Mixing
Equivalent metrics generate the same topology
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Let $X$ be a set. Two metrics $d_1, d_2: Xtimes XtomathbbR$ are equivalent, if constants $alpha,beta > 0$ exist such that for all $x,yin X$ holds:
$alpha d_1(x,y)leq d_2(x,y)leq beta d_1(x,y)$
Show, that equivalent metrics generate the same topology
Proof:
Let $d_1, d_2: Xtimes XtomathbbR$ be metrics and $tau_1, tau_2$ the induced topolgies.
We have to show, that $tau_1=tau_2$.
- $tau_1subseteqtau_2$.
Let $Uintau_1$ open. Then exists for every $xin U$ a $epsilon >0$ such that $B_d_1(x,epsilon)subseteq U$.
It is $B_d_1(x,epsilon)=yin X$.
Since $d_1$ and $d_2$ are equivalent, there are constants $alpha,beta >0$ such that $alpha d_1(x,y)leq d_2(x,y)leq beta d_1(x,y)$.
Take $epsilon':=beta^-1epsilon >0$.
We get $d_2(x,y)leq beta d_1(x,y)<betacdot beta^-1cdotepsilon=epsilon$.
Hence $Uintau_2checkmark$.
The other inclusion $tau_1supseteqtau_2$ works analogously.
Is this proof correct?
Thanks in advance.
general-topology proof-verification metric-spaces
asked Jul 15 at 1:38
Cornman
2,53921128
add a comment |Â
up vote
0
down vote
favorite
Let $X$ be a set. Two metrics $d_1, d_2: Xtimes XtomathbbR$ are equivalent, if constants $alpha,beta > 0$ exist such that for all $x,yin X$ holds:
$alpha d_1(x,y)leq d_2(x,y)leq beta d_1(x,y)$
Show, that equivalent metrics generate the same topology
Proof:
Let $d_1, d_2: Xtimes XtomathbbR$ be metrics and $tau_1, tau_2$ the induced topolgies.
We have to show, that $tau_1=tau_2$.
- $tau_1subseteqtau_2$.
Let $Uintau_1$ open. Then exists for every $xin U$ a $epsilon >0$ such that $B_d_1(x,epsilon)subseteq U$.
It is $B_d_1(x,epsilon)=yin X$.
Since $d_1$ and $d_2$ are equivalent, there are constants $alpha,beta >0$ such that $alpha d_1(x,y)leq d_2(x,y)leq beta d_1(x,y)$.
Take $epsilon':=beta^-1epsilon >0$.
We get $d_2(x,y)leq beta d_1(x,y)<betacdot beta^-1cdotepsilon=epsilon$.
Hence $Uintau_2checkmark$.
The other inclusion $tau_1supseteqtau_2$ works analogously.
Is this proof correct?
Thanks in advance.
general-topology proof-verification metric-spaces
asked Jul 15 at 1:38
Cornman
2,53921128
I'd say so. You basically need to show that open balls in one topology correspond to open balls in the other. If you can always fit an open ball in one topology into the other, you're done.
– Adrian Keister
Jul 15 at 2:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a set. Two metrics $d_1, d_2: Xtimes XtomathbbR$ are equivalent, if constants $alpha,beta > 0$ exist such that for all $x,yin X$ holds:
$alpha d_1(x,y)leq d_2(x,y)leq beta d_1(x,y)$
Show, that equivalent metrics generate the same topology
Proof:
Let $d_1, d_2: Xtimes XtomathbbR$ be metrics and $tau_1, tau_2$ the induced topolgies.
We have to show, that $tau_1=tau_2$.
- $tau_1subseteqtau_2$.
Let $Uintau_1$ open. Then exists for every $xin U$ a $epsilon >0$ such that $B_d_1(x,epsilon)subseteq U$.
It is $B_d_1(x,epsilon)=yin X$.
Since $d_1$ and $d_2$ are equivalent, there are constants $alpha,beta >0$ such that $alpha d_1(x,y)leq d_2(x,y)leq beta d_1(x,y)$.
Take $epsilon':=beta^-1epsilon >0$.
We get $d_2(x,y)leq beta d_1(x,y)<betacdot beta^-1cdotepsilon=epsilon$.
Hence $Uintau_2checkmark$.
The other inclusion $tau_1supseteqtau_2$ works analogously.
Is this proof correct?
Thanks in advance.
general-topology proof-verification metric-spaces
asked Jul 15 at 1:38
Cornman
2,53921128
Let $X$ be a set. Two metrics $d_1, d_2: Xtimes XtomathbbR$ are equivalent, if constants $alpha,beta > 0$ exist such that for all $x,yin X$ holds:
$alpha d_1(x,y)leq d_2(x,y)leq beta d_1(x,y)$
Show, that equivalent metrics generate the same topology
Proof:
Let $d_1, d_2: Xtimes XtomathbbR$ be metrics and $tau_1, tau_2$ the induced topolgies.
We have to show, that $tau_1=tau_2$.
- $tau_1subseteqtau_2$.
Let $Uintau_1$ open. Then exists for every $xin U$ a $epsilon >0$ such that $B_d_1(x,epsilon)subseteq U$.
It is $B_d_1(x,epsilon)=yin X$.
Since $d_1$ and $d_2$ are equivalent, there are constants $alpha,beta >0$ such that $alpha d_1(x,y)leq d_2(x,y)leq beta d_1(x,y)$.
Take $epsilon':=beta^-1epsilon >0$.
We get $d_2(x,y)leq beta d_1(x,y)<betacdot beta^-1cdotepsilon=epsilon$.
Hence $Uintau_2checkmark$.
The other inclusion $tau_1supseteqtau_2$ works analogously.
Is this proof correct?
Thanks in advance.
general-topology proof-verification metric-spaces
asked Jul 15 at 1:38
Cornman
2,53921128
asked Jul 15 at 1:38
Cornman
2,53921128
asked Jul 15 at 1:38
Cornman
2,53921128
asked Jul 15 at 1:38
Cornman
2,53921128
2,53921128
I'd say so. You basically need to show that open balls in one topology correspond to open balls in the other. If you can always fit an open ball in one topology into the other, you're done.
– Adrian Keister
Jul 15 at 2:01
add a comment |Â
I'd say so. You basically need to show that open balls in one topology correspond to open balls in the other. If you can always fit an open ball in one topology into the other, you're done.
– Adrian Keister
Jul 15 at 2:01
I'd say so. You basically need to show that open balls in one topology correspond to open balls in the other. If you can always fit an open ball in one topology into the other, you're done.
– Adrian Keister
Jul 15 at 2:01
I'd say so. You basically need to show that open balls in one topology correspond to open balls in the other. If you can always fit an open ball in one topology into the other, you're done.
– Adrian Keister
Jul 15 at 2:01
add a comment |Â
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Why does $d_2(x,y) < varepsilon$ imply $U in tau_2$? This does not follow, you're leaving out parts of the argument, and going the wrong way with inequalities:
So at the start you know that we have $alpha, beta>0$ such that for all $x,y$ we have $$alpha d_1(x,y) le d_2(x,y) le beta d_1(x,y)$$
It's OK to start with $U in tau_1$. We want to show $U in tau_2$, so let $x in U$. As $U in tau_1$ there exists an $varepsilon > 0$ such that $B_d_1(x,varepsilon) subseteq U$. Then I claim that
$$B_d_2(x,alpha varepsilon) subseteq B_d_1(x,varepsilon)$$
Suppose $y in B_d_2(x,alphavarepsilon)$, then $d_2(x,y) < alphavarepsilon$ and so $d_1(x,y) le frac1alpha d_2(x,y) < frac1alphaalphavarepsilon = varepsilon$ and so $y in B_d_1(x,varepsilon)$.
Then as $B_d_1(x,varepsilon) subseteq U$ we see that also $B_d_2(x,varepsilonalpha) subseteq U$ and thus $x$ is an interior point of $U$ wrt $d_2$. As $x in U$ was arbitrary, $U in tau_2$ and so $tau_1 subseteq tau_2$. As an exercise, I'd suggest to do the reverse inclusion in the same detail as I did above, not be lazy and say "it's analogous".
answered Jul 15 at 7:17
Henno Brandsma
91.6k342100
How do you get the inequality $d_1(x,y)leq alpha d_2(x,y)<alphacdotfracepsilonalpha$? Should you know take $B_d_2(x,alphaepsilon)$? Since you get $d_1(x,y)leq alpha^-1d_2(x,y)<alpha^-1alphaepsilon=epsilon$. If you divide $alpha$ you get $d_1(x,y)leqalpha^-1d_2(x,y)$. Thats why I am confused right now.
– Cornman
Jul 16 at 5:42
@Cornman Yes, I was mistaken; I edited. It should be OK now.
– Henno Brandsma
Jul 16 at 15:31
Thank you very much!
– Cornman
Jul 16 at 15:36
@Cornman you’re welcome.
– Henno Brandsma
Jul 16 at 16:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Why does $d_2(x,y) < varepsilon$ imply $U in tau_2$? This does not follow, you're leaving out parts of the argument, and going the wrong way with inequalities:
So at the start you know that we have $alpha, beta>0$ such that for all $x,y$ we have $$alpha d_1(x,y) le d_2(x,y) le beta d_1(x,y)$$
It's OK to start with $U in tau_1$. We want to show $U in tau_2$, so let $x in U$. As $U in tau_1$ there exists an $varepsilon > 0$ such that $B_d_1(x,varepsilon) subseteq U$. Then I claim that
$$B_d_2(x,alpha varepsilon) subseteq B_d_1(x,varepsilon)$$
Suppose $y in B_d_2(x,alphavarepsilon)$, then $d_2(x,y) < alphavarepsilon$ and so $d_1(x,y) le frac1alpha d_2(x,y) < frac1alphaalphavarepsilon = varepsilon$ and so $y in B_d_1(x,varepsilon)$.
Then as $B_d_1(x,varepsilon) subseteq U$ we see that also $B_d_2(x,varepsilonalpha) subseteq U$ and thus $x$ is an interior point of $U$ wrt $d_2$. As $x in U$ was arbitrary, $U in tau_2$ and so $tau_1 subseteq tau_2$. As an exercise, I'd suggest to do the reverse inclusion in the same detail as I did above, not be lazy and say "it's analogous".
answered Jul 15 at 7:17
Henno Brandsma
91.6k342100
How do you get the inequality $d_1(x,y)leq alpha d_2(x,y)<alphacdotfracepsilonalpha$? Should you know take $B_d_2(x,alphaepsilon)$? Since you get $d_1(x,y)leq alpha^-1d_2(x,y)<alpha^-1alphaepsilon=epsilon$. If you divide $alpha$ you get $d_1(x,y)leqalpha^-1d_2(x,y)$. Thats why I am confused right now.
– Cornman
Jul 16 at 5:42
@Cornman Yes, I was mistaken; I edited. It should be OK now.
– Henno Brandsma
Jul 16 at 15:31
Thank you very much!
– Cornman
Jul 16 at 15:36
@Cornman you’re welcome.
– Henno Brandsma
Jul 16 at 16:40
add a comment |Â
up vote
1
down vote
accepted
Why does $d_2(x,y) < varepsilon$ imply $U in tau_2$? This does not follow, you're leaving out parts of the argument, and going the wrong way with inequalities:
So at the start you know that we have $alpha, beta>0$ such that for all $x,y$ we have $$alpha d_1(x,y) le d_2(x,y) le beta d_1(x,y)$$
It's OK to start with $U in tau_1$. We want to show $U in tau_2$, so let $x in U$. As $U in tau_1$ there exists an $varepsilon > 0$ such that $B_d_1(x,varepsilon) subseteq U$. Then I claim that
$$B_d_2(x,alpha varepsilon) subseteq B_d_1(x,varepsilon)$$
Suppose $y in B_d_2(x,alphavarepsilon)$, then $d_2(x,y) < alphavarepsilon$ and so $d_1(x,y) le frac1alpha d_2(x,y) < frac1alphaalphavarepsilon = varepsilon$ and so $y in B_d_1(x,varepsilon)$.
Then as $B_d_1(x,varepsilon) subseteq U$ we see that also $B_d_2(x,varepsilonalpha) subseteq U$ and thus $x$ is an interior point of $U$ wrt $d_2$. As $x in U$ was arbitrary, $U in tau_2$ and so $tau_1 subseteq tau_2$. As an exercise, I'd suggest to do the reverse inclusion in the same detail as I did above, not be lazy and say "it's analogous".
answered Jul 15 at 7:17
Henno Brandsma
91.6k342100
How do you get the inequality $d_1(x,y)leq alpha d_2(x,y)<alphacdotfracepsilonalpha$? Should you know take $B_d_2(x,alphaepsilon)$? Since you get $d_1(x,y)leq alpha^-1d_2(x,y)<alpha^-1alphaepsilon=epsilon$. If you divide $alpha$ you get $d_1(x,y)leqalpha^-1d_2(x,y)$. Thats why I am confused right now.
– Cornman
Jul 16 at 5:42
@Cornman Yes, I was mistaken; I edited. It should be OK now.
– Henno Brandsma
Jul 16 at 15:31
Thank you very much!
– Cornman
Jul 16 at 15:36
@Cornman you’re welcome.
– Henno Brandsma
Jul 16 at 16:40
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Why does $d_2(x,y) < varepsilon$ imply $U in tau_2$? This does not follow, you're leaving out parts of the argument, and going the wrong way with inequalities:
So at the start you know that we have $alpha, beta>0$ such that for all $x,y$ we have $$alpha d_1(x,y) le d_2(x,y) le beta d_1(x,y)$$
It's OK to start with $U in tau_1$. We want to show $U in tau_2$, so let $x in U$. As $U in tau_1$ there exists an $varepsilon > 0$ such that $B_d_1(x,varepsilon) subseteq U$. Then I claim that
$$B_d_2(x,alpha varepsilon) subseteq B_d_1(x,varepsilon)$$
Suppose $y in B_d_2(x,alphavarepsilon)$, then $d_2(x,y) < alphavarepsilon$ and so $d_1(x,y) le frac1alpha d_2(x,y) < frac1alphaalphavarepsilon = varepsilon$ and so $y in B_d_1(x,varepsilon)$.
Then as $B_d_1(x,varepsilon) subseteq U$ we see that also $B_d_2(x,varepsilonalpha) subseteq U$ and thus $x$ is an interior point of $U$ wrt $d_2$. As $x in U$ was arbitrary, $U in tau_2$ and so $tau_1 subseteq tau_2$. As an exercise, I'd suggest to do the reverse inclusion in the same detail as I did above, not be lazy and say "it's analogous".
answered Jul 15 at 7:17
Henno Brandsma
91.6k342100
Why does $d_2(x,y) < varepsilon$ imply $U in tau_2$? This does not follow, you're leaving out parts of the argument, and going the wrong way with inequalities:
So at the start you know that we have $alpha, beta>0$ such that for all $x,y$ we have $$alpha d_1(x,y) le d_2(x,y) le beta d_1(x,y)$$
It's OK to start with $U in tau_1$. We want to show $U in tau_2$, so let $x in U$. As $U in tau_1$ there exists an $varepsilon > 0$ such that $B_d_1(x,varepsilon) subseteq U$. Then I claim that
$$B_d_2(x,alpha varepsilon) subseteq B_d_1(x,varepsilon)$$
Suppose $y in B_d_2(x,alphavarepsilon)$, then $d_2(x,y) < alphavarepsilon$ and so $d_1(x,y) le frac1alpha d_2(x,y) < frac1alphaalphavarepsilon = varepsilon$ and so $y in B_d_1(x,varepsilon)$.
Then as $B_d_1(x,varepsilon) subseteq U$ we see that also $B_d_2(x,varepsilonalpha) subseteq U$ and thus $x$ is an interior point of $U$ wrt $d_2$. As $x in U$ was arbitrary, $U in tau_2$ and so $tau_1 subseteq tau_2$. As an exercise, I'd suggest to do the reverse inclusion in the same detail as I did above, not be lazy and say "it's analogous".
answered Jul 15 at 7:17
Henno Brandsma
91.6k342100
edited Jul 16 at 15:30
answered Jul 15 at 7:17
Henno Brandsma
91.6k342100
answered Jul 15 at 7:17
Henno Brandsma
91.6k342100
answered Jul 15 at 7:17
Henno Brandsma
91.6k342100
91.6k342100
How do you get the inequality $d_1(x,y)leq alpha d_2(x,y)<alphacdotfracepsilonalpha$? Should you know take $B_d_2(x,alphaepsilon)$? Since you get $d_1(x,y)leq alpha^-1d_2(x,y)<alpha^-1alphaepsilon=epsilon$. If you divide $alpha$ you get $d_1(x,y)leqalpha^-1d_2(x,y)$. Thats why I am confused right now.
– Cornman
Jul 16 at 5:42
@Cornman Yes, I was mistaken; I edited. It should be OK now.
– Henno Brandsma
Jul 16 at 15:31
Thank you very much!
– Cornman
Jul 16 at 15:36
@Cornman you’re welcome.
– Henno Brandsma
Jul 16 at 16:40
add a comment |Â
How do you get the inequality $d_1(x,y)leq alpha d_2(x,y)<alphacdotfracepsilonalpha$? Should you know take $B_d_2(x,alphaepsilon)$? Since you get $d_1(x,y)leq alpha^-1d_2(x,y)<alpha^-1alphaepsilon=epsilon$. If you divide $alpha$ you get $d_1(x,y)leqalpha^-1d_2(x,y)$. Thats why I am confused right now.
– Cornman
Jul 16 at 5:42
@Cornman Yes, I was mistaken; I edited. It should be OK now.
– Henno Brandsma
Jul 16 at 15:31
Thank you very much!
– Cornman
Jul 16 at 15:36
@Cornman you’re welcome.
– Henno Brandsma
Jul 16 at 16:40
How do you get the inequality $d_1(x,y)leq alpha d_2(x,y)<alphacdotfracepsilonalpha$? Should you know take $B_d_2(x,alphaepsilon)$? Since you get $d_1(x,y)leq alpha^-1d_2(x,y)<alpha^-1alphaepsilon=epsilon$. If you divide $alpha$ you get $d_1(x,y)leqalpha^-1d_2(x,y)$. Thats why I am confused right now.
– Cornman
Jul 16 at 5:42
How do you get the inequality $d_1(x,y)leq alpha d_2(x,y)<alphacdotfracepsilonalpha$? Should you know take $B_d_2(x,alphaepsilon)$? Since you get $d_1(x,y)leq alpha^-1d_2(x,y)<alpha^-1alphaepsilon=epsilon$. If you divide $alpha$ you get $d_1(x,y)leqalpha^-1d_2(x,y)$. Thats why I am confused right now.
– Cornman
Jul 16 at 5:42
@Cornman Yes, I was mistaken; I edited. It should be OK now.
– Henno Brandsma
Jul 16 at 15:31
@Cornman Yes, I was mistaken; I edited. It should be OK now.
– Henno Brandsma
Jul 16 at 15:31
Thank you very much!
– Cornman
Jul 16 at 15:36
Thank you very much!
– Cornman
Jul 16 at 15:36
@Cornman you’re welcome.
– Henno Brandsma
Jul 16 at 16:40
@Cornman you’re welcome.
– Henno Brandsma
Jul 16 at 16:40
add a comment |Â
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I'd say so. You basically need to show that open balls in one topology correspond to open balls in the other. If you can always fit an open ball in one topology into the other, you're done.
– Adrian Keister
Jul 15 at 2:01