Mixing
Continuous function problem involving distance of sets in $mathbbR^n$.
Clash Royale CLAN TAG#URR8PPP
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Given $a,b in mathbbR$ and $A,B$ disjoints non-empty closed sets, show that there is a function $f: mathbbR^n to mathbbR$ continous satisfying the following properties:
$$f(x) = a;; forall x in A,quad f(x) = b;; forall x in B.$$
$$mina,b leq f(x) leq maxa,b;; forall x in mathbbR^n.$$
Define
$$x longmapsto aleft(fracoperatornamed(x,B)operatornamed(x,B) + operatornamed(x,A)right) + bleft(fracoperatornamed(x,A)operatornamed(x,A) + operatornamed(x,B)right).$$
I proved almost all properties, It remains to be shown that
$$mina,b leq f(x) leq maxa,b;; forall x in left(mathbbR^nsetminus (A cup B)right).$$
How do I do this?
Notation. $operatornamed(x,A)$ is the distance of $x$ to $A$.
real-analysis continuity
asked Jul 25 at 18:51
Lucas Corrêa
1,101319
add a comment |Â
up vote
0
down vote
favorite
Given $a,b in mathbbR$ and $A,B$ disjoints non-empty closed sets, show that there is a function $f: mathbbR^n to mathbbR$ continous satisfying the following properties:
$$f(x) = a;; forall x in A,quad f(x) = b;; forall x in B.$$
$$mina,b leq f(x) leq maxa,b;; forall x in mathbbR^n.$$
Define
$$x longmapsto aleft(fracoperatornamed(x,B)operatornamed(x,B) + operatornamed(x,A)right) + bleft(fracoperatornamed(x,A)operatornamed(x,A) + operatornamed(x,B)right).$$
I proved almost all properties, It remains to be shown that
$$mina,b leq f(x) leq maxa,b;; forall x in left(mathbbR^nsetminus (A cup B)right).$$
How do I do this?
Notation. $operatornamed(x,A)$ is the distance of $x$ to $A$.
real-analysis continuity
asked Jul 25 at 18:51
Lucas Corrêa
1,101319
$f$ has domain $mathbb R^n$ but $x in A$, so how does $f(x)$ make any sense?
– Lee Mosher
Jul 25 at 18:58
2
$f(x)$ is a convex combination of $a$ and $b$.
– Daniel Fischer♦
Jul 25 at 18:59
1
Oh, I see. $A$ and $B$ are subsets of $mathbb R^n$. That should definitely be said in the question.
– Lee Mosher
Jul 25 at 19:01
@LeeMosher, yes! In fact, the question is poorly written, but I also assumed $A, B subset mathbbR^n$
– Lucas Corrêa
Jul 25 at 19:03
@DanielFischer, Of course! Thank you!
– Lucas Corrêa
Jul 25 at 19:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $a,b in mathbbR$ and $A,B$ disjoints non-empty closed sets, show that there is a function $f: mathbbR^n to mathbbR$ continous satisfying the following properties:
$$f(x) = a;; forall x in A,quad f(x) = b;; forall x in B.$$
$$mina,b leq f(x) leq maxa,b;; forall x in mathbbR^n.$$
Define
$$x longmapsto aleft(fracoperatornamed(x,B)operatornamed(x,B) + operatornamed(x,A)right) + bleft(fracoperatornamed(x,A)operatornamed(x,A) + operatornamed(x,B)right).$$
I proved almost all properties, It remains to be shown that
$$mina,b leq f(x) leq maxa,b;; forall x in left(mathbbR^nsetminus (A cup B)right).$$
How do I do this?
Notation. $operatornamed(x,A)$ is the distance of $x$ to $A$.
real-analysis continuity
asked Jul 25 at 18:51
Lucas Corrêa
1,101319
Given $a,b in mathbbR$ and $A,B$ disjoints non-empty closed sets, show that there is a function $f: mathbbR^n to mathbbR$ continous satisfying the following properties:
$$f(x) = a;; forall x in A,quad f(x) = b;; forall x in B.$$
$$mina,b leq f(x) leq maxa,b;; forall x in mathbbR^n.$$
Define
$$x longmapsto aleft(fracoperatornamed(x,B)operatornamed(x,B) + operatornamed(x,A)right) + bleft(fracoperatornamed(x,A)operatornamed(x,A) + operatornamed(x,B)right).$$
I proved almost all properties, It remains to be shown that
$$mina,b leq f(x) leq maxa,b;; forall x in left(mathbbR^nsetminus (A cup B)right).$$
How do I do this?
Notation. $operatornamed(x,A)$ is the distance of $x$ to $A$.
real-analysis continuity
asked Jul 25 at 18:51
Lucas Corrêa
1,101319
asked Jul 25 at 18:51
Lucas Corrêa
1,101319
asked Jul 25 at 18:51
Lucas Corrêa
1,101319
asked Jul 25 at 18:51
Lucas Corrêa
1,101319
1,101319
$f$ has domain $mathbb R^n$ but $x in A$, so how does $f(x)$ make any sense?
– Lee Mosher
Jul 25 at 18:58
2
$f(x)$ is a convex combination of $a$ and $b$.
– Daniel Fischer♦
Jul 25 at 18:59
1
Oh, I see. $A$ and $B$ are subsets of $mathbb R^n$. That should definitely be said in the question.
– Lee Mosher
Jul 25 at 19:01
@LeeMosher, yes! In fact, the question is poorly written, but I also assumed $A, B subset mathbbR^n$
– Lucas Corrêa
Jul 25 at 19:03
@DanielFischer, Of course! Thank you!
– Lucas Corrêa
Jul 25 at 19:04
add a comment |Â
$f$ has domain $mathbb R^n$ but $x in A$, so how does $f(x)$ make any sense?
– Lee Mosher
Jul 25 at 18:58
2
$f(x)$ is a convex combination of $a$ and $b$.
– Daniel Fischer♦
Jul 25 at 18:59
1
Oh, I see. $A$ and $B$ are subsets of $mathbb R^n$. That should definitely be said in the question.
– Lee Mosher
Jul 25 at 19:01
@LeeMosher, yes! In fact, the question is poorly written, but I also assumed $A, B subset mathbbR^n$
– Lucas Corrêa
Jul 25 at 19:03
@DanielFischer, Of course! Thank you!
– Lucas Corrêa
Jul 25 at 19:04
$f$ has domain $mathbb R^n$ but $x in A$, so how does $f(x)$ make any sense?
– Lee Mosher
Jul 25 at 18:58
$f$ has domain $mathbb R^n$ but $x in A$, so how does $f(x)$ make any sense?
– Lee Mosher
Jul 25 at 18:58
2
2
$f(x)$ is a convex combination of $a$ and $b$.
– Daniel Fischer♦
Jul 25 at 18:59
$f(x)$ is a convex combination of $a$ and $b$.
– Daniel Fischer♦
Jul 25 at 18:59
1
1
Oh, I see. $A$ and $B$ are subsets of $mathbb R^n$. That should definitely be said in the question.
– Lee Mosher
Jul 25 at 19:01
Oh, I see. $A$ and $B$ are subsets of $mathbb R^n$. That should definitely be said in the question.
– Lee Mosher
Jul 25 at 19:01
@LeeMosher, yes! In fact, the question is poorly written, but I also assumed $A, B subset mathbbR^n$
– Lucas Corrêa
Jul 25 at 19:03
@LeeMosher, yes! In fact, the question is poorly written, but I also assumed $A, B subset mathbbR^n$
– Lucas Corrêa
Jul 25 at 19:03
@DanielFischer, Of course! Thank you!
– Lucas Corrêa
Jul 25 at 19:04
@DanielFischer, Of course! Thank you!
– Lucas Corrêa
Jul 25 at 19:04
add a comment |Â
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$f$ has domain $mathbb R^n$ but $x in A$, so how does $f(x)$ make any sense?
– Lee Mosher
Jul 25 at 18:58
2
$f(x)$ is a convex combination of $a$ and $b$.
– Daniel Fischer♦
Jul 25 at 18:59
1
Oh, I see. $A$ and $B$ are subsets of $mathbb R^n$. That should definitely be said in the question.
– Lee Mosher
Jul 25 at 19:01
@LeeMosher, yes! In fact, the question is poorly written, but I also assumed $A, B subset mathbbR^n$
– Lucas Corrêa
Jul 25 at 19:03
@DanielFischer, Of course! Thank you!
– Lucas Corrêa
Jul 25 at 19:04