Decomposing $f + g > t$ The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraMinimal dense subset of $mathbbQ cap [0,1]$A classic example in measure theory, hold the measure theoryStep-by-step help using the distributive law in set theoryProving set function is measureCheck the identity $Acap (B - C)=(Acap B) - (Acap C)$Classification of an open set in realSuppose $A^text' $ is the set of limit point of $A$. Prove $A^text' $ is closed set.Show finite unions of rectangles form an algebraSymmetric difference as a equivalent relation-equivalence classesReformulation of intersection of two unions
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Decomposing $f + g > t$
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraMinimal dense subset of $mathbbQ cap [0,1]$A classic example in measure theory, hold the measure theoryStep-by-step help using the distributive law in set theoryProving set function is measureCheck the identity $Acap (B - C)=(Acap B) - (Acap C)$Classification of an open set in realSuppose $A^text' $ is the set of limit point of $A$. Prove $A^text' $ is closed set.Show finite unions of rectangles form an algebraSymmetric difference as a equivalent relation-equivalence classesReformulation of intersection of two unions
$begingroup$
For this purpose of this question, define
$$
leftf > tright = left f(x) > tright
$$
I was told that
$$
leftf + g > t right = cup_r in mathbbQ leftf > r right cap leftg > t - r right
$$
I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.
$$
leftf + g > t right = cup_r in mathbbR leftf > r right cap leftg > t - r right
$$
Any hint?
real-analysis general-topology elementary-set-theory
$endgroup$
add a comment |
$begingroup$
For this purpose of this question, define
$$
leftf > tright = left f(x) > tright
$$
I was told that
$$
leftf + g > t right = cup_r in mathbbQ leftf > r right cap leftg > t - r right
$$
I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.
$$
leftf + g > t right = cup_r in mathbbR leftf > r right cap leftg > t - r right
$$
Any hint?
real-analysis general-topology elementary-set-theory
$endgroup$
add a comment |
$begingroup$
For this purpose of this question, define
$$
leftf > tright = left f(x) > tright
$$
I was told that
$$
leftf + g > t right = cup_r in mathbbQ leftf > r right cap leftg > t - r right
$$
I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.
$$
leftf + g > t right = cup_r in mathbbR leftf > r right cap leftg > t - r right
$$
Any hint?
real-analysis general-topology elementary-set-theory
$endgroup$
For this purpose of this question, define
$$
leftf > tright = left f(x) > tright
$$
I was told that
$$
leftf + g > t right = cup_r in mathbbQ leftf > r right cap leftg > t - r right
$$
I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.
$$
leftf + g > t right = cup_r in mathbbR leftf > r right cap leftg > t - r right
$$
Any hint?
real-analysis general-topology elementary-set-theory
real-analysis general-topology elementary-set-theory
asked Apr 8 at 6:34
nalzoknalzok
291217
291217
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?
$endgroup$
$begingroup$
Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
$endgroup$
– nalzok
Apr 8 at 6:39
add a comment |
Your Answer
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?
$endgroup$
$begingroup$
Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
$endgroup$
– nalzok
Apr 8 at 6:39
add a comment |
$begingroup$
Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?
$endgroup$
$begingroup$
Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
$endgroup$
– nalzok
Apr 8 at 6:39
add a comment |
$begingroup$
Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?
$endgroup$
Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?
answered Apr 8 at 6:38
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
74.4k53270
$begingroup$
Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
$endgroup$
– nalzok
Apr 8 at 6:39
add a comment |
$begingroup$
Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
$endgroup$
– nalzok
Apr 8 at 6:39
$begingroup$
Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
$endgroup$
– nalzok
Apr 8 at 6:39
$begingroup$
Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
$endgroup$
– nalzok
Apr 8 at 6:39
add a comment |
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