Shrinking Topologist's Sine Curve
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2
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The Topologist's Sine Curve is well-known:
The set $S = {(0,0)} cup {(x, sin(1/x)) | x in left]0,1right]}$, as a subspace of $mathbb{R}^n$, is connected but not path-connected.
An intuitive reason is that no path from $S - {(0,0)}$ can reach $(0,0)$ in a finite amount of time. However, what if I shrink the distances involved? That is, consider
$$
S' = {(0,0)} cup {(x, xsin(1/x)) | x in left]0,1right]}.
$$
Is $S'$ path-connected?
general-topology examples-counterexamples connectedness
asked Mar 17 '13 at 5:01
Herng Yi
1,3941023
add a comment |
up vote
2
down vote
favorite
The Topologist's Sine Curve is well-known:
The set $S = {(0,0)} cup {(x, sin(1/x)) | x in left]0,1right]}$, as a subspace of $mathbb{R}^n$, is connected but not path-connected.
An intuitive reason is that no path from $S - {(0,0)}$ can reach $(0,0)$ in a finite amount of time. However, what if I shrink the distances involved? That is, consider
$$
S' = {(0,0)} cup {(x, xsin(1/x)) | x in left]0,1right]}.
$$
Is $S'$ path-connected?
general-topology examples-counterexamples connectedness
asked Mar 17 '13 at 5:01
Herng Yi
1,3941023
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The Topologist's Sine Curve is well-known:
The set $S = {(0,0)} cup {(x, sin(1/x)) | x in left]0,1right]}$, as a subspace of $mathbb{R}^n$, is connected but not path-connected.
An intuitive reason is that no path from $S - {(0,0)}$ can reach $(0,0)$ in a finite amount of time. However, what if I shrink the distances involved? That is, consider
$$
S' = {(0,0)} cup {(x, xsin(1/x)) | x in left]0,1right]}.
$$
Is $S'$ path-connected?
general-topology examples-counterexamples connectedness
asked Mar 17 '13 at 5:01
Herng Yi
1,3941023
The Topologist's Sine Curve is well-known:
The set $S = {(0,0)} cup {(x, sin(1/x)) | x in left]0,1right]}$, as a subspace of $mathbb{R}^n$, is connected but not path-connected.
An intuitive reason is that no path from $S - {(0,0)}$ can reach $(0,0)$ in a finite amount of time. However, what if I shrink the distances involved? That is, consider
$$
S' = {(0,0)} cup {(x, xsin(1/x)) | x in left]0,1right]}.
$$
Is $S'$ path-connected?
general-topology examples-counterexamples connectedness
general-topology examples-counterexamples connectedness
asked Mar 17 '13 at 5:01
Herng Yi
1,3941023
asked Mar 17 '13 at 5:01
Herng Yi
1,3941023
asked Mar 17 '13 at 5:01
Herng Yi
1,3941023
asked Mar 17 '13 at 5:01
Herng Yi
1,3941023
asked Mar 17 '13 at 5:01
Herng Yi
1,3941023
1,3941023
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Let $$f:[0,1]toBbb R:xmapstobegin{cases}xsinfrac1x,&text{if }0<xle 1\\0,&text{if }x=0;;end{cases}$$
then $f$ is a path from which paths connecting any two points of $S'$ can easily be derived.
answered Mar 17 '13 at 5:05
Brian M. Scott
453k38503902
Indeed, as $lim_{xto 0} xsin{(frac{1}{x})} = 0 = f(0)$
– Rustyn
Mar 17 '13 at 5:09
@Rustyn: Exactly. (But I figured that that was pretty well known.)
– Brian M. Scott
Mar 17 '13 at 5:10
+1 nice answer@ Brian sir
– user525416
May 19 at 1:13
add a comment |
up vote
1
down vote
I should point out some subtlety here. The problem is never about infinite length of these curves. Remember that we have space filling (continuous) curves into $R^2$. So, why is it possible to turn the shrinking sine cure into a path, but the same is not possible for the usual sine curve?
The real answer is the non-/existence of the limit at zero. In both cases we already have a curve defined from $(o,1]$ into your space. In one case we can assign a value at $0$, and extend the function's domain continuously to the closed $[0,1]$, while in the other we cannot, since there is no limit at zero.
When can we extend a continuous function from $(0,1]$ to all of $[0,1]$? Precisely, if and only if the (original) function is uniformly continuous.
answered Nov 15 at 22:27
Behnam Esmayli
1,913515
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Let $$f:[0,1]toBbb R:xmapstobegin{cases}xsinfrac1x,&text{if }0<xle 1\\0,&text{if }x=0;;end{cases}$$
then $f$ is a path from which paths connecting any two points of $S'$ can easily be derived.
answered Mar 17 '13 at 5:05
Brian M. Scott
453k38503902
Indeed, as $lim_{xto 0} xsin{(frac{1}{x})} = 0 = f(0)$
– Rustyn
Mar 17 '13 at 5:09
@Rustyn: Exactly. (But I figured that that was pretty well known.)
– Brian M. Scott
Mar 17 '13 at 5:10
+1 nice answer@ Brian sir
– user525416
May 19 at 1:13
add a comment |
up vote
5
down vote
accepted
Let $$f:[0,1]toBbb R:xmapstobegin{cases}xsinfrac1x,&text{if }0<xle 1\\0,&text{if }x=0;;end{cases}$$
then $f$ is a path from which paths connecting any two points of $S'$ can easily be derived.
answered Mar 17 '13 at 5:05
Brian M. Scott
453k38503902
Indeed, as $lim_{xto 0} xsin{(frac{1}{x})} = 0 = f(0)$
– Rustyn
Mar 17 '13 at 5:09
@Rustyn: Exactly. (But I figured that that was pretty well known.)
– Brian M. Scott
Mar 17 '13 at 5:10
+1 nice answer@ Brian sir
– user525416
May 19 at 1:13
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Let $$f:[0,1]toBbb R:xmapstobegin{cases}xsinfrac1x,&text{if }0<xle 1\\0,&text{if }x=0;;end{cases}$$
then $f$ is a path from which paths connecting any two points of $S'$ can easily be derived.
answered Mar 17 '13 at 5:05
Brian M. Scott
453k38503902
Let $$f:[0,1]toBbb R:xmapstobegin{cases}xsinfrac1x,&text{if }0<xle 1\\0,&text{if }x=0;;end{cases}$$
then $f$ is a path from which paths connecting any two points of $S'$ can easily be derived.
answered Mar 17 '13 at 5:05
Brian M. Scott
453k38503902
answered Mar 17 '13 at 5:05
Brian M. Scott
453k38503902
answered Mar 17 '13 at 5:05
Brian M. Scott
453k38503902
answered Mar 17 '13 at 5:05
Brian M. Scott
453k38503902
453k38503902
Indeed, as $lim_{xto 0} xsin{(frac{1}{x})} = 0 = f(0)$
– Rustyn
Mar 17 '13 at 5:09
@Rustyn: Exactly. (But I figured that that was pretty well known.)
– Brian M. Scott
Mar 17 '13 at 5:10
+1 nice answer@ Brian sir
– user525416
May 19 at 1:13
add a comment |
Indeed, as $lim_{xto 0} xsin{(frac{1}{x})} = 0 = f(0)$
– Rustyn
Mar 17 '13 at 5:09
@Rustyn: Exactly. (But I figured that that was pretty well known.)
– Brian M. Scott
Mar 17 '13 at 5:10
+1 nice answer@ Brian sir
– user525416
May 19 at 1:13
Indeed, as $lim_{xto 0} xsin{(frac{1}{x})} = 0 = f(0)$
– Rustyn
Mar 17 '13 at 5:09
Indeed, as $lim_{xto 0} xsin{(frac{1}{x})} = 0 = f(0)$
– Rustyn
Mar 17 '13 at 5:09
@Rustyn: Exactly. (But I figured that that was pretty well known.)
– Brian M. Scott
Mar 17 '13 at 5:10
@Rustyn: Exactly. (But I figured that that was pretty well known.)
– Brian M. Scott
Mar 17 '13 at 5:10
+1 nice answer@ Brian sir
– user525416
May 19 at 1:13
+1 nice answer@ Brian sir
– user525416
May 19 at 1:13
add a comment |
up vote
1
down vote
I should point out some subtlety here. The problem is never about infinite length of these curves. Remember that we have space filling (continuous) curves into $R^2$. So, why is it possible to turn the shrinking sine cure into a path, but the same is not possible for the usual sine curve?
The real answer is the non-/existence of the limit at zero. In both cases we already have a curve defined from $(o,1]$ into your space. In one case we can assign a value at $0$, and extend the function's domain continuously to the closed $[0,1]$, while in the other we cannot, since there is no limit at zero.
When can we extend a continuous function from $(0,1]$ to all of $[0,1]$? Precisely, if and only if the (original) function is uniformly continuous.
answered Nov 15 at 22:27
Behnam Esmayli
1,913515
add a comment |
up vote
1
down vote
I should point out some subtlety here. The problem is never about infinite length of these curves. Remember that we have space filling (continuous) curves into $R^2$. So, why is it possible to turn the shrinking sine cure into a path, but the same is not possible for the usual sine curve?
The real answer is the non-/existence of the limit at zero. In both cases we already have a curve defined from $(o,1]$ into your space. In one case we can assign a value at $0$, and extend the function's domain continuously to the closed $[0,1]$, while in the other we cannot, since there is no limit at zero.
When can we extend a continuous function from $(0,1]$ to all of $[0,1]$? Precisely, if and only if the (original) function is uniformly continuous.
answered Nov 15 at 22:27
Behnam Esmayli
1,913515
add a comment |
up vote
1
down vote
up vote
1
down vote
I should point out some subtlety here. The problem is never about infinite length of these curves. Remember that we have space filling (continuous) curves into $R^2$. So, why is it possible to turn the shrinking sine cure into a path, but the same is not possible for the usual sine curve?
The real answer is the non-/existence of the limit at zero. In both cases we already have a curve defined from $(o,1]$ into your space. In one case we can assign a value at $0$, and extend the function's domain continuously to the closed $[0,1]$, while in the other we cannot, since there is no limit at zero.
When can we extend a continuous function from $(0,1]$ to all of $[0,1]$? Precisely, if and only if the (original) function is uniformly continuous.
answered Nov 15 at 22:27
Behnam Esmayli
1,913515
I should point out some subtlety here. The problem is never about infinite length of these curves. Remember that we have space filling (continuous) curves into $R^2$. So, why is it possible to turn the shrinking sine cure into a path, but the same is not possible for the usual sine curve?
The real answer is the non-/existence of the limit at zero. In both cases we already have a curve defined from $(o,1]$ into your space. In one case we can assign a value at $0$, and extend the function's domain continuously to the closed $[0,1]$, while in the other we cannot, since there is no limit at zero.
When can we extend a continuous function from $(0,1]$ to all of $[0,1]$? Precisely, if and only if the (original) function is uniformly continuous.
answered Nov 15 at 22:27
Behnam Esmayli
1,913515
answered Nov 15 at 22:27
Behnam Esmayli
1,913515
answered Nov 15 at 22:27
Behnam Esmayli
1,913515
answered Nov 15 at 22:27
Behnam Esmayli
1,913515
1,913515
add a comment |
add a comment |
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