Is my intuition correct?
$begingroup$
Let $f : X rightarrow [0,infty)$ and $A$ , $B,$ two subsets of $X$ such that $A cap B neq emptyset$.
If I have that $ inf limits_{X setminus A} f ,>, inf limits_B f $, does it imply that $inf limits_B f = inf limits_{A cap B} f $ ??
Intuitevely it makes sense to me but I can't prove it. Any thoughts ? Or counterexapmple ?
supremum-and-infimum
asked Dec 9 '18 at 15:15
vl.athvl.ath
929
$endgroup$
add a comment |
$begingroup$
Let $f : X rightarrow [0,infty)$ and $A$ , $B,$ two subsets of $X$ such that $A cap B neq emptyset$.
If I have that $ inf limits_{X setminus A} f ,>, inf limits_B f $, does it imply that $inf limits_B f = inf limits_{A cap B} f $ ??
Intuitevely it makes sense to me but I can't prove it. Any thoughts ? Or counterexapmple ?
supremum-and-infimum
asked Dec 9 '18 at 15:15
vl.athvl.ath
929
$endgroup$
add a comment |
$begingroup$
Let $f : X rightarrow [0,infty)$ and $A$ , $B,$ two subsets of $X$ such that $A cap B neq emptyset$.
If I have that $ inf limits_{X setminus A} f ,>, inf limits_B f $, does it imply that $inf limits_B f = inf limits_{A cap B} f $ ??
Intuitevely it makes sense to me but I can't prove it. Any thoughts ? Or counterexapmple ?
supremum-and-infimum
asked Dec 9 '18 at 15:15
vl.athvl.ath
929
$endgroup$
Let $f : X rightarrow [0,infty)$ and $A$ , $B,$ two subsets of $X$ such that $A cap B neq emptyset$.
If I have that $ inf limits_{X setminus A} f ,>, inf limits_B f $, does it imply that $inf limits_B f = inf limits_{A cap B} f $ ??
Intuitevely it makes sense to me but I can't prove it. Any thoughts ? Or counterexapmple ?
supremum-and-infimum
supremum-and-infimum
asked Dec 9 '18 at 15:15
vl.athvl.ath
929
asked Dec 9 '18 at 15:15
vl.athvl.ath
929
asked Dec 9 '18 at 15:15
vl.athvl.ath
929
asked Dec 9 '18 at 15:15
vl.athvl.ath
929
asked Dec 9 '18 at 15:15
vl.athvl.ath
929
929
add a comment |
add a comment |
1 Answer
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$begingroup$
If $inf_B f neq inf_{Acap B} f$ then $inf_B f < inf_{Acap B} f$.
On the other hand
$$
inf_B f = min { inf_{Acap B} f , , inf_{Bsetminus A} f}.
$$
Hence $inf_B f = inf_{Bsetminus A} f$.
However $inf_{Bsetminus A} f>inf_{Xsetminus A} f >inf_B f$, contradiction.
answered Dec 9 '18 at 15:34
AnguepaAnguepa
1,401819
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1 Answer
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1 Answer
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$begingroup$
If $inf_B f neq inf_{Acap B} f$ then $inf_B f < inf_{Acap B} f$.
On the other hand
$$
inf_B f = min { inf_{Acap B} f , , inf_{Bsetminus A} f}.
$$
Hence $inf_B f = inf_{Bsetminus A} f$.
However $inf_{Bsetminus A} f>inf_{Xsetminus A} f >inf_B f$, contradiction.
answered Dec 9 '18 at 15:34
AnguepaAnguepa
1,401819
$endgroup$
add a comment |
$begingroup$
If $inf_B f neq inf_{Acap B} f$ then $inf_B f < inf_{Acap B} f$.
On the other hand
$$
inf_B f = min { inf_{Acap B} f , , inf_{Bsetminus A} f}.
$$
Hence $inf_B f = inf_{Bsetminus A} f$.
However $inf_{Bsetminus A} f>inf_{Xsetminus A} f >inf_B f$, contradiction.
answered Dec 9 '18 at 15:34
AnguepaAnguepa
1,401819
$endgroup$
add a comment |
$begingroup$
If $inf_B f neq inf_{Acap B} f$ then $inf_B f < inf_{Acap B} f$.
On the other hand
$$
inf_B f = min { inf_{Acap B} f , , inf_{Bsetminus A} f}.
$$
Hence $inf_B f = inf_{Bsetminus A} f$.
However $inf_{Bsetminus A} f>inf_{Xsetminus A} f >inf_B f$, contradiction.
answered Dec 9 '18 at 15:34
AnguepaAnguepa
1,401819
$endgroup$
If $inf_B f neq inf_{Acap B} f$ then $inf_B f < inf_{Acap B} f$.
On the other hand
$$
inf_B f = min { inf_{Acap B} f , , inf_{Bsetminus A} f}.
$$
Hence $inf_B f = inf_{Bsetminus A} f$.
However $inf_{Bsetminus A} f>inf_{Xsetminus A} f >inf_B f$, contradiction.
answered Dec 9 '18 at 15:34
AnguepaAnguepa
1,401819
answered Dec 9 '18 at 15:34
AnguepaAnguepa
1,401819
answered Dec 9 '18 at 15:34
AnguepaAnguepa
1,401819
answered Dec 9 '18 at 15:34
AnguepaAnguepa
1,401819
1,401819
add a comment |
add a comment |
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