Mixing
Two neighbourhoods $V_delta(a)$ and $V_epsilon(b)$ are disjoint iff $epsilon+deltale|b-a|$
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I'm beginning studying real analysis, and this came up in the book, without proof:
($V_epsilon(a)$ denotes the $epsilon$-neighbourhood of $a$)
For all $a,b in mathbbR$; $delta,epsilon >0$ :
$epsilon+delta le lvert b-a rvert iff V_delta(a) cap V_epsilon(b) = emptyset$
I easily proved that the left side implies the right, by contradiction, but I can't find how to prove the rest. I tried to show that there's an $x$ for which $|x-a|<delta$ and $|x-b|<epsilon$ when the left side is false, thinking that $x=(a+delta+b-epsilon)/2$ works (when $ale b$), but it is either wrong, or I can't get to the desired result.
Thanks in advance.
real-analysis
edited Jul 26 at 17:49
Andrés E. Caicedo
63.1k7151235
asked Jul 26 at 17:22
Ruben a
62
 |Â
show 1 more comment
up vote
0
down vote
favorite
I'm beginning studying real analysis, and this came up in the book, without proof:
($V_epsilon(a)$ denotes the $epsilon$-neighbourhood of $a$)
For all $a,b in mathbbR$; $delta,epsilon >0$ :
$epsilon+delta le lvert b-a rvert iff V_delta(a) cap V_epsilon(b) = emptyset$
I easily proved that the left side implies the right, by contradiction, but I can't find how to prove the rest. I tried to show that there's an $x$ for which $|x-a|<delta$ and $|x-b|<epsilon$ when the left side is false, thinking that $x=(a+delta+b-epsilon)/2$ works (when $ale b$), but it is either wrong, or I can't get to the desired result.
Thanks in advance.
real-analysis
edited Jul 26 at 17:49
Andrés E. Caicedo
63.1k7151235
asked Jul 26 at 17:22
Ruben a
62
What book is that? What is $V$? Please include more context to the question.
– amWhy
Jul 26 at 17:24
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 26 at 17:24
Vε(a) denotes the ε-neighbourhood of a
– Ruben a
Jul 26 at 17:27
$x=fraca+delta ;+;b-varepsilon2$ should indeed work. Which trouble do you run into when trying to prove it is in both of the balls?
– Henning Makholm
Jul 26 at 17:33
Note that you can also assume that $varepsilon$ and $delta$ are both $le|b-a|$, because if not then either $a$ or $b$ will be in the intersection and you're done. That just leaves a case where $a le b-varepsilon < a+delta le b$, where it should be easy to prove that your $x$ works.
– Henning Makholm
Jul 26 at 17:37
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm beginning studying real analysis, and this came up in the book, without proof:
($V_epsilon(a)$ denotes the $epsilon$-neighbourhood of $a$)
For all $a,b in mathbbR$; $delta,epsilon >0$ :
$epsilon+delta le lvert b-a rvert iff V_delta(a) cap V_epsilon(b) = emptyset$
I easily proved that the left side implies the right, by contradiction, but I can't find how to prove the rest. I tried to show that there's an $x$ for which $|x-a|<delta$ and $|x-b|<epsilon$ when the left side is false, thinking that $x=(a+delta+b-epsilon)/2$ works (when $ale b$), but it is either wrong, or I can't get to the desired result.
Thanks in advance.
real-analysis
edited Jul 26 at 17:49
Andrés E. Caicedo
63.1k7151235
asked Jul 26 at 17:22
Ruben a
62
I'm beginning studying real analysis, and this came up in the book, without proof:
($V_epsilon(a)$ denotes the $epsilon$-neighbourhood of $a$)
For all $a,b in mathbbR$; $delta,epsilon >0$ :
$epsilon+delta le lvert b-a rvert iff V_delta(a) cap V_epsilon(b) = emptyset$
I easily proved that the left side implies the right, by contradiction, but I can't find how to prove the rest. I tried to show that there's an $x$ for which $|x-a|<delta$ and $|x-b|<epsilon$ when the left side is false, thinking that $x=(a+delta+b-epsilon)/2$ works (when $ale b$), but it is either wrong, or I can't get to the desired result.
Thanks in advance.
real-analysis
edited Jul 26 at 17:49
Andrés E. Caicedo
63.1k7151235
asked Jul 26 at 17:22
Ruben a
62
edited Jul 26 at 17:49
Andrés E. Caicedo
63.1k7151235
edited Jul 26 at 17:49
Andrés E. Caicedo
63.1k7151235
edited Jul 26 at 17:49
Andrés E. Caicedo
63.1k7151235
63.1k7151235
asked Jul 26 at 17:22
Ruben a
62
asked Jul 26 at 17:22
Ruben a
62
asked Jul 26 at 17:22
Ruben a
62
62
What book is that? What is $V$? Please include more context to the question.
– amWhy
Jul 26 at 17:24
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 26 at 17:24
Vε(a) denotes the ε-neighbourhood of a
– Ruben a
Jul 26 at 17:27
$x=fraca+delta ;+;b-varepsilon2$ should indeed work. Which trouble do you run into when trying to prove it is in both of the balls?
– Henning Makholm
Jul 26 at 17:33
Note that you can also assume that $varepsilon$ and $delta$ are both $le|b-a|$, because if not then either $a$ or $b$ will be in the intersection and you're done. That just leaves a case where $a le b-varepsilon < a+delta le b$, where it should be easy to prove that your $x$ works.
– Henning Makholm
Jul 26 at 17:37
 |Â
show 1 more comment
What book is that? What is $V$? Please include more context to the question.
– amWhy
Jul 26 at 17:24
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 26 at 17:24
Vε(a) denotes the ε-neighbourhood of a
– Ruben a
Jul 26 at 17:27
$x=fraca+delta ;+;b-varepsilon2$ should indeed work. Which trouble do you run into when trying to prove it is in both of the balls?
– Henning Makholm
Jul 26 at 17:33
Note that you can also assume that $varepsilon$ and $delta$ are both $le|b-a|$, because if not then either $a$ or $b$ will be in the intersection and you're done. That just leaves a case where $a le b-varepsilon < a+delta le b$, where it should be easy to prove that your $x$ works.
– Henning Makholm
Jul 26 at 17:37
What book is that? What is $V$? Please include more context to the question.
– amWhy
Jul 26 at 17:24
What book is that? What is $V$? Please include more context to the question.
– amWhy
Jul 26 at 17:24
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 26 at 17:24
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 26 at 17:24
Vε(a) denotes the ε-neighbourhood of a
– Ruben a
Jul 26 at 17:27
Vε(a) denotes the ε-neighbourhood of a
– Ruben a
Jul 26 at 17:27
$x=fraca+delta ;+;b-varepsilon2$ should indeed work. Which trouble do you run into when trying to prove it is in both of the balls?
– Henning Makholm
Jul 26 at 17:33
$x=fraca+delta ;+;b-varepsilon2$ should indeed work. Which trouble do you run into when trying to prove it is in both of the balls?
– Henning Makholm
Jul 26 at 17:33
Note that you can also assume that $varepsilon$ and $delta$ are both $le|b-a|$, because if not then either $a$ or $b$ will be in the intersection and you're done. That just leaves a case where $a le b-varepsilon < a+delta le b$, where it should be easy to prove that your $x$ works.
– Henning Makholm
Jul 26 at 17:37
Note that you can also assume that $varepsilon$ and $delta$ are both $le|b-a|$, because if not then either $a$ or $b$ will be in the intersection and you're done. That just leaves a case where $a le b-varepsilon < a+delta le b$, where it should be easy to prove that your $x$ works.
– Henning Makholm
Jul 26 at 17:37
 |Â
show 1 more comment
2 Answers
2
active
oldest
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up vote
0
down vote
Without loss of generality, lets assume that $a<b.$ Let $V_delta(a) cap V_epsilon(b) = emptyset,$ so there is an $xin(a,b)$ s.t. $|x-a|geq delta$ and $|x-b|geq epsilon.$ Since $xin (a,b),$ we have $x-a>0$ and $b-x>0.$ Then: $$delta +epsilon leq |x-a|+|x-b|= x-a-(x-b)=b-a=|b-a|.$$
answered Jul 26 at 17:49
Emin
1,27621330
add a comment |Â
up vote
0
down vote
Without loss of generality assume that $a<b$
There are four numbers to look at:
$$a-delta, a+delta, b-epsilon, b+epsilon$$
The intersection of $$ (
a-delta, a+delta)$$ and $$(b-epsilon, b+epsilon)$$ is empty if and only if $$
a+delta le b-epsilon$$
Which is equivalent to $$ delta +epsilon le b-a$$
Thus the intersection is not empty if and only if $$ delta +epsilon >b-a$$
answered Jul 26 at 18:03
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Without loss of generality, lets assume that $a<b.$ Let $V_delta(a) cap V_epsilon(b) = emptyset,$ so there is an $xin(a,b)$ s.t. $|x-a|geq delta$ and $|x-b|geq epsilon.$ Since $xin (a,b),$ we have $x-a>0$ and $b-x>0.$ Then: $$delta +epsilon leq |x-a|+|x-b|= x-a-(x-b)=b-a=|b-a|.$$
answered Jul 26 at 17:49
Emin
1,27621330
add a comment |Â
up vote
0
down vote
Without loss of generality, lets assume that $a<b.$ Let $V_delta(a) cap V_epsilon(b) = emptyset,$ so there is an $xin(a,b)$ s.t. $|x-a|geq delta$ and $|x-b|geq epsilon.$ Since $xin (a,b),$ we have $x-a>0$ and $b-x>0.$ Then: $$delta +epsilon leq |x-a|+|x-b|= x-a-(x-b)=b-a=|b-a|.$$
answered Jul 26 at 17:49
Emin
1,27621330
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Without loss of generality, lets assume that $a<b.$ Let $V_delta(a) cap V_epsilon(b) = emptyset,$ so there is an $xin(a,b)$ s.t. $|x-a|geq delta$ and $|x-b|geq epsilon.$ Since $xin (a,b),$ we have $x-a>0$ and $b-x>0.$ Then: $$delta +epsilon leq |x-a|+|x-b|= x-a-(x-b)=b-a=|b-a|.$$
answered Jul 26 at 17:49
Emin
1,27621330
Without loss of generality, lets assume that $a<b.$ Let $V_delta(a) cap V_epsilon(b) = emptyset,$ so there is an $xin(a,b)$ s.t. $|x-a|geq delta$ and $|x-b|geq epsilon.$ Since $xin (a,b),$ we have $x-a>0$ and $b-x>0.$ Then: $$delta +epsilon leq |x-a|+|x-b|= x-a-(x-b)=b-a=|b-a|.$$
answered Jul 26 at 17:49
Emin
1,27621330
answered Jul 26 at 17:49
Emin
1,27621330
answered Jul 26 at 17:49
Emin
1,27621330
answered Jul 26 at 17:49
Emin
1,27621330
1,27621330
add a comment |Â
add a comment |Â
up vote
0
down vote
Without loss of generality assume that $a<b$
There are four numbers to look at:
$$a-delta, a+delta, b-epsilon, b+epsilon$$
The intersection of $$ (
a-delta, a+delta)$$ and $$(b-epsilon, b+epsilon)$$ is empty if and only if $$
a+delta le b-epsilon$$
Which is equivalent to $$ delta +epsilon le b-a$$
Thus the intersection is not empty if and only if $$ delta +epsilon >b-a$$
answered Jul 26 at 18:03
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
Without loss of generality assume that $a<b$
There are four numbers to look at:
$$a-delta, a+delta, b-epsilon, b+epsilon$$
The intersection of $$ (
a-delta, a+delta)$$ and $$(b-epsilon, b+epsilon)$$ is empty if and only if $$
a+delta le b-epsilon$$
Which is equivalent to $$ delta +epsilon le b-a$$
Thus the intersection is not empty if and only if $$ delta +epsilon >b-a$$
answered Jul 26 at 18:03
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Without loss of generality assume that $a<b$
There are four numbers to look at:
$$a-delta, a+delta, b-epsilon, b+epsilon$$
The intersection of $$ (
a-delta, a+delta)$$ and $$(b-epsilon, b+epsilon)$$ is empty if and only if $$
a+delta le b-epsilon$$
Which is equivalent to $$ delta +epsilon le b-a$$
Thus the intersection is not empty if and only if $$ delta +epsilon >b-a$$
answered Jul 26 at 18:03
Mohammad Riazi-Kermani
27.3k41851
Without loss of generality assume that $a<b$
There are four numbers to look at:
$$a-delta, a+delta, b-epsilon, b+epsilon$$
The intersection of $$ (
a-delta, a+delta)$$ and $$(b-epsilon, b+epsilon)$$ is empty if and only if $$
a+delta le b-epsilon$$
Which is equivalent to $$ delta +epsilon le b-a$$
Thus the intersection is not empty if and only if $$ delta +epsilon >b-a$$
answered Jul 26 at 18:03
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:03
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:03
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:03
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
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What book is that? What is $V$? Please include more context to the question.
– amWhy
Jul 26 at 17:24
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 26 at 17:24
Vε(a) denotes the ε-neighbourhood of a
– Ruben a
Jul 26 at 17:27
$x=fraca+delta ;+;b-varepsilon2$ should indeed work. Which trouble do you run into when trying to prove it is in both of the balls?
– Henning Makholm
Jul 26 at 17:33
Note that you can also assume that $varepsilon$ and $delta$ are both $le|b-a|$, because if not then either $a$ or $b$ will be in the intersection and you're done. That just leaves a case where $a le b-varepsilon < a+delta le b$, where it should be easy to prove that your $x$ works.
– Henning Makholm
Jul 26 at 17:37