Some problems about compactness, connectivity amd arcwise-connectivity
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I have some home work about justifying why some sets are compact, connected or arcwise-connected.
The sets are:
$C={(x,y)in mathbb{R}^2 mid xin (0,1)}$
$D={(x,y)inmathbb{R}^2 mid 0leq x^2+y^2leq 1}$
Now I've already proved that $C$ is not compact (it's open, and by Heine-Borell must be closed to be compact in $mathbb{R}^2$), and if I can proof that it's arcwise-connected, then is connected, and I think this is true, but I can't find the curve $alpha : [a,b]mapsto C$ such that, for every $x,yin C$:
a) $alpha(a)=x$
b) $alpha(b)=y$
c) if $t_kmapsto t$ then $alpha(t_k)mapstoalpha(t)$
d) $alpha([a,b])subset C$.
For $D$, I don't know even if it's compact.
I could use some help here.
real-analysis general-topology analysis
edited Dec 4 '18 at 18:48
Federico
5,124514
asked Dec 4 '18 at 18:45
Armando RosasArmando Rosas
525
$endgroup$
add a comment |
$begingroup$
I have some home work about justifying why some sets are compact, connected or arcwise-connected.
The sets are:
$C={(x,y)in mathbb{R}^2 mid xin (0,1)}$
$D={(x,y)inmathbb{R}^2 mid 0leq x^2+y^2leq 1}$
Now I've already proved that $C$ is not compact (it's open, and by Heine-Borell must be closed to be compact in $mathbb{R}^2$), and if I can proof that it's arcwise-connected, then is connected, and I think this is true, but I can't find the curve $alpha : [a,b]mapsto C$ such that, for every $x,yin C$:
a) $alpha(a)=x$
b) $alpha(b)=y$
c) if $t_kmapsto t$ then $alpha(t_k)mapstoalpha(t)$
d) $alpha([a,b])subset C$.
For $D$, I don't know even if it's compact.
I could use some help here.
real-analysis general-topology analysis
edited Dec 4 '18 at 18:48
Federico
5,124514
asked Dec 4 '18 at 18:45
Armando RosasArmando Rosas
525
$endgroup$
$begingroup$
Draw a picture. Place two points in $C$ and link them by a segment. Give a function to describe this segment. For $D$, draw a picture and you will see $D$ is a closed disk.
$endgroup$
– Syuizen
Dec 4 '18 at 18:51
add a comment |
$begingroup$
I have some home work about justifying why some sets are compact, connected or arcwise-connected.
The sets are:
$C={(x,y)in mathbb{R}^2 mid xin (0,1)}$
$D={(x,y)inmathbb{R}^2 mid 0leq x^2+y^2leq 1}$
Now I've already proved that $C$ is not compact (it's open, and by Heine-Borell must be closed to be compact in $mathbb{R}^2$), and if I can proof that it's arcwise-connected, then is connected, and I think this is true, but I can't find the curve $alpha : [a,b]mapsto C$ such that, for every $x,yin C$:
a) $alpha(a)=x$
b) $alpha(b)=y$
c) if $t_kmapsto t$ then $alpha(t_k)mapstoalpha(t)$
d) $alpha([a,b])subset C$.
For $D$, I don't know even if it's compact.
I could use some help here.
real-analysis general-topology analysis
edited Dec 4 '18 at 18:48
Federico
5,124514
asked Dec 4 '18 at 18:45
Armando RosasArmando Rosas
525
$endgroup$
I have some home work about justifying why some sets are compact, connected or arcwise-connected.
The sets are:
$C={(x,y)in mathbb{R}^2 mid xin (0,1)}$
$D={(x,y)inmathbb{R}^2 mid 0leq x^2+y^2leq 1}$
Now I've already proved that $C$ is not compact (it's open, and by Heine-Borell must be closed to be compact in $mathbb{R}^2$), and if I can proof that it's arcwise-connected, then is connected, and I think this is true, but I can't find the curve $alpha : [a,b]mapsto C$ such that, for every $x,yin C$:
a) $alpha(a)=x$
b) $alpha(b)=y$
c) if $t_kmapsto t$ then $alpha(t_k)mapstoalpha(t)$
d) $alpha([a,b])subset C$.
For $D$, I don't know even if it's compact.
I could use some help here.
real-analysis general-topology analysis
real-analysis general-topology analysis
edited Dec 4 '18 at 18:48
Federico
5,124514
asked Dec 4 '18 at 18:45
Armando RosasArmando Rosas
525
edited Dec 4 '18 at 18:48
Federico
5,124514
asked Dec 4 '18 at 18:45
Armando RosasArmando Rosas
525
edited Dec 4 '18 at 18:48
Federico
5,124514
edited Dec 4 '18 at 18:48
Federico
5,124514
edited Dec 4 '18 at 18:48
Federico
5,124514
5,124514
asked Dec 4 '18 at 18:45
Armando RosasArmando Rosas
525
asked Dec 4 '18 at 18:45
Armando RosasArmando Rosas
525
asked Dec 4 '18 at 18:45
Armando RosasArmando Rosas
525
525
$begingroup$
Draw a picture. Place two points in $C$ and link them by a segment. Give a function to describe this segment. For $D$, draw a picture and you will see $D$ is a closed disk.
$endgroup$
– Syuizen
Dec 4 '18 at 18:51
add a comment |
$begingroup$
Draw a picture. Place two points in $C$ and link them by a segment. Give a function to describe this segment. For $D$, draw a picture and you will see $D$ is a closed disk.
$endgroup$
– Syuizen
Dec 4 '18 at 18:51
$begingroup$
Draw a picture. Place two points in $C$ and link them by a segment. Give a function to describe this segment. For $D$, draw a picture and you will see $D$ is a closed disk.
$endgroup$
– Syuizen
Dec 4 '18 at 18:51
$begingroup$
Draw a picture. Place two points in $C$ and link them by a segment. Give a function to describe this segment. For $D$, draw a picture and you will see $D$ is a closed disk.
$endgroup$
– Syuizen
Dec 4 '18 at 18:51
add a comment |
1 Answer
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$C$ is convex, hence it is connected.
$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.
answered Dec 4 '18 at 18:51
FedericoFederico
5,124514
$endgroup$
$begingroup$
but i haven't seen convexity nor bounded properties yet, i don't know how to use that.
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:55
$begingroup$
What is compact for you? Closed and bounded? Something else?
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
Convexity is just a very special case of being connected in which the path from $x$ to $y$ is a straight segment
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
sorry, i forgot Heine-Borell's theorem, so D is compact because of that? Can i bound it with the circumference of radius 1?
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:59
$begingroup$
Exactly. Being bounded is precisely saying that the function $x^2+y^2$ is bounded on your set. In the case of $D$, this is by definition.
$endgroup$
– Federico
Dec 4 '18 at 19:02
|
show 6 more comments
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1 Answer
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$begingroup$
$C$ is convex, hence it is connected.
$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.
answered Dec 4 '18 at 18:51
FedericoFederico
5,124514
$endgroup$
$begingroup$
but i haven't seen convexity nor bounded properties yet, i don't know how to use that.
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:55
$begingroup$
What is compact for you? Closed and bounded? Something else?
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
Convexity is just a very special case of being connected in which the path from $x$ to $y$ is a straight segment
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
sorry, i forgot Heine-Borell's theorem, so D is compact because of that? Can i bound it with the circumference of radius 1?
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:59
$begingroup$
Exactly. Being bounded is precisely saying that the function $x^2+y^2$ is bounded on your set. In the case of $D$, this is by definition.
$endgroup$
– Federico
Dec 4 '18 at 19:02
|
show 6 more comments
$begingroup$
$C$ is convex, hence it is connected.
$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.
answered Dec 4 '18 at 18:51
FedericoFederico
5,124514
$endgroup$
$begingroup$
but i haven't seen convexity nor bounded properties yet, i don't know how to use that.
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:55
$begingroup$
What is compact for you? Closed and bounded? Something else?
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
Convexity is just a very special case of being connected in which the path from $x$ to $y$ is a straight segment
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
sorry, i forgot Heine-Borell's theorem, so D is compact because of that? Can i bound it with the circumference of radius 1?
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:59
$begingroup$
Exactly. Being bounded is precisely saying that the function $x^2+y^2$ is bounded on your set. In the case of $D$, this is by definition.
$endgroup$
– Federico
Dec 4 '18 at 19:02
|
show 6 more comments
$begingroup$
$C$ is convex, hence it is connected.
$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.
answered Dec 4 '18 at 18:51
FedericoFederico
5,124514
$endgroup$
$C$ is convex, hence it is connected.
$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.
answered Dec 4 '18 at 18:51
FedericoFederico
5,124514
answered Dec 4 '18 at 18:51
FedericoFederico
5,124514
answered Dec 4 '18 at 18:51
FedericoFederico
5,124514
answered Dec 4 '18 at 18:51
FedericoFederico
5,124514
5,124514
$begingroup$
but i haven't seen convexity nor bounded properties yet, i don't know how to use that.
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:55
$begingroup$
What is compact for you? Closed and bounded? Something else?
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
Convexity is just a very special case of being connected in which the path from $x$ to $y$ is a straight segment
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
sorry, i forgot Heine-Borell's theorem, so D is compact because of that? Can i bound it with the circumference of radius 1?
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:59
$begingroup$
Exactly. Being bounded is precisely saying that the function $x^2+y^2$ is bounded on your set. In the case of $D$, this is by definition.
$endgroup$
– Federico
Dec 4 '18 at 19:02
|
show 6 more comments
$begingroup$
but i haven't seen convexity nor bounded properties yet, i don't know how to use that.
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:55
$begingroup$
What is compact for you? Closed and bounded? Something else?
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
Convexity is just a very special case of being connected in which the path from $x$ to $y$ is a straight segment
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
sorry, i forgot Heine-Borell's theorem, so D is compact because of that? Can i bound it with the circumference of radius 1?
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:59
$begingroup$
Exactly. Being bounded is precisely saying that the function $x^2+y^2$ is bounded on your set. In the case of $D$, this is by definition.
$endgroup$
– Federico
Dec 4 '18 at 19:02
$begingroup$
but i haven't seen convexity nor bounded properties yet, i don't know how to use that.
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:55
$begingroup$
but i haven't seen convexity nor bounded properties yet, i don't know how to use that.
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:55
$begingroup$
What is compact for you? Closed and bounded? Something else?
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
What is compact for you? Closed and bounded? Something else?
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
Convexity is just a very special case of being connected in which the path from $x$ to $y$ is a straight segment
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
Convexity is just a very special case of being connected in which the path from $x$ to $y$ is a straight segment
$endgroup$
– Federico
Dec 4 '18 at 18:56
$begingroup$
sorry, i forgot Heine-Borell's theorem, so D is compact because of that? Can i bound it with the circumference of radius 1?
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:59
$begingroup$
sorry, i forgot Heine-Borell's theorem, so D is compact because of that? Can i bound it with the circumference of radius 1?
$endgroup$
– Armando Rosas
Dec 4 '18 at 18:59
$begingroup$
Exactly. Being bounded is precisely saying that the function $x^2+y^2$ is bounded on your set. In the case of $D$, this is by definition.
$endgroup$
– Federico
Dec 4 '18 at 19:02
$begingroup$
Exactly. Being bounded is precisely saying that the function $x^2+y^2$ is bounded on your set. In the case of $D$, this is by definition.
$endgroup$
– Federico
Dec 4 '18 at 19:02
|
show 6 more comments
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$begingroup$
Draw a picture. Place two points in $C$ and link them by a segment. Give a function to describe this segment. For $D$, draw a picture and you will see $D$ is a closed disk.
$endgroup$
– Syuizen
Dec 4 '18 at 18:51