jan 6 at 15:29
1
$begingroup$
@HenningMakholm Thank you for saving my time. You are right, of course.
$endgroup$
– José Carlos Santos
Jan 6 at 15:34
$begingroup$
Is the function uniformly continuous on $X times X$?@HenningMakholm
$endgroup$
– cmi
Jan 6 at 16:26
1
$begingroup$
@cmi Yes, it is.
$endgroup$
– José Carlos Santos
Jan 6 at 16:34
add a comment |
$begingroup$
Your attempt is interesting, but bot correct. Consider the set$$S={(x,y)in Xtimes X,|,d(x,y)=a}.$$You claimed that$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)=bigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$This would indeed prove that $f$ is continuous, but actually you only have$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$I sugest that you prove that $f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)$ is a neighborhood of all of its points.
answered Jan 6 at 15:20
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
$begingroup$
I can not understand why you are saying $$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$ ? Can you give an element which is only in the bigger set@Jose Carlos Santos
$endgroup$
– cmi
Jan 6 at 15:28
2
$begingroup$
@cmi: It might even be that $S$ is empty such that the entire right-hand-side is empty too. For example, this is the case if if $X$ is $mathbb Q$ with the usual distance and $a=sqrt2$. Then for every $varepsilon>0$, the left hand side will contain plenty of pairs.
$endgroup$
– Henning Makholm
Jan 6 at 15:29
1
$begingroup$
@HenningMakholm Thank you for saving my time. You are right, of course.
$endgroup$
– José Carlos Santos
Jan 6 at 15:34
$begingroup$
Is the function uniformly continuous on $X times X$?@HenningMakholm
$endgroup$
– cmi
Jan 6 at 16:26
1
$begingroup$
@cmi Yes, it is.
$endgroup$
– José Carlos Santos
Jan 6 at 16:34
add a comment |
$begingroup$
Your attempt is interesting, but bot correct. Consider the set$$S={(x,y)in Xtimes X,|,d(x,y)=a}.$$You claimed that$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)=bigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$This would indeed prove that $f$ is continuous, but actually you only have$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$I sugest that you prove that $f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)$ is a neighborhood of all of its points.
answered Jan 6 at 15:20
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
Your attempt is interesting, but bot correct. Consider the set$$S={(x,y)in Xtimes X,|,d(x,y)=a}.$$You claimed that$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)=bigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$This would indeed prove that $f$ is continuous, but actually you only have$$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$I sugest that you prove that $f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)$ is a neighborhood of all of its points.
answered Jan 6 at 15:20
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 6 at 15:20
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 6 at 15:20
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 6 at 15:20
José Carlos SantosJosé Carlos Santos
156k22126227
156k22126227
$begingroup$
I can not understand why you are saying $$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$ ? Can you give an element which is only in the bigger set@Jose Carlos Santos
$endgroup$
– cmi
Jan 6 at 15:28
2
$begingroup$
@cmi: It might even be that $S$ is empty such that the entire right-hand-side is empty too. For example, this is the case if if $X$ is $mathbb Q$ with the usual distance and $a=sqrt2$. Then for every $varepsilon>0$, the left hand side will contain plenty of pairs.
$endgroup$
– Henning Makholm
Jan 6 at 15:29
1
$begingroup$
@HenningMakholm Thank you for saving my time. You are right, of course.
$endgroup$
– José Carlos Santos
Jan 6 at 15:34
$begingroup$
Is the function uniformly continuous on $X times X$?@HenningMakholm
$endgroup$
– cmi
Jan 6 at 16:26
1
$begingroup$
@cmi Yes, it is.
$endgroup$
– José Carlos Santos
Jan 6 at 16:34
add a comment |
$begingroup$
I can not understand why you are saying $$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$ ? Can you give an element which is only in the bigger set@Jose Carlos Santos
$endgroup$
– cmi
Jan 6 at 15:28
2
$begingroup$
@cmi: It might even be that $S$ is empty such that the entire right-hand-side is empty too. For example, this is the case if if $X$ is $mathbb Q$ with the usual distance and $a=sqrt2$. Then for every $varepsilon>0$, the left hand side will contain plenty of pairs.
$endgroup$
– Henning Makholm
Jan 6 at 15:29
1
$begingroup$
@HenningMakholm Thank you for saving my time. You are right, of course.
$endgroup$
– José Carlos Santos
Jan 6 at 15:34
$begingroup$
Is the function uniformly continuous on $X times X$?@HenningMakholm
$endgroup$
– cmi
Jan 6 at 16:26
1
$begingroup$
@cmi Yes, it is.
$endgroup$
– José Carlos Santos
Jan 6 at 16:34
$begingroup$
I can not understand why you are saying $$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$ ? Can you give an element which is only in the bigger set@Jose Carlos Santos
$endgroup$
– cmi
Jan 6 at 15:28
$begingroup$
I can not understand why you are saying $$f^{-1}bigl((a-varepsilon,a+varepsilon)bigr)supsetbigcup_{(x,y)in S}B_{fracvarepsilon2}(x)times B_{fracvarepsilon2}(y).$$ ? Can you give an element which is only in the bigger set@Jose Carlos Santos
$endgroup$
– cmi
Jan 6 at 15:28
2
2
$begingroup$
@cmi: It might even be that $S$ is empty such that the entire right-hand-side is empty too. For example, this is the case if if $X$ is $mathbb Q$ with the usual distance and $a=sqrt2$. Then for every $varepsilon>0$, the left hand side will contain plenty of pairs.
$endgroup$
– Henning Makholm
Jan 6 at 15:29
$begingroup$
@cmi: It might even be that $S$ is empty such that the entire right-hand-side is empty too. For example, this is the case if if $X$ is $mathbb Q$ with the usual distance and $a=sqrt2$. Then for every $varepsilon>0$, the left hand side will contain plenty of pairs.
$endgroup$
– Henning Makholm
Jan 6 at 15:29
1
1
$begingroup$
@HenningMakholm Thank you for saving my time. You are right, of course.
$endgroup$
– José Carlos Santos
Jan 6 at 15:34
$begingroup$
@HenningMakholm Thank you for saving my time. You are right, of course.
$endgroup$
– José Carlos Santos
Jan 6 at 15:34
$begingroup$
Is the function uniformly continuous on $X times X$?@HenningMakholm
$endgroup$
– cmi
Jan 6 at 16:26
$begingroup$
Is the function uniformly continuous on $X times X$?@HenningMakholm
$endgroup$
– cmi
Jan 6 at 16:26
1
1
$begingroup$
@cmi Yes, it is.
$endgroup$
– José Carlos Santos
Jan 6 at 16:34
$begingroup$
@cmi Yes, it is.
$endgroup$
– José Carlos Santos
Jan 6 at 16:34
add a comment |
$begingroup$
You want to start from $(x,y)in Xtimes X$ and prove that $d$ is continuous at $(x,y)$.
Let $varepsilon>0$; take $x'in B(x,varepsilon/2)$ and $y'in B(x,varepsilon/2)$; then
$$
d(x',y')le d(x',x)+d(x,y')le d(x',x)+d(x,y)+d(y,y')le d(x,y)+varepsilon
$$
and therefore
$$
d(x',y')-d(x,y)levarepsilon
$$
Can you finish?
Your idea is good as well, but it should be better justified.
answered Jan 6 at 15:15
egregegreg
180k1485202
$endgroup$
$begingroup$
Is my attempt wrong?@egreg
$endgroup$
– cmi
Jan 6 at 15:16
$begingroup$
Yea I can conclude by the sequential argument.@egreg
$endgroup$
– cmi
Jan 6 at 15:17
1
$begingroup$
@cmi The idea is good, but it should be more detailed.
$endgroup$
– egreg
Jan 6 at 15:17
1
$begingroup$
@cmi You can do similarly to show that $d(x',y')-d(x,y)ge-varepsilon$, no sequences.
$endgroup$
– egreg
Jan 6 at 15:18
1
$begingroup$
@cmi Certainly so.
$endgroup$
– egreg
Jan 6 at 16:27
|
show 1 more comment
$begingroup$
You want to start from $(x,y)in Xtimes X$ and prove that $d$ is continuous at $(x,y)$.
Let $varepsilon>0$; take $x'in B(x,varepsilon/2)$ and $y'in B(x,varepsilon/2)$; then
$$
d(x',y')le d(x',x)+d(x,y')le d(x',x)+d(x,y)+d(y,y')le d(x,y)+varepsilon
$$
and therefore
$$
d(x',y')-d(x,y)levarepsilon
$$
Can you finish?
Your idea is good as well, but it should be better justified.
answered Jan 6 at 15:15
egregegreg
180k1485202
$endgroup$
$begingroup$
Is my attempt wrong?@egreg
$endgroup$
– cmi
Jan 6 at 15:16
$begingroup$
Yea I can conclude by the sequential argument.@egreg
$endgroup$
– cmi
Jan 6 at 15:17
1
$begingroup$
@cmi The idea is good, but it should be more detailed.
$endgroup$
– egreg
Jan 6 at 15:17
1
$begingroup$
@cmi You can do similarly to show that $d(x',y')-d(x,y)ge-varepsilon$, no sequences.
$endgroup$
– egreg
Jan 6 at 15:18
1
$begingroup$
@cmi Certainly so.
$endgroup$
– egreg
Jan 6 at 16:27
|
show 1 more comment
$begingroup$
You want to start from $(x,y)in Xtimes X$ and prove that $d$ is continuous at $(x,y)$.
Let $varepsilon>0$; take $x'in B(x,varepsilon/2)$ and $y'in B(x,varepsilon/2)$; then
$$
d(x',y')le d(x',x)+d(x,y')le d(x',x)+d(x,y)+d(y,y')le d(x,y)+varepsilon
$$
and therefore
$$
d(x',y')-d(x,y)levarepsilon
$$
Can you finish?
Your idea is good as well, but it should be better justified.
answered Jan 6 at 15:15
egregegreg
180k1485202
$endgroup$
You want to start from $(x,y)in Xtimes X$ and prove that $d$ is continuous at $(x,y)$.
Let $varepsilon>0$; take $x'in B(x,varepsilon/2)$ and $y'in B(x,varepsilon/2)$; then
$$
d(x',y')le d(x',x)+d(x,y')le d(x',x)+d(x,y)+d(y,y')le d(x,y)+varepsilon
$$
and therefore
$$
d(x',y')-d(x,y)levarepsilon
$$
Can you finish?
Your idea is good as well, but it should be better justified.
answered Jan 6 at 15:15
egregegreg
180k1485202
edited Jan 6 at 15:16
answered Jan 6 at 15:15
egregegreg
180k1485202
answered Jan 6 at 15:15
egregegreg
180k1485202
answered Jan 6 at 15:15
egregegreg
180k1485202
180k1485202
$begingroup$
Is my attempt wrong?@egreg
$endgroup$
– cmi
Jan 6 at 15:16
$begingroup$
Yea I can conclude by the sequential argument.@egreg
$endgroup$
– cmi
Jan 6 at 15:17
1
$begingroup$
@cmi The idea is good, but it should be more detailed.
$endgroup$
– egreg
Jan 6 at 15:17
1
$begingroup$
@cmi You can do similarly to show that $d(x',y')-d(x,y)ge-varepsilon$, no sequences.
$endgroup$
– egreg
Jan 6 at 15:18
1
$begingroup$
@cmi Certainly so.
$endgroup$
– egreg
Jan 6 at 16:27
|
show 1 more comment
$begingroup$
Is my attempt wrong?@egreg
$endgroup$
– cmi
Jan 6 at 15:16
$begingroup$
Yea I can conclude by the sequential argument.@egreg
$endgroup$
– cmi
Jan 6 at 15:17
1
$begingroup$
@cmi The idea is good, but it should be more detailed.
$endgroup$
– egreg
Jan 6 at 15:17
1
$begingroup$
@cmi You can do similarly to show that $d(x',y')-d(x,y)ge-varepsilon$, no sequences.
$endgroup$
– egreg
Jan 6 at 15:18
1
$begingroup$
@cmi Certainly so.
$endgroup$
– egreg
Jan 6 at 16:27
$begingroup$
Is my attempt wrong?@egreg
$endgroup$
– cmi
Jan 6 at 15:16
$begingroup$
Is my attempt wrong?@egreg
$endgroup$
– cmi
Jan 6 at 15:16
$begingroup$
Yea I can conclude by the sequential argument.@egreg
$endgroup$
– cmi
Jan 6 at 15:17
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Yea I can conclude by the sequential argument.@egreg
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– cmi
Jan 6 at 15:17
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1
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@cmi The idea is good, but it should be more detailed.
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– egreg
Jan 6 at 15:17
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@cmi The idea is good, but it should be more detailed.
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– egreg
Jan 6 at 15:17
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1
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@cmi You can do similarly to show that $d(x',y')-d(x,y)ge-varepsilon$, no sequences.
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– egreg
Jan 6 at 15:18
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@cmi You can do similarly to show that $d(x',y')-d(x,y)ge-varepsilon$, no sequences.
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– egreg
Jan 6 at 15:18
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1
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@cmi Certainly so.
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– egreg
Jan 6 at 16:27
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@cmi Certainly so.
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– egreg
Jan 6 at 16:27
|
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$begingroup$
Possible duplicate of How to prove the continuity of the metric function?
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– Cardioid_Ass_22
Jan 6 at 15:24