Compact spaces and Urysohn Lemma
$begingroup$
Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x in X$ a point. We define the distance $d(x, A)$ via
$$
d(x,A)=inf({ d(x,a)mid a in A})
$$
(a) Show that $dcolon X to mathbb{R}$ is a continuous function and that $xin overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f colon X to mathbb{R}$
defined by
$$
f(x)=frac{d(x, A)}{d(x, A) + d(x, B)}
$$
Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as
in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for
all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a)
Can we say that the function is continuous if we have $x,yin X$ so, that
$d(x,A)leq d(x,a)leq d(x,y) + d(y,a)$
for $a in A$
$d(x,A)-d(x,y) leq inf{d(y,a)}=d(y,A)$
so
$d(x,A)-d(y,A)leq d(x,y)$
Is this right or am I missing something? And how can I show that $xinoverline{A}$ if and only if $d(x,A)=0$?
Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
general-topology
edited Jan 10 at 5:14
Henno Brandsma
111k348118
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
$endgroup$
add a comment |
$begingroup$
Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x in X$ a point. We define the distance $d(x, A)$ via
$$
d(x,A)=inf({ d(x,a)mid a in A})
$$
(a) Show that $dcolon X to mathbb{R}$ is a continuous function and that $xin overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f colon X to mathbb{R}$
defined by
$$
f(x)=frac{d(x, A)}{d(x, A) + d(x, B)}
$$
Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as
in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for
all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a)
Can we say that the function is continuous if we have $x,yin X$ so, that
$d(x,A)leq d(x,a)leq d(x,y) + d(y,a)$
for $a in A$
$d(x,A)-d(x,y) leq inf{d(y,a)}=d(y,A)$
so
$d(x,A)-d(y,A)leq d(x,y)$
Is this right or am I missing something? And how can I show that $xinoverline{A}$ if and only if $d(x,A)=0$?
Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
general-topology
edited Jan 10 at 5:14
Henno Brandsma
111k348118
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
$endgroup$
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
add a comment |
$begingroup$
Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x in X$ a point. We define the distance $d(x, A)$ via
$$
d(x,A)=inf({ d(x,a)mid a in A})
$$
(a) Show that $dcolon X to mathbb{R}$ is a continuous function and that $xin overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f colon X to mathbb{R}$
defined by
$$
f(x)=frac{d(x, A)}{d(x, A) + d(x, B)}
$$
Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as
in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for
all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a)
Can we say that the function is continuous if we have $x,yin X$ so, that
$d(x,A)leq d(x,a)leq d(x,y) + d(y,a)$
for $a in A$
$d(x,A)-d(x,y) leq inf{d(y,a)}=d(y,A)$
so
$d(x,A)-d(y,A)leq d(x,y)$
Is this right or am I missing something? And how can I show that $xinoverline{A}$ if and only if $d(x,A)=0$?
Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
general-topology
edited Jan 10 at 5:14
Henno Brandsma
111k348118
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
$endgroup$
Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x in X$ a point. We define the distance $d(x, A)$ via
$$
d(x,A)=inf({ d(x,a)mid a in A})
$$
(a) Show that $dcolon X to mathbb{R}$ is a continuous function and that $xin overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f colon X to mathbb{R}$
defined by
$$
f(x)=frac{d(x, A)}{d(x, A) + d(x, B)}
$$
Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as
in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for
all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a)
Can we say that the function is continuous if we have $x,yin X$ so, that
$d(x,A)leq d(x,a)leq d(x,y) + d(y,a)$
for $a in A$
$d(x,A)-d(x,y) leq inf{d(y,a)}=d(y,A)$
so
$d(x,A)-d(y,A)leq d(x,y)$
Is this right or am I missing something? And how can I show that $xinoverline{A}$ if and only if $d(x,A)=0$?
Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
general-topology
general-topology
edited Jan 10 at 5:14
Henno Brandsma
111k348118
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
edited Jan 10 at 5:14
Henno Brandsma
111k348118
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
edited Jan 10 at 5:14
Henno Brandsma
111k348118
edited Jan 10 at 5:14
Henno Brandsma
111k348118
edited Jan 10 at 5:14
Henno Brandsma
111k348118
111k348118
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
asked Jan 9 at 11:00
SlowCatMichelangeloSlowCatMichelangelo
111
111
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
add a comment |
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
8,3122720
$endgroup$
add a comment |
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$begingroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
8,3122720
$endgroup$
add a comment |
$begingroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
8,3122720
$endgroup$
add a comment |
$begingroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
8,3122720
$endgroup$
a is incorrect.
Prove instead: x in $overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.
answered Jan 9 at 12:52
William ElliotWilliam Elliot
8,3122720
edited Jan 9 at 13:01
answered Jan 9 at 12:52
William ElliotWilliam Elliot
8,3122720
answered Jan 9 at 12:52
William ElliotWilliam Elliot
8,3122720
answered Jan 9 at 12:52
William ElliotWilliam Elliot
8,3122720
8,3122720
add a comment |
add a comment |
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$begingroup$
I think you're missing the hypothesis that $X$ is compact.
$endgroup$
– egreg
Jan 9 at 13:03
$begingroup$
@egreg This works in any metric space. Compactness is not needed, despite the title.
$endgroup$
– Henno Brandsma
Jan 10 at 5:12