Proof verification: $d$ and $tilde{d}$ are topologically equivalent
$begingroup$
Given $X = (0,1]$ a metric space with $tilde{d}$ defined as $$tilde{d}(x,y) = bigl|frac{1}{x} - frac{1}{y}bigr| text{for} x,y in X$$
I'm trying to prove that $tilde{d}$ and $d$ the standard metric on $(0,1]$ are topologically equivalent, using the open sets criteria and would like to know if it's correct.
Proof:
$U$ is open in $(X,tilde{d}) implies U$ is open in $(X,d)$
Let $x in U$ and $epsilon > 0$ such that $B_{epsilon}(x,tilde{d}) subseteq U$
If $y in B_{epsilon}(x,tilde{d})$ which means $ tilde{d}(x,y) =bigl|frac{1}{x} - frac{1}{y}bigr| = bigl|frac{y-x}{xy}bigr| = frac{|x-y|}{|xy|} < epsilon$ and that's equivalent to $d(x,y) = |x-y| < |xy|epsilon$, then $y in U$
Therefore, for any $x in U$ we can find an epsilon $tilde{epsilon} = |xy|epsilon$ such that: $y in B_{|xy|epsilon}(x, d) implies y in U$
$Longleftarrow$
Similarly, If $y in B_{epsilon}(x,d)$, it follows that $d(x,y) = |x-y| < epsilon$ which is equivalent to $tilde{d}(x,y) =bigl|frac{1}{x} - frac{1}{y}bigr| < frac{epsilon}{|xy|}$
Therefore, for any $x in U$ we can find an epsilon $tilde{epsilon} = frac{epsilon}{|xy|}$ such that: $y in B_{frac{epsilon}{|xy|}}(x, tilde{d}) implies y in U$
real-analysis general-topology proof-verification
edited Dec 25 '18 at 22:10
Math1000
19.3k31745
asked Dec 25 '18 at 20:16
mikemike
564
$endgroup$
add a comment |
$begingroup$
Given $X = (0,1]$ a metric space with $tilde{d}$ defined as $$tilde{d}(x,y) = bigl|frac{1}{x} - frac{1}{y}bigr| text{for} x,y in X$$
I'm trying to prove that $tilde{d}$ and $d$ the standard metric on $(0,1]$ are topologically equivalent, using the open sets criteria and would like to know if it's correct.
Proof:
$U$ is open in $(X,tilde{d}) implies U$ is open in $(X,d)$
Let $x in U$ and $epsilon > 0$ such that $B_{epsilon}(x,tilde{d}) subseteq U$
If $y in B_{epsilon}(x,tilde{d})$ which means $ tilde{d}(x,y) =bigl|frac{1}{x} - frac{1}{y}bigr| = bigl|frac{y-x}{xy}bigr| = frac{|x-y|}{|xy|} < epsilon$ and that's equivalent to $d(x,y) = |x-y| < |xy|epsilon$, then $y in U$
Therefore, for any $x in U$ we can find an epsilon $tilde{epsilon} = |xy|epsilon$ such that: $y in B_{|xy|epsilon}(x, d) implies y in U$
$Longleftarrow$
Similarly, If $y in B_{epsilon}(x,d)$, it follows that $d(x,y) = |x-y| < epsilon$ which is equivalent to $tilde{d}(x,y) =bigl|frac{1}{x} - frac{1}{y}bigr| < frac{epsilon}{|xy|}$
Therefore, for any $x in U$ we can find an epsilon $tilde{epsilon} = frac{epsilon}{|xy|}$ such that: $y in B_{frac{epsilon}{|xy|}}(x, tilde{d}) implies y in U$
real-analysis general-topology proof-verification
edited Dec 25 '18 at 22:10
Math1000
19.3k31745
asked Dec 25 '18 at 20:16
mikemike
564
$endgroup$
$begingroup$
There is a mistake right here $d(x,y) = |x-y| < |xy|epsilon$, then $tilde{epsilon} = |xy|epsilon$. You have to gain an $tilde{epsilon}$ such that it does not depend on $y$.
$endgroup$
– S.S.Danyal
Dec 25 '18 at 23:47
$begingroup$
Yes, that's true. "$y in B_{|xy|epsilon}(x, d)$" doesn't make sense at all
$endgroup$
– mike
Dec 25 '18 at 23:52
add a comment |
$begingroup$
Given $X = (0,1]$ a metric space with $tilde{d}$ defined as $$tilde{d}(x,y) = bigl|frac{1}{x} - frac{1}{y}bigr| text{for} x,y in X$$
I'm trying to prove that $tilde{d}$ and $d$ the standard metric on $(0,1]$ are topologically equivalent, using the open sets criteria and would like to know if it's correct.
Proof:
$U$ is open in $(X,tilde{d}) implies U$ is open in $(X,d)$
Let $x in U$ and $epsilon > 0$ such that $B_{epsilon}(x,tilde{d}) subseteq U$
If $y in B_{epsilon}(x,tilde{d})$ which means $ tilde{d}(x,y) =bigl|frac{1}{x} - frac{1}{y}bigr| = bigl|frac{y-x}{xy}bigr| = frac{|x-y|}{|xy|} < epsilon$ and that's equivalent to $d(x,y) = |x-y| < |xy|epsilon$, then $y in U$
Therefore, for any $x in U$ we can find an epsilon $tilde{epsilon} = |xy|epsilon$ such that: $y in B_{|xy|epsilon}(x, d) implies y in U$
$Longleftarrow$
Similarly, If $y in B_{epsilon}(x,d)$, it follows that $d(x,y) = |x-y| < epsilon$ which is equivalent to $tilde{d}(x,y) =bigl|frac{1}{x} - frac{1}{y}bigr| < frac{epsilon}{|xy|}$
Therefore, for any $x in U$ we can find an epsilon $tilde{epsilon} = frac{epsilon}{|xy|}$ such that: $y in B_{frac{epsilon}{|xy|}}(x, tilde{d}) implies y in U$
real-analysis general-topology proof-verification
edited Dec 25 '18 at 22:10
Math1000
19.3k31745
asked Dec 25 '18 at 20:16
mikemike
564
$endgroup$
Given $X = (0,1]$ a metric space with $tilde{d}$ defined as $$tilde{d}(x,y) = bigl|frac{1}{x} - frac{1}{y}bigr| text{for} x,y in X$$
I'm trying to prove that $tilde{d}$ and $d$ the standard metric on $(0,1]$ are topologically equivalent, using the open sets criteria and would like to know if it's correct.
Proof:
$U$ is open in $(X,tilde{d}) implies U$ is open in $(X,d)$
Let $x in U$ and $epsilon > 0$ such that $B_{epsilon}(x,tilde{d}) subseteq U$
If $y in B_{epsilon}(x,tilde{d})$ which means $ tilde{d}(x,y) =bigl|frac{1}{x} - frac{1}{y}bigr| = bigl|frac{y-x}{xy}bigr| = frac{|x-y|}{|xy|} < epsilon$ and that's equivalent to $d(x,y) = |x-y| < |xy|epsilon$, then $y in U$
Therefore, for any $x in U$ we can find an epsilon $tilde{epsilon} = |xy|epsilon$ such that: $y in B_{|xy|epsilon}(x, d) implies y in U$
$Longleftarrow$
Similarly, If $y in B_{epsilon}(x,d)$, it follows that $d(x,y) = |x-y| < epsilon$ which is equivalent to $tilde{d}(x,y) =bigl|frac{1}{x} - frac{1}{y}bigr| < frac{epsilon}{|xy|}$
Therefore, for any $x in U$ we can find an epsilon $tilde{epsilon} = frac{epsilon}{|xy|}$ such that: $y in B_{frac{epsilon}{|xy|}}(x, tilde{d}) implies y in U$
real-analysis general-topology proof-verification
real-analysis general-topology proof-verification
edited Dec 25 '18 at 22:10
Math1000
19.3k31745
asked Dec 25 '18 at 20:16
mikemike
564
edited Dec 25 '18 at 22:10
Math1000
19.3k31745
asked Dec 25 '18 at 20:16
mikemike
564
edited Dec 25 '18 at 22:10
Math1000
19.3k31745
edited Dec 25 '18 at 22:10
Math1000
19.3k31745
edited Dec 25 '18 at 22:10
Math1000
19.3k31745
19.3k31745
asked Dec 25 '18 at 20:16
mikemike
564
asked Dec 25 '18 at 20:16
mikemike
564
asked Dec 25 '18 at 20:16
mikemike
564
564
$begingroup$
There is a mistake right here $d(x,y) = |x-y| < |xy|epsilon$, then $tilde{epsilon} = |xy|epsilon$. You have to gain an $tilde{epsilon}$ such that it does not depend on $y$.
$endgroup$
– S.S.Danyal
Dec 25 '18 at 23:47
$begingroup$
Yes, that's true. "$y in B_{|xy|epsilon}(x, d)$" doesn't make sense at all
$endgroup$
– mike
Dec 25 '18 at 23:52
add a comment |
$begingroup$
There is a mistake right here $d(x,y) = |x-y| < |xy|epsilon$, then $tilde{epsilon} = |xy|epsilon$. You have to gain an $tilde{epsilon}$ such that it does not depend on $y$.
$endgroup$
– S.S.Danyal
Dec 25 '18 at 23:47
$begingroup$
Yes, that's true. "$y in B_{|xy|epsilon}(x, d)$" doesn't make sense at all
$endgroup$
– mike
Dec 25 '18 at 23:52
$begingroup$
There is a mistake right here $d(x,y) = |x-y| < |xy|epsilon$, then $tilde{epsilon} = |xy|epsilon$. You have to gain an $tilde{epsilon}$ such that it does not depend on $y$.
$endgroup$
– S.S.Danyal
Dec 25 '18 at 23:47
$begingroup$
There is a mistake right here $d(x,y) = |x-y| < |xy|epsilon$, then $tilde{epsilon} = |xy|epsilon$. You have to gain an $tilde{epsilon}$ such that it does not depend on $y$.
$endgroup$
– S.S.Danyal
Dec 25 '18 at 23:47
$begingroup$
Yes, that's true. "$y in B_{|xy|epsilon}(x, d)$" doesn't make sense at all
$endgroup$
– mike
Dec 25 '18 at 23:52
$begingroup$
Yes, that's true. "$y in B_{|xy|epsilon}(x, d)$" doesn't make sense at all
$endgroup$
– mike
Dec 25 '18 at 23:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your argument is not valid. You cannot make $tilde {epsilon}$ depend on $y$. Suppose $B_{epsilon} (x,tilde {d}) subset U$ with $0<epsilon <1$. Choose $ tilde {epsilon} >0$ such that $tilde {epsilon} < min {frac x 2,frac {epsilon x^{2}}2}$. (Note that $x >0$. This is crucial here). Then $|y-x| < tilde {epsilon}$ implies $|frac 1 y -frac 1 x|=frac {|y-x|} {xy} <epsilon$ because $epsilon xy >epsilon x(x-tilde {epsilon})=epsilon x^{2}-epsilon x{tilde {epsilon}} >2tilde {epsilon}-epsilon x{tilde {epsilon}} >{tilde {epsilon}}$ (since $epsilon x <1$). Hence $B_{tilde {epsilon}} (x,d) subset B_{epsilon} (x,tilde {d}) subset U$. I leave the other part to you.
answered Dec 25 '18 at 23:52
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
$begingroup$
Thank you. I am still trying to work out your answer. If you don't mind, could you please explain more how you picked such $tilde {epsilon}$?
$endgroup$
– mike
Dec 26 '18 at 0:12
add a comment |
$begingroup$
There is a fault in your proof. Kindly see the proof below for the other side
begin{align}bar{d}(x,y)<r_2 &iff left| dfrac{1}{x}-dfrac{1}{y} right| <r_2 iff dfrac{1}{x}-r_2<dfrac{1}{y}<dfrac{1}{x}+r_2\&
iff dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}
end{align}
Choose $r_2<dfrac{1}{x},$ then
begin{align}
dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}&iff dfrac{x }{1+x r_2}<y<dfrac{x }{1-xr_2}\& iff - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}.
end{align}
As $r_2to 0,$ then $dfrac{x^2 r_2 }{1+x r_2}to 0$ and $dfrac{x^2 r_2 }{1-xr_2}to 0.$ Hence, $forall , r>0,$ and $forall ,xin (0,1],$ there exists $r_2in (0,1]$ such that $r_2<(r/2x^2).$ So,
begin{align}
y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Also, begin{align}
r_2<dfrac{r}{2x^2}iff -r_2>-dfrac{r}{2x^2}iff -x^2 r_2>-dfrac{r}{2}.
end{align}
But, $r_2in (0,1]implies r_2>-dfrac{1}{2x},$ implies $x r_2>-dfrac{1}{2} iff 1+x r_2>dfrac{1}{2}iff dfrac{1}{1+x r_2}<2.$ Hence,
begin{align}
y-x> - dfrac{x^2 r_2 }{1+x r_2}>-dfrac{r}{2(1+x r_2)} > -dfrac{2r}{2}=-r
end{align}
Hence, we have that
begin{align}
-r=-dfrac{2r}{2}<-dfrac{r}{2(1+x r_2)}< - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Therefore, begin{align}
|x-y|<r iff d(x,y)<r
end{align}
And you are done. Kindly get back if you have some questions. Remember that they don't necessarily need to have the same radius.
answered Dec 26 '18 at 1:29
Omojola MichealOmojola Micheal
1,969324
$endgroup$
$begingroup$
Thank you so much for your detailed answer! :) It's super helpful. I just got abit stuck in the part you applied the Archimedean principle. How does it relate to $dfrac{x^2 r_2 }{1+x r_2}$ and $dfrac{x^2 r_2 }{1-xr_2}$ tending to zero, as $r_2to 0$?
$endgroup$
– mike
Dec 26 '18 at 23:51
$begingroup$
@mike: You can either use the fact that $r_2to 0$ or Archimedean principle to establish your claim. Anyway, I'll edit for clarity.
$endgroup$
– Omojola Micheal
Dec 27 '18 at 7:55
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Your argument is not valid. You cannot make $tilde {epsilon}$ depend on $y$. Suppose $B_{epsilon} (x,tilde {d}) subset U$ with $0<epsilon <1$. Choose $ tilde {epsilon} >0$ such that $tilde {epsilon} < min {frac x 2,frac {epsilon x^{2}}2}$. (Note that $x >0$. This is crucial here). Then $|y-x| < tilde {epsilon}$ implies $|frac 1 y -frac 1 x|=frac {|y-x|} {xy} <epsilon$ because $epsilon xy >epsilon x(x-tilde {epsilon})=epsilon x^{2}-epsilon x{tilde {epsilon}} >2tilde {epsilon}-epsilon x{tilde {epsilon}} >{tilde {epsilon}}$ (since $epsilon x <1$). Hence $B_{tilde {epsilon}} (x,d) subset B_{epsilon} (x,tilde {d}) subset U$. I leave the other part to you.
answered Dec 25 '18 at 23:52
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
$begingroup$
Thank you. I am still trying to work out your answer. If you don't mind, could you please explain more how you picked such $tilde {epsilon}$?
$endgroup$
– mike
Dec 26 '18 at 0:12
add a comment |
$begingroup$
Your argument is not valid. You cannot make $tilde {epsilon}$ depend on $y$. Suppose $B_{epsilon} (x,tilde {d}) subset U$ with $0<epsilon <1$. Choose $ tilde {epsilon} >0$ such that $tilde {epsilon} < min {frac x 2,frac {epsilon x^{2}}2}$. (Note that $x >0$. This is crucial here). Then $|y-x| < tilde {epsilon}$ implies $|frac 1 y -frac 1 x|=frac {|y-x|} {xy} <epsilon$ because $epsilon xy >epsilon x(x-tilde {epsilon})=epsilon x^{2}-epsilon x{tilde {epsilon}} >2tilde {epsilon}-epsilon x{tilde {epsilon}} >{tilde {epsilon}}$ (since $epsilon x <1$). Hence $B_{tilde {epsilon}} (x,d) subset B_{epsilon} (x,tilde {d}) subset U$. I leave the other part to you.
answered Dec 25 '18 at 23:52
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
$begingroup$
Thank you. I am still trying to work out your answer. If you don't mind, could you please explain more how you picked such $tilde {epsilon}$?
$endgroup$
– mike
Dec 26 '18 at 0:12
add a comment |
$begingroup$
Your argument is not valid. You cannot make $tilde {epsilon}$ depend on $y$. Suppose $B_{epsilon} (x,tilde {d}) subset U$ with $0<epsilon <1$. Choose $ tilde {epsilon} >0$ such that $tilde {epsilon} < min {frac x 2,frac {epsilon x^{2}}2}$. (Note that $x >0$. This is crucial here). Then $|y-x| < tilde {epsilon}$ implies $|frac 1 y -frac 1 x|=frac {|y-x|} {xy} <epsilon$ because $epsilon xy >epsilon x(x-tilde {epsilon})=epsilon x^{2}-epsilon x{tilde {epsilon}} >2tilde {epsilon}-epsilon x{tilde {epsilon}} >{tilde {epsilon}}$ (since $epsilon x <1$). Hence $B_{tilde {epsilon}} (x,d) subset B_{epsilon} (x,tilde {d}) subset U$. I leave the other part to you.
answered Dec 25 '18 at 23:52
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
Your argument is not valid. You cannot make $tilde {epsilon}$ depend on $y$. Suppose $B_{epsilon} (x,tilde {d}) subset U$ with $0<epsilon <1$. Choose $ tilde {epsilon} >0$ such that $tilde {epsilon} < min {frac x 2,frac {epsilon x^{2}}2}$. (Note that $x >0$. This is crucial here). Then $|y-x| < tilde {epsilon}$ implies $|frac 1 y -frac 1 x|=frac {|y-x|} {xy} <epsilon$ because $epsilon xy >epsilon x(x-tilde {epsilon})=epsilon x^{2}-epsilon x{tilde {epsilon}} >2tilde {epsilon}-epsilon x{tilde {epsilon}} >{tilde {epsilon}}$ (since $epsilon x <1$). Hence $B_{tilde {epsilon}} (x,d) subset B_{epsilon} (x,tilde {d}) subset U$. I leave the other part to you.
answered Dec 25 '18 at 23:52
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered Dec 25 '18 at 23:52
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered Dec 25 '18 at 23:52
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered Dec 25 '18 at 23:52
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
66.3k42867
$begingroup$
Thank you. I am still trying to work out your answer. If you don't mind, could you please explain more how you picked such $tilde {epsilon}$?
$endgroup$
– mike
Dec 26 '18 at 0:12
add a comment |
$begingroup$
Thank you. I am still trying to work out your answer. If you don't mind, could you please explain more how you picked such $tilde {epsilon}$?
$endgroup$
– mike
Dec 26 '18 at 0:12
$begingroup$
Thank you. I am still trying to work out your answer. If you don't mind, could you please explain more how you picked such $tilde {epsilon}$?
$endgroup$
– mike
Dec 26 '18 at 0:12
$begingroup$
Thank you. I am still trying to work out your answer. If you don't mind, could you please explain more how you picked such $tilde {epsilon}$?
$endgroup$
– mike
Dec 26 '18 at 0:12
add a comment |
$begingroup$
There is a fault in your proof. Kindly see the proof below for the other side
begin{align}bar{d}(x,y)<r_2 &iff left| dfrac{1}{x}-dfrac{1}{y} right| <r_2 iff dfrac{1}{x}-r_2<dfrac{1}{y}<dfrac{1}{x}+r_2\&
iff dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}
end{align}
Choose $r_2<dfrac{1}{x},$ then
begin{align}
dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}&iff dfrac{x }{1+x r_2}<y<dfrac{x }{1-xr_2}\& iff - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}.
end{align}
As $r_2to 0,$ then $dfrac{x^2 r_2 }{1+x r_2}to 0$ and $dfrac{x^2 r_2 }{1-xr_2}to 0.$ Hence, $forall , r>0,$ and $forall ,xin (0,1],$ there exists $r_2in (0,1]$ such that $r_2<(r/2x^2).$ So,
begin{align}
y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Also, begin{align}
r_2<dfrac{r}{2x^2}iff -r_2>-dfrac{r}{2x^2}iff -x^2 r_2>-dfrac{r}{2}.
end{align}
But, $r_2in (0,1]implies r_2>-dfrac{1}{2x},$ implies $x r_2>-dfrac{1}{2} iff 1+x r_2>dfrac{1}{2}iff dfrac{1}{1+x r_2}<2.$ Hence,
begin{align}
y-x> - dfrac{x^2 r_2 }{1+x r_2}>-dfrac{r}{2(1+x r_2)} > -dfrac{2r}{2}=-r
end{align}
Hence, we have that
begin{align}
-r=-dfrac{2r}{2}<-dfrac{r}{2(1+x r_2)}< - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Therefore, begin{align}
|x-y|<r iff d(x,y)<r
end{align}
And you are done. Kindly get back if you have some questions. Remember that they don't necessarily need to have the same radius.
answered Dec 26 '18 at 1:29
Omojola MichealOmojola Micheal
1,969324
$endgroup$
$begingroup$
Thank you so much for your detailed answer! :) It's super helpful. I just got abit stuck in the part you applied the Archimedean principle. How does it relate to $dfrac{x^2 r_2 }{1+x r_2}$ and $dfrac{x^2 r_2 }{1-xr_2}$ tending to zero, as $r_2to 0$?
$endgroup$
– mike
Dec 26 '18 at 23:51
$begingroup$
@mike: You can either use the fact that $r_2to 0$ or Archimedean principle to establish your claim. Anyway, I'll edit for clarity.
$endgroup$
– Omojola Micheal
Dec 27 '18 at 7:55
add a comment |
$begingroup$
There is a fault in your proof. Kindly see the proof below for the other side
begin{align}bar{d}(x,y)<r_2 &iff left| dfrac{1}{x}-dfrac{1}{y} right| <r_2 iff dfrac{1}{x}-r_2<dfrac{1}{y}<dfrac{1}{x}+r_2\&
iff dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}
end{align}
Choose $r_2<dfrac{1}{x},$ then
begin{align}
dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}&iff dfrac{x }{1+x r_2}<y<dfrac{x }{1-xr_2}\& iff - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}.
end{align}
As $r_2to 0,$ then $dfrac{x^2 r_2 }{1+x r_2}to 0$ and $dfrac{x^2 r_2 }{1-xr_2}to 0.$ Hence, $forall , r>0,$ and $forall ,xin (0,1],$ there exists $r_2in (0,1]$ such that $r_2<(r/2x^2).$ So,
begin{align}
y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Also, begin{align}
r_2<dfrac{r}{2x^2}iff -r_2>-dfrac{r}{2x^2}iff -x^2 r_2>-dfrac{r}{2}.
end{align}
But, $r_2in (0,1]implies r_2>-dfrac{1}{2x},$ implies $x r_2>-dfrac{1}{2} iff 1+x r_2>dfrac{1}{2}iff dfrac{1}{1+x r_2}<2.$ Hence,
begin{align}
y-x> - dfrac{x^2 r_2 }{1+x r_2}>-dfrac{r}{2(1+x r_2)} > -dfrac{2r}{2}=-r
end{align}
Hence, we have that
begin{align}
-r=-dfrac{2r}{2}<-dfrac{r}{2(1+x r_2)}< - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Therefore, begin{align}
|x-y|<r iff d(x,y)<r
end{align}
And you are done. Kindly get back if you have some questions. Remember that they don't necessarily need to have the same radius.
answered Dec 26 '18 at 1:29
Omojola MichealOmojola Micheal
1,969324
$endgroup$
$begingroup$
Thank you so much for your detailed answer! :) It's super helpful. I just got abit stuck in the part you applied the Archimedean principle. How does it relate to $dfrac{x^2 r_2 }{1+x r_2}$ and $dfrac{x^2 r_2 }{1-xr_2}$ tending to zero, as $r_2to 0$?
$endgroup$
– mike
Dec 26 '18 at 23:51
$begingroup$
@mike: You can either use the fact that $r_2to 0$ or Archimedean principle to establish your claim. Anyway, I'll edit for clarity.
$endgroup$
– Omojola Micheal
Dec 27 '18 at 7:55
add a comment |
$begingroup$
There is a fault in your proof. Kindly see the proof below for the other side
begin{align}bar{d}(x,y)<r_2 &iff left| dfrac{1}{x}-dfrac{1}{y} right| <r_2 iff dfrac{1}{x}-r_2<dfrac{1}{y}<dfrac{1}{x}+r_2\&
iff dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}
end{align}
Choose $r_2<dfrac{1}{x},$ then
begin{align}
dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}&iff dfrac{x }{1+x r_2}<y<dfrac{x }{1-xr_2}\& iff - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}.
end{align}
As $r_2to 0,$ then $dfrac{x^2 r_2 }{1+x r_2}to 0$ and $dfrac{x^2 r_2 }{1-xr_2}to 0.$ Hence, $forall , r>0,$ and $forall ,xin (0,1],$ there exists $r_2in (0,1]$ such that $r_2<(r/2x^2).$ So,
begin{align}
y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Also, begin{align}
r_2<dfrac{r}{2x^2}iff -r_2>-dfrac{r}{2x^2}iff -x^2 r_2>-dfrac{r}{2}.
end{align}
But, $r_2in (0,1]implies r_2>-dfrac{1}{2x},$ implies $x r_2>-dfrac{1}{2} iff 1+x r_2>dfrac{1}{2}iff dfrac{1}{1+x r_2}<2.$ Hence,
begin{align}
y-x> - dfrac{x^2 r_2 }{1+x r_2}>-dfrac{r}{2(1+x r_2)} > -dfrac{2r}{2}=-r
end{align}
Hence, we have that
begin{align}
-r=-dfrac{2r}{2}<-dfrac{r}{2(1+x r_2)}< - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Therefore, begin{align}
|x-y|<r iff d(x,y)<r
end{align}
And you are done. Kindly get back if you have some questions. Remember that they don't necessarily need to have the same radius.
answered Dec 26 '18 at 1:29
Omojola MichealOmojola Micheal
1,969324
$endgroup$
There is a fault in your proof. Kindly see the proof below for the other side
begin{align}bar{d}(x,y)<r_2 &iff left| dfrac{1}{x}-dfrac{1}{y} right| <r_2 iff dfrac{1}{x}-r_2<dfrac{1}{y}<dfrac{1}{x}+r_2\&
iff dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}
end{align}
Choose $r_2<dfrac{1}{x},$ then
begin{align}
dfrac{1-x r_2 }{x}<dfrac{1}{y}<dfrac{1+xr_2 }{x}&iff dfrac{x }{1+x r_2}<y<dfrac{x }{1-xr_2}\& iff - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}.
end{align}
As $r_2to 0,$ then $dfrac{x^2 r_2 }{1+x r_2}to 0$ and $dfrac{x^2 r_2 }{1-xr_2}to 0.$ Hence, $forall , r>0,$ and $forall ,xin (0,1],$ there exists $r_2in (0,1]$ such that $r_2<(r/2x^2).$ So,
begin{align}
y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Also, begin{align}
r_2<dfrac{r}{2x^2}iff -r_2>-dfrac{r}{2x^2}iff -x^2 r_2>-dfrac{r}{2}.
end{align}
But, $r_2in (0,1]implies r_2>-dfrac{1}{2x},$ implies $x r_2>-dfrac{1}{2} iff 1+x r_2>dfrac{1}{2}iff dfrac{1}{1+x r_2}<2.$ Hence,
begin{align}
y-x> - dfrac{x^2 r_2 }{1+x r_2}>-dfrac{r}{2(1+x r_2)} > -dfrac{2r}{2}=-r
end{align}
Hence, we have that
begin{align}
-r=-dfrac{2r}{2}<-dfrac{r}{2(1+x r_2)}< - dfrac{x^2 r_2 }{1+x r_2}<y-x<dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <dfrac{r }{2}<r.
end{align}
Therefore, begin{align}
|x-y|<r iff d(x,y)<r
end{align}
And you are done. Kindly get back if you have some questions. Remember that they don't necessarily need to have the same radius.
answered Dec 26 '18 at 1:29
Omojola MichealOmojola Micheal
1,969324
edited Dec 27 '18 at 7:55
answered Dec 26 '18 at 1:29
Omojola MichealOmojola Micheal
1,969324
answered Dec 26 '18 at 1:29
Omojola MichealOmojola Micheal
1,969324
answered Dec 26 '18 at 1:29
Omojola MichealOmojola Micheal
1,969324
1,969324
$begingroup$
Thank you so much for your detailed answer! :) It's super helpful. I just got abit stuck in the part you applied the Archimedean principle. How does it relate to $dfrac{x^2 r_2 }{1+x r_2}$ and $dfrac{x^2 r_2 }{1-xr_2}$ tending to zero, as $r_2to 0$?
$endgroup$
– mike
Dec 26 '18 at 23:51
$begingroup$
@mike: You can either use the fact that $r_2to 0$ or Archimedean principle to establish your claim. Anyway, I'll edit for clarity.
$endgroup$
– Omojola Micheal
Dec 27 '18 at 7:55
add a comment |
$begingroup$
Thank you so much for your detailed answer! :) It's super helpful. I just got abit stuck in the part you applied the Archimedean principle. How does it relate to $dfrac{x^2 r_2 }{1+x r_2}$ and $dfrac{x^2 r_2 }{1-xr_2}$ tending to zero, as $r_2to 0$?
$endgroup$
– mike
Dec 26 '18 at 23:51
$begingroup$
@mike: You can either use the fact that $r_2to 0$ or Archimedean principle to establish your claim. Anyway, I'll edit for clarity.
$endgroup$
– Omojola Micheal
Dec 27 '18 at 7:55
$begingroup$
Thank you so much for your detailed answer! :) It's super helpful. I just got abit stuck in the part you applied the Archimedean principle. How does it relate to $dfrac{x^2 r_2 }{1+x r_2}$ and $dfrac{x^2 r_2 }{1-xr_2}$ tending to zero, as $r_2to 0$?
$endgroup$
– mike
Dec 26 '18 at 23:51
$begingroup$
Thank you so much for your detailed answer! :) It's super helpful. I just got abit stuck in the part you applied the Archimedean principle. How does it relate to $dfrac{x^2 r_2 }{1+x r_2}$ and $dfrac{x^2 r_2 }{1-xr_2}$ tending to zero, as $r_2to 0$?
$endgroup$
– mike
Dec 26 '18 at 23:51
$begingroup$
@mike: You can either use the fact that $r_2to 0$ or Archimedean principle to establish your claim. Anyway, I'll edit for clarity.
$endgroup$
– Omojola Micheal
Dec 27 '18 at 7:55
$begingroup$
@mike: You can either use the fact that $r_2to 0$ or Archimedean principle to establish your claim. Anyway, I'll edit for clarity.
$endgroup$
– Omojola Micheal
Dec 27 '18 at 7:55
add a comment |
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There is a mistake right here $d(x,y) = |x-y| < |xy|epsilon$, then $tilde{epsilon} = |xy|epsilon$. You have to gain an $tilde{epsilon}$ such that it does not depend on $y$.
$endgroup$
– S.S.Danyal
Dec 25 '18 at 23:47
$begingroup$
Yes, that's true. "$y in B_{|xy|epsilon}(x, d)$" doesn't make sense at all
$endgroup$
– mike
Dec 25 '18 at 23:52