Topologist Sine Curve, connected but not path connected
$begingroup$
So I have to show that $X=U∪V$,
$U={(0,y),yin[-1,1]}$,
$V={(x,sin({1 over x}),x>0}$
is connected but not path connected
My proof:
Assume that $X $ is path connected and let $f:[0,1]rightarrow X$ with $f(0)=({1over pi},0)$ and $f(1)=(0,0)$. Let $c=inf{tin[0,1]|f(t)in U}$
So $f([0,c])$ must at most contain 1 element from U where the closure of $f([0,c])$ must contain all of U. So since $f([0,c])ne cls(f([0,c])$ then $f([0,c])$ is not closed and since $X$ is a Euclidean Space it is not compact. But since f is continous and $[0,c]$ is compact then $f([0,c])$ is compact as well and we have a contradiction and X is not path connected.
X is connected since any open set of U must contain elements of V and the fact that U and V are connected, then X must be connected.
Is this proof valid or are there some things I need to expand on?
Thank you for your time!
general-topology connectedness path-connected
asked Dec 27 '18 at 17:10
NikolajNikolaj
578
$endgroup$
add a comment |
$begingroup$
So I have to show that $X=U∪V$,
$U={(0,y),yin[-1,1]}$,
$V={(x,sin({1 over x}),x>0}$
is connected but not path connected
My proof:
Assume that $X $ is path connected and let $f:[0,1]rightarrow X$ with $f(0)=({1over pi},0)$ and $f(1)=(0,0)$. Let $c=inf{tin[0,1]|f(t)in U}$
So $f([0,c])$ must at most contain 1 element from U where the closure of $f([0,c])$ must contain all of U. So since $f([0,c])ne cls(f([0,c])$ then $f([0,c])$ is not closed and since $X$ is a Euclidean Space it is not compact. But since f is continous and $[0,c]$ is compact then $f([0,c])$ is compact as well and we have a contradiction and X is not path connected.
X is connected since any open set of U must contain elements of V and the fact that U and V are connected, then X must be connected.
Is this proof valid or are there some things I need to expand on?
Thank you for your time!
general-topology connectedness path-connected
asked Dec 27 '18 at 17:10
NikolajNikolaj
578
$endgroup$
1
$begingroup$
This has been asked quite often in this forum. See for example math.stackexchange.com/q/35054 and math.stackexchange.com/q/317125.
$endgroup$
– Paul Frost
Dec 27 '18 at 18:21
add a comment |
$begingroup$
So I have to show that $X=U∪V$,
$U={(0,y),yin[-1,1]}$,
$V={(x,sin({1 over x}),x>0}$
is connected but not path connected
My proof:
Assume that $X $ is path connected and let $f:[0,1]rightarrow X$ with $f(0)=({1over pi},0)$ and $f(1)=(0,0)$. Let $c=inf{tin[0,1]|f(t)in U}$
So $f([0,c])$ must at most contain 1 element from U where the closure of $f([0,c])$ must contain all of U. So since $f([0,c])ne cls(f([0,c])$ then $f([0,c])$ is not closed and since $X$ is a Euclidean Space it is not compact. But since f is continous and $[0,c]$ is compact then $f([0,c])$ is compact as well and we have a contradiction and X is not path connected.
X is connected since any open set of U must contain elements of V and the fact that U and V are connected, then X must be connected.
Is this proof valid or are there some things I need to expand on?
Thank you for your time!
general-topology connectedness path-connected
asked Dec 27 '18 at 17:10
NikolajNikolaj
578
$endgroup$
So I have to show that $X=U∪V$,
$U={(0,y),yin[-1,1]}$,
$V={(x,sin({1 over x}),x>0}$
is connected but not path connected
My proof:
Assume that $X $ is path connected and let $f:[0,1]rightarrow X$ with $f(0)=({1over pi},0)$ and $f(1)=(0,0)$. Let $c=inf{tin[0,1]|f(t)in U}$
So $f([0,c])$ must at most contain 1 element from U where the closure of $f([0,c])$ must contain all of U. So since $f([0,c])ne cls(f([0,c])$ then $f([0,c])$ is not closed and since $X$ is a Euclidean Space it is not compact. But since f is continous and $[0,c]$ is compact then $f([0,c])$ is compact as well and we have a contradiction and X is not path connected.
X is connected since any open set of U must contain elements of V and the fact that U and V are connected, then X must be connected.
Is this proof valid or are there some things I need to expand on?
Thank you for your time!
general-topology connectedness path-connected
general-topology connectedness path-connected
asked Dec 27 '18 at 17:10
NikolajNikolaj
578
asked Dec 27 '18 at 17:10
NikolajNikolaj
578
edited Dec 27 '18 at 17:54
Nikolaj
asked Dec 27 '18 at 17:10
NikolajNikolaj
578
asked Dec 27 '18 at 17:10
NikolajNikolaj
578
asked Dec 27 '18 at 17:10
NikolajNikolaj
578
578
1
$begingroup$
This has been asked quite often in this forum. See for example math.stackexchange.com/q/35054 and math.stackexchange.com/q/317125.
$endgroup$
– Paul Frost
Dec 27 '18 at 18:21
add a comment |
1
$begingroup$
This has been asked quite often in this forum. See for example math.stackexchange.com/q/35054 and math.stackexchange.com/q/317125.
$endgroup$
– Paul Frost
Dec 27 '18 at 18:21
1
1
$begingroup$
This has been asked quite often in this forum. See for example math.stackexchange.com/q/35054 and math.stackexchange.com/q/317125.
$endgroup$
– Paul Frost
Dec 27 '18 at 18:21
$begingroup$
This has been asked quite often in this forum. See for example math.stackexchange.com/q/35054 and math.stackexchange.com/q/317125.
$endgroup$
– Paul Frost
Dec 27 '18 at 18:21
add a comment |
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$begingroup$
This has been asked quite often in this forum. See for example math.stackexchange.com/q/35054 and math.stackexchange.com/q/317125.
$endgroup$
– Paul Frost
Dec 27 '18 at 18:21