Some problems about compactness, connectivity amd arcwise-connectivity












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 18:48









Federico

5,124514




5,124514










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

$C$ is convex, hence it is connected.



$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

$C$ is convex, hence it is connected.



$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

$C$ is convex, hence it is connected.



$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

$C$ is convex, hence it is connected.



$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.






share|cite|improve this answer









$endgroup$



$C$ is convex, hence it is connected.



$D$ is closed and bounded, hence compact. Moreover it is convex, hence connected.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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