Mixing
Show that, for every metric $d$, the metrics $d/(1+d)$ and $min1,d$ are equivalent
Clash Royale CLAN TAG#URR8PPP
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The actual problem looks like-
Let, $(X,d)$ be a metric space where $Xneemptyset$. Define $d_1$ and
$d_2$ on $Xtimes X$ by $d_1(x,y)=fracd(x,y)1+d(x,y)$,
$d_2(x,y)=min1, d(x,y)quadforall(x,y)in Xtimes X.$(It is easy
to verify $d_1,d_2$ are two metrics on $X$.) Show that, $d_1, d_2$ are
equivalent metrics on $X$.
Firstly, I'll use the standard theorem to prove two metrics on a non-empty set are equivalent-
Two metrics $d_1$ and $d_2$ on a non-empty set $X$ are equivalent if and only if for each $xin X$ and each open ball $B_d_i (x,r)$ in $(X,d_i)$(for some $r>0$), there is an $s>0$ such that $B_d_j (x,s)subseteq B_d_i (x,r)$, for $i, j=1,2$ and $ine j$.
Now, to solve the problem, I break it into to parts- If I can show that both of $d_1,d_2$ are equivalent with the metric $d$, then we are done.
I'm able to show $d$ and $d_2$ are equivalent in the following way-
Let, $xin X$ and $B_d_2 (x,r)$ be an open ball in $(X, d_2)$ for some $r>0$. Then $B_d (x,r)subseteq B_d_2 (x,r)$.
Again, if we take an open ball $B_d (x,t)$ in $(X,d)$ for some $t>0$. Then $B_d_2 (x,s)subseteq B_d (x,t)$, where $s=min 1,t$.
But how to show $d_1$ and $d$ are equivalent metrics in $X$?
Can anyone assist me to prove $d_1$ and $d$ are equivalent metrics on $X$? Thanks for your help in advance.
metric-spaces equivalent-metrics
edited Jul 26 at 6:58
Did
242k23208441
asked Jul 26 at 6:38
Biswarup Saha
2298
add a comment |Â
up vote
1
down vote
favorite
The actual problem looks like-
Let, $(X,d)$ be a metric space where $Xneemptyset$. Define $d_1$ and
$d_2$ on $Xtimes X$ by $d_1(x,y)=fracd(x,y)1+d(x,y)$,
$d_2(x,y)=min1, d(x,y)quadforall(x,y)in Xtimes X.$(It is easy
to verify $d_1,d_2$ are two metrics on $X$.) Show that, $d_1, d_2$ are
equivalent metrics on $X$.
Firstly, I'll use the standard theorem to prove two metrics on a non-empty set are equivalent-
Two metrics $d_1$ and $d_2$ on a non-empty set $X$ are equivalent if and only if for each $xin X$ and each open ball $B_d_i (x,r)$ in $(X,d_i)$(for some $r>0$), there is an $s>0$ such that $B_d_j (x,s)subseteq B_d_i (x,r)$, for $i, j=1,2$ and $ine j$.
Now, to solve the problem, I break it into to parts- If I can show that both of $d_1,d_2$ are equivalent with the metric $d$, then we are done.
I'm able to show $d$ and $d_2$ are equivalent in the following way-
Let, $xin X$ and $B_d_2 (x,r)$ be an open ball in $(X, d_2)$ for some $r>0$. Then $B_d (x,r)subseteq B_d_2 (x,r)$.
Again, if we take an open ball $B_d (x,t)$ in $(X,d)$ for some $t>0$. Then $B_d_2 (x,s)subseteq B_d (x,t)$, where $s=min 1,t$.
But how to show $d_1$ and $d$ are equivalent metrics in $X$?
Can anyone assist me to prove $d_1$ and $d$ are equivalent metrics on $X$? Thanks for your help in advance.
metric-spaces equivalent-metrics
edited Jul 26 at 6:58
Did
242k23208441
asked Jul 26 at 6:38
Biswarup Saha
2298
1
As an alternative, show that $$d/(1+d)leqslantmin1,dleqslant2d/(1+d)$$
– Did
Jul 26 at 7:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The actual problem looks like-
Let, $(X,d)$ be a metric space where $Xneemptyset$. Define $d_1$ and
$d_2$ on $Xtimes X$ by $d_1(x,y)=fracd(x,y)1+d(x,y)$,
$d_2(x,y)=min1, d(x,y)quadforall(x,y)in Xtimes X.$(It is easy
to verify $d_1,d_2$ are two metrics on $X$.) Show that, $d_1, d_2$ are
equivalent metrics on $X$.
Firstly, I'll use the standard theorem to prove two metrics on a non-empty set are equivalent-
Two metrics $d_1$ and $d_2$ on a non-empty set $X$ are equivalent if and only if for each $xin X$ and each open ball $B_d_i (x,r)$ in $(X,d_i)$(for some $r>0$), there is an $s>0$ such that $B_d_j (x,s)subseteq B_d_i (x,r)$, for $i, j=1,2$ and $ine j$.
Now, to solve the problem, I break it into to parts- If I can show that both of $d_1,d_2$ are equivalent with the metric $d$, then we are done.
I'm able to show $d$ and $d_2$ are equivalent in the following way-
Let, $xin X$ and $B_d_2 (x,r)$ be an open ball in $(X, d_2)$ for some $r>0$. Then $B_d (x,r)subseteq B_d_2 (x,r)$.
Again, if we take an open ball $B_d (x,t)$ in $(X,d)$ for some $t>0$. Then $B_d_2 (x,s)subseteq B_d (x,t)$, where $s=min 1,t$.
But how to show $d_1$ and $d$ are equivalent metrics in $X$?
Can anyone assist me to prove $d_1$ and $d$ are equivalent metrics on $X$? Thanks for your help in advance.
metric-spaces equivalent-metrics
edited Jul 26 at 6:58
Did
242k23208441
asked Jul 26 at 6:38
Biswarup Saha
2298
The actual problem looks like-
Let, $(X,d)$ be a metric space where $Xneemptyset$. Define $d_1$ and
$d_2$ on $Xtimes X$ by $d_1(x,y)=fracd(x,y)1+d(x,y)$,
$d_2(x,y)=min1, d(x,y)quadforall(x,y)in Xtimes X.$(It is easy
to verify $d_1,d_2$ are two metrics on $X$.) Show that, $d_1, d_2$ are
equivalent metrics on $X$.
Firstly, I'll use the standard theorem to prove two metrics on a non-empty set are equivalent-
Two metrics $d_1$ and $d_2$ on a non-empty set $X$ are equivalent if and only if for each $xin X$ and each open ball $B_d_i (x,r)$ in $(X,d_i)$(for some $r>0$), there is an $s>0$ such that $B_d_j (x,s)subseteq B_d_i (x,r)$, for $i, j=1,2$ and $ine j$.
Now, to solve the problem, I break it into to parts- If I can show that both of $d_1,d_2$ are equivalent with the metric $d$, then we are done.
I'm able to show $d$ and $d_2$ are equivalent in the following way-
Let, $xin X$ and $B_d_2 (x,r)$ be an open ball in $(X, d_2)$ for some $r>0$. Then $B_d (x,r)subseteq B_d_2 (x,r)$.
Again, if we take an open ball $B_d (x,t)$ in $(X,d)$ for some $t>0$. Then $B_d_2 (x,s)subseteq B_d (x,t)$, where $s=min 1,t$.
But how to show $d_1$ and $d$ are equivalent metrics in $X$?
Can anyone assist me to prove $d_1$ and $d$ are equivalent metrics on $X$? Thanks for your help in advance.
metric-spaces equivalent-metrics
edited Jul 26 at 6:58
Did
242k23208441
asked Jul 26 at 6:38
Biswarup Saha
2298
edited Jul 26 at 6:58
Did
242k23208441
edited Jul 26 at 6:58
Did
242k23208441
edited Jul 26 at 6:58
Did
242k23208441
242k23208441
asked Jul 26 at 6:38
Biswarup Saha
2298
asked Jul 26 at 6:38
Biswarup Saha
2298
asked Jul 26 at 6:38
Biswarup Saha
2298
2298
1
As an alternative, show that $$d/(1+d)leqslantmin1,dleqslant2d/(1+d)$$
– Did
Jul 26 at 7:01
add a comment |Â
1
As an alternative, show that $$d/(1+d)leqslantmin1,dleqslant2d/(1+d)$$
– Did
Jul 26 at 7:01
1
1
As an alternative, show that $$d/(1+d)leqslantmin1,dleqslant2d/(1+d)$$
– Did
Jul 26 at 7:01
As an alternative, show that $$d/(1+d)leqslantmin1,dleqslant2d/(1+d)$$
– Did
Jul 26 at 7:01
add a comment |Â
3 Answers
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Just verify that $B_d(x,r)$ contains $B_d_1(x,s)$ where $s=
frac r 1+r$ and $B_d_1(x,r)$ contains $B_d(x,s)$ where $s=
frac r 1-r$ Note: it is enough to consider the case when $r<1$: if $rgeq 1$ first find $s$ such that $B_d_1(x,frac 1 2)$ contains $B_d(x,s)$ and note that $B_d(x,s)subset B_d_1(x,frac 1 2) subset B_d_1(x,r)$.
answered Jul 26 at 6:46
Kavi Rama Murthy
20k2829
Hmmm, for $r=2$, say?
– Did
Jul 26 at 7:02
Kavi Rama Murthy, if $r>1$, then $s<0$??
– Biswarup Saha
Jul 26 at 7:13
@BiswarupSaha to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:27
@Did to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:28
Sorry but I wonder what is the point of your comment. It is good that you have added some explanations to the answer though.
– Did
Jul 26 at 10:36
add a comment |Â
up vote
0
down vote
For proving that $d$ and $d_1$ are equivalent metrics it is enough to prove that for any $1>r>0$ there exist $s$ and $t$ positive real numbers s.t. for any $xin X$ there holds $B_d_1(x,s)subseteq B_d(x,r)subseteq B_d_1(x,t).$
Let suppose that $s=fracr1+r$. Let $yin B_d_1(x,s),$ so $d_1(x,y)<sRightarrow fracd(x,y)1+d(x,y)<sRightarrow d(x,y)<fracs1-s=r.$ So we proved that $yin B_d_1(x,s) Rightarrow yin B_d(x,r)$ or $B_d_1(x,s)subseteq B_d(x,r).$
Now, let $t=fracr1+r$ and $yin B_d(x,r),$ so $d(x,y)<rRightarrow frac1d(x,y)>frac1rRightarrow 1+frac1d(x,y)>1+frac1rRightarrow frac1+d(x,y)d(x,y)>frac1+rrRightarrow fracd(x,y)1+d(x,y)<fracr1+r=t.$ So we proved that $yin B_d(x,r) Rightarrow yin B_d_1(x,t)$ or $B_d(x,r)subseteq B_d_1(x,t).$
answered Jul 26 at 8:18
Emin
1,27621330
'there exist $r,s$ and $t$ ...' is not enough.
– Kavi Rama Murthy
Jul 26 at 8:21
Look at the edited version.
– Emin
Jul 26 at 8:24
add a comment |Â
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0
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Note that $d_1(x,y)leq d(x,y)$ for all $x,yin X$, since $1+d(x,y)geq 1$. Thus, it follows that $B_varepsilon^d(x)subseteq B_varepsilon^d_1(x)$ for all $xin X$ and $varepsilon>0$. Hence, if $Usubseteq X$ is $d_1$-open, it is also $d$-open.
Conversely, suppose that $U$ is $d$-open and let $xin U$. By definition of an open set in a metric space, there exists $varepsilon>0$ such that $B_varepsilon^d(x)subseteq U$. Now let $delta=minvarepsilon/2,1/2$. If $d_1(x,y)<delta$ then $d(x,y)<varepsilon$, because:
$$d(x,y)=fracd_1(x,y)1-d_1(x,y)leq 2d_1(x,y)< epsilon$$
Hence, $B_delta^d_1(x)subseteq B_varepsilon^d(x)subseteq U$, which implies that $U$ is $d_1$-open.
Thus, by the above, $(X,d)$ and $(X,d_1)$ are topologically equivalent.
Since you have shown that $(X,d)$ and $(X,d_2)$ are topologically equivalent, it follows from the fact that topological equivalence is a equivalence relation that $(X,d_1)$ and $(X,d_2)$ are topologically equivalent.
answered Jul 26 at 8:08
projectilemotion
11k61941
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Just verify that $B_d(x,r)$ contains $B_d_1(x,s)$ where $s=
frac r 1+r$ and $B_d_1(x,r)$ contains $B_d(x,s)$ where $s=
frac r 1-r$ Note: it is enough to consider the case when $r<1$: if $rgeq 1$ first find $s$ such that $B_d_1(x,frac 1 2)$ contains $B_d(x,s)$ and note that $B_d(x,s)subset B_d_1(x,frac 1 2) subset B_d_1(x,r)$.
answered Jul 26 at 6:46
Kavi Rama Murthy
20k2829
Hmmm, for $r=2$, say?
– Did
Jul 26 at 7:02
Kavi Rama Murthy, if $r>1$, then $s<0$??
– Biswarup Saha
Jul 26 at 7:13
@BiswarupSaha to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:27
@Did to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:28
Sorry but I wonder what is the point of your comment. It is good that you have added some explanations to the answer though.
– Did
Jul 26 at 10:36
add a comment |Â
up vote
0
down vote
Just verify that $B_d(x,r)$ contains $B_d_1(x,s)$ where $s=
frac r 1+r$ and $B_d_1(x,r)$ contains $B_d(x,s)$ where $s=
frac r 1-r$ Note: it is enough to consider the case when $r<1$: if $rgeq 1$ first find $s$ such that $B_d_1(x,frac 1 2)$ contains $B_d(x,s)$ and note that $B_d(x,s)subset B_d_1(x,frac 1 2) subset B_d_1(x,r)$.
answered Jul 26 at 6:46
Kavi Rama Murthy
20k2829
Hmmm, for $r=2$, say?
– Did
Jul 26 at 7:02
Kavi Rama Murthy, if $r>1$, then $s<0$??
– Biswarup Saha
Jul 26 at 7:13
@BiswarupSaha to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:27
@Did to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:28
Sorry but I wonder what is the point of your comment. It is good that you have added some explanations to the answer though.
– Did
Jul 26 at 10:36
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Just verify that $B_d(x,r)$ contains $B_d_1(x,s)$ where $s=
frac r 1+r$ and $B_d_1(x,r)$ contains $B_d(x,s)$ where $s=
frac r 1-r$ Note: it is enough to consider the case when $r<1$: if $rgeq 1$ first find $s$ such that $B_d_1(x,frac 1 2)$ contains $B_d(x,s)$ and note that $B_d(x,s)subset B_d_1(x,frac 1 2) subset B_d_1(x,r)$.
answered Jul 26 at 6:46
Kavi Rama Murthy
20k2829
Just verify that $B_d(x,r)$ contains $B_d_1(x,s)$ where $s=
frac r 1+r$ and $B_d_1(x,r)$ contains $B_d(x,s)$ where $s=
frac r 1-r$ Note: it is enough to consider the case when $r<1$: if $rgeq 1$ first find $s$ such that $B_d_1(x,frac 1 2)$ contains $B_d(x,s)$ and note that $B_d(x,s)subset B_d_1(x,frac 1 2) subset B_d_1(x,r)$.
answered Jul 26 at 6:46
Kavi Rama Murthy
20k2829
edited Jul 26 at 7:25
answered Jul 26 at 6:46
Kavi Rama Murthy
20k2829
answered Jul 26 at 6:46
Kavi Rama Murthy
20k2829
answered Jul 26 at 6:46
Kavi Rama Murthy
20k2829
20k2829
Hmmm, for $r=2$, say?
– Did
Jul 26 at 7:02
Kavi Rama Murthy, if $r>1$, then $s<0$??
– Biswarup Saha
Jul 26 at 7:13
@BiswarupSaha to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:27
@Did to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:28
Sorry but I wonder what is the point of your comment. It is good that you have added some explanations to the answer though.
– Did
Jul 26 at 10:36
add a comment |Â
Hmmm, for $r=2$, say?
– Did
Jul 26 at 7:02
Kavi Rama Murthy, if $r>1$, then $s<0$??
– Biswarup Saha
Jul 26 at 7:13
@BiswarupSaha to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:27
@Did to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:28
Sorry but I wonder what is the point of your comment. It is good that you have added some explanations to the answer though.
– Did
Jul 26 at 10:36
Hmmm, for $r=2$, say?
– Did
Jul 26 at 7:02
Hmmm, for $r=2$, say?
– Did
Jul 26 at 7:02
Kavi Rama Murthy, if $r>1$, then $s<0$??
– Biswarup Saha
Jul 26 at 7:13
Kavi Rama Murthy, if $r>1$, then $s<0$??
– Biswarup Saha
Jul 26 at 7:13
@BiswarupSaha to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:27
@BiswarupSaha to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:27
@Did to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:28
@Did to show that two metrics are equivalent it is enough to consider small balls. This is very elementary and down-voting is not justified for this. Anyway I have edited the answer now.
– Kavi Rama Murthy
Jul 26 at 7:28
Sorry but I wonder what is the point of your comment. It is good that you have added some explanations to the answer though.
– Did
Jul 26 at 10:36
Sorry but I wonder what is the point of your comment. It is good that you have added some explanations to the answer though.
– Did
Jul 26 at 10:36
add a comment |Â
up vote
0
down vote
For proving that $d$ and $d_1$ are equivalent metrics it is enough to prove that for any $1>r>0$ there exist $s$ and $t$ positive real numbers s.t. for any $xin X$ there holds $B_d_1(x,s)subseteq B_d(x,r)subseteq B_d_1(x,t).$
Let suppose that $s=fracr1+r$. Let $yin B_d_1(x,s),$ so $d_1(x,y)<sRightarrow fracd(x,y)1+d(x,y)<sRightarrow d(x,y)<fracs1-s=r.$ So we proved that $yin B_d_1(x,s) Rightarrow yin B_d(x,r)$ or $B_d_1(x,s)subseteq B_d(x,r).$
Now, let $t=fracr1+r$ and $yin B_d(x,r),$ so $d(x,y)<rRightarrow frac1d(x,y)>frac1rRightarrow 1+frac1d(x,y)>1+frac1rRightarrow frac1+d(x,y)d(x,y)>frac1+rrRightarrow fracd(x,y)1+d(x,y)<fracr1+r=t.$ So we proved that $yin B_d(x,r) Rightarrow yin B_d_1(x,t)$ or $B_d(x,r)subseteq B_d_1(x,t).$
answered Jul 26 at 8:18
Emin
1,27621330
'there exist $r,s$ and $t$ ...' is not enough.
– Kavi Rama Murthy
Jul 26 at 8:21
Look at the edited version.
– Emin
Jul 26 at 8:24
add a comment |Â
up vote
0
down vote
For proving that $d$ and $d_1$ are equivalent metrics it is enough to prove that for any $1>r>0$ there exist $s$ and $t$ positive real numbers s.t. for any $xin X$ there holds $B_d_1(x,s)subseteq B_d(x,r)subseteq B_d_1(x,t).$
Let suppose that $s=fracr1+r$. Let $yin B_d_1(x,s),$ so $d_1(x,y)<sRightarrow fracd(x,y)1+d(x,y)<sRightarrow d(x,y)<fracs1-s=r.$ So we proved that $yin B_d_1(x,s) Rightarrow yin B_d(x,r)$ or $B_d_1(x,s)subseteq B_d(x,r).$
Now, let $t=fracr1+r$ and $yin B_d(x,r),$ so $d(x,y)<rRightarrow frac1d(x,y)>frac1rRightarrow 1+frac1d(x,y)>1+frac1rRightarrow frac1+d(x,y)d(x,y)>frac1+rrRightarrow fracd(x,y)1+d(x,y)<fracr1+r=t.$ So we proved that $yin B_d(x,r) Rightarrow yin B_d_1(x,t)$ or $B_d(x,r)subseteq B_d_1(x,t).$
answered Jul 26 at 8:18
Emin
1,27621330
'there exist $r,s$ and $t$ ...' is not enough.
– Kavi Rama Murthy
Jul 26 at 8:21
Look at the edited version.
– Emin
Jul 26 at 8:24
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For proving that $d$ and $d_1$ are equivalent metrics it is enough to prove that for any $1>r>0$ there exist $s$ and $t$ positive real numbers s.t. for any $xin X$ there holds $B_d_1(x,s)subseteq B_d(x,r)subseteq B_d_1(x,t).$
Let suppose that $s=fracr1+r$. Let $yin B_d_1(x,s),$ so $d_1(x,y)<sRightarrow fracd(x,y)1+d(x,y)<sRightarrow d(x,y)<fracs1-s=r.$ So we proved that $yin B_d_1(x,s) Rightarrow yin B_d(x,r)$ or $B_d_1(x,s)subseteq B_d(x,r).$
Now, let $t=fracr1+r$ and $yin B_d(x,r),$ so $d(x,y)<rRightarrow frac1d(x,y)>frac1rRightarrow 1+frac1d(x,y)>1+frac1rRightarrow frac1+d(x,y)d(x,y)>frac1+rrRightarrow fracd(x,y)1+d(x,y)<fracr1+r=t.$ So we proved that $yin B_d(x,r) Rightarrow yin B_d_1(x,t)$ or $B_d(x,r)subseteq B_d_1(x,t).$
answered Jul 26 at 8:18
Emin
1,27621330
For proving that $d$ and $d_1$ are equivalent metrics it is enough to prove that for any $1>r>0$ there exist $s$ and $t$ positive real numbers s.t. for any $xin X$ there holds $B_d_1(x,s)subseteq B_d(x,r)subseteq B_d_1(x,t).$
Let suppose that $s=fracr1+r$. Let $yin B_d_1(x,s),$ so $d_1(x,y)<sRightarrow fracd(x,y)1+d(x,y)<sRightarrow d(x,y)<fracs1-s=r.$ So we proved that $yin B_d_1(x,s) Rightarrow yin B_d(x,r)$ or $B_d_1(x,s)subseteq B_d(x,r).$
Now, let $t=fracr1+r$ and $yin B_d(x,r),$ so $d(x,y)<rRightarrow frac1d(x,y)>frac1rRightarrow 1+frac1d(x,y)>1+frac1rRightarrow frac1+d(x,y)d(x,y)>frac1+rrRightarrow fracd(x,y)1+d(x,y)<fracr1+r=t.$ So we proved that $yin B_d(x,r) Rightarrow yin B_d_1(x,t)$ or $B_d(x,r)subseteq B_d_1(x,t).$
answered Jul 26 at 8:18
Emin
1,27621330
edited Jul 26 at 8:31
answered Jul 26 at 8:18
Emin
1,27621330
answered Jul 26 at 8:18
Emin
1,27621330
answered Jul 26 at 8:18
Emin
1,27621330
1,27621330
'there exist $r,s$ and $t$ ...' is not enough.
– Kavi Rama Murthy
Jul 26 at 8:21
Look at the edited version.
– Emin
Jul 26 at 8:24
add a comment |Â
'there exist $r,s$ and $t$ ...' is not enough.
– Kavi Rama Murthy
Jul 26 at 8:21
Look at the edited version.
– Emin
Jul 26 at 8:24
'there exist $r,s$ and $t$ ...' is not enough.
– Kavi Rama Murthy
Jul 26 at 8:21
'there exist $r,s$ and $t$ ...' is not enough.
– Kavi Rama Murthy
Jul 26 at 8:21
Look at the edited version.
– Emin
Jul 26 at 8:24
Look at the edited version.
– Emin
Jul 26 at 8:24
add a comment |Â
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Note that $d_1(x,y)leq d(x,y)$ for all $x,yin X$, since $1+d(x,y)geq 1$. Thus, it follows that $B_varepsilon^d(x)subseteq B_varepsilon^d_1(x)$ for all $xin X$ and $varepsilon>0$. Hence, if $Usubseteq X$ is $d_1$-open, it is also $d$-open.
Conversely, suppose that $U$ is $d$-open and let $xin U$. By definition of an open set in a metric space, there exists $varepsilon>0$ such that $B_varepsilon^d(x)subseteq U$. Now let $delta=minvarepsilon/2,1/2$. If $d_1(x,y)<delta$ then $d(x,y)<varepsilon$, because:
$$d(x,y)=fracd_1(x,y)1-d_1(x,y)leq 2d_1(x,y)< epsilon$$
Hence, $B_delta^d_1(x)subseteq B_varepsilon^d(x)subseteq U$, which implies that $U$ is $d_1$-open.
Thus, by the above, $(X,d)$ and $(X,d_1)$ are topologically equivalent.
Since you have shown that $(X,d)$ and $(X,d_2)$ are topologically equivalent, it follows from the fact that topological equivalence is a equivalence relation that $(X,d_1)$ and $(X,d_2)$ are topologically equivalent.
answered Jul 26 at 8:08
projectilemotion
11k61941
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Note that $d_1(x,y)leq d(x,y)$ for all $x,yin X$, since $1+d(x,y)geq 1$. Thus, it follows that $B_varepsilon^d(x)subseteq B_varepsilon^d_1(x)$ for all $xin X$ and $varepsilon>0$. Hence, if $Usubseteq X$ is $d_1$-open, it is also $d$-open.
Conversely, suppose that $U$ is $d$-open and let $xin U$. By definition of an open set in a metric space, there exists $varepsilon>0$ such that $B_varepsilon^d(x)subseteq U$. Now let $delta=minvarepsilon/2,1/2$. If $d_1(x,y)<delta$ then $d(x,y)<varepsilon$, because:
$$d(x,y)=fracd_1(x,y)1-d_1(x,y)leq 2d_1(x,y)< epsilon$$
Hence, $B_delta^d_1(x)subseteq B_varepsilon^d(x)subseteq U$, which implies that $U$ is $d_1$-open.
Thus, by the above, $(X,d)$ and $(X,d_1)$ are topologically equivalent.
Since you have shown that $(X,d)$ and $(X,d_2)$ are topologically equivalent, it follows from the fact that topological equivalence is a equivalence relation that $(X,d_1)$ and $(X,d_2)$ are topologically equivalent.
answered Jul 26 at 8:08
projectilemotion
11k61941
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up vote
0
down vote
up vote
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down vote
Note that $d_1(x,y)leq d(x,y)$ for all $x,yin X$, since $1+d(x,y)geq 1$. Thus, it follows that $B_varepsilon^d(x)subseteq B_varepsilon^d_1(x)$ for all $xin X$ and $varepsilon>0$. Hence, if $Usubseteq X$ is $d_1$-open, it is also $d$-open.
Conversely, suppose that $U$ is $d$-open and let $xin U$. By definition of an open set in a metric space, there exists $varepsilon>0$ such that $B_varepsilon^d(x)subseteq U$. Now let $delta=minvarepsilon/2,1/2$. If $d_1(x,y)<delta$ then $d(x,y)<varepsilon$, because:
$$d(x,y)=fracd_1(x,y)1-d_1(x,y)leq 2d_1(x,y)< epsilon$$
Hence, $B_delta^d_1(x)subseteq B_varepsilon^d(x)subseteq U$, which implies that $U$ is $d_1$-open.
Thus, by the above, $(X,d)$ and $(X,d_1)$ are topologically equivalent.
Since you have shown that $(X,d)$ and $(X,d_2)$ are topologically equivalent, it follows from the fact that topological equivalence is a equivalence relation that $(X,d_1)$ and $(X,d_2)$ are topologically equivalent.
answered Jul 26 at 8:08
projectilemotion
11k61941
Note that $d_1(x,y)leq d(x,y)$ for all $x,yin X$, since $1+d(x,y)geq 1$. Thus, it follows that $B_varepsilon^d(x)subseteq B_varepsilon^d_1(x)$ for all $xin X$ and $varepsilon>0$. Hence, if $Usubseteq X$ is $d_1$-open, it is also $d$-open.
Conversely, suppose that $U$ is $d$-open and let $xin U$. By definition of an open set in a metric space, there exists $varepsilon>0$ such that $B_varepsilon^d(x)subseteq U$. Now let $delta=minvarepsilon/2,1/2$. If $d_1(x,y)<delta$ then $d(x,y)<varepsilon$, because:
$$d(x,y)=fracd_1(x,y)1-d_1(x,y)leq 2d_1(x,y)< epsilon$$
Hence, $B_delta^d_1(x)subseteq B_varepsilon^d(x)subseteq U$, which implies that $U$ is $d_1$-open.
Thus, by the above, $(X,d)$ and $(X,d_1)$ are topologically equivalent.
Since you have shown that $(X,d)$ and $(X,d_2)$ are topologically equivalent, it follows from the fact that topological equivalence is a equivalence relation that $(X,d_1)$ and $(X,d_2)$ are topologically equivalent.
answered Jul 26 at 8:08
projectilemotion
11k61941
edited Jul 26 at 8:46
answered Jul 26 at 8:08
projectilemotion
11k61941
answered Jul 26 at 8:08
projectilemotion
11k61941
answered Jul 26 at 8:08
projectilemotion
11k61941
11k61941
add a comment |Â
add a comment |Â
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1
As an alternative, show that $$d/(1+d)leqslantmin1,dleqslant2d/(1+d)$$
– Did
Jul 26 at 7:01