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Let $A$ be a countable subset of $\mathbb{R}^2$. Show that $\mathbb{R}^2-A$ is path connected.

These are my steps:

  1. Let $x$ and $y$ be arbitrary points of $R^2$

  2. Let $f^r:[0,1] \to \mathbb{R}^2$ be given by $f^r(t)=(1-t^r)x+t^ry$ for $r \in\mathbb{R}$.

  3. Show that $f^r$ is a continuous path between $x$ and $y$ and that non of them intersect.

  4. Show that $f^r$ is bijective with $\mathbb{R}$ through $F(f^r)=r$ and conclude that there are uncountable non intersecting paths.

  5. Assume for the sake of contradiction that $A$ intersects all of the paths $f^r$ and deduce that $A$ must be uncountable (since it must intersect every $f^r$ at a different point).

  6. Contradiction! ($A$ is countable) Hence there exist a paths $f^r$ for some $r \in\mathbb{R}$ which doesn't intersect $A$.

  7. $\mathbb{R}^2-A$ is path-connected.

Is this proof sound?

Saal Hardali
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    Yes, it is. See also the answers to this question. – Brian M. Scott Dec 02 '13 at 20:14
  • The first half of 4 is formulated a bit oddly, but apart from that it is fine. – Carsten S Dec 02 '13 at 20:16
  • @Dimitris $A$ having measure $0$ is not sufficient. For example, if we take $A = {(x,y): x^2 + y^2 = 1}$, the resulting set is not path connected, even though $m(A) = 0$ – Ben Grossmann Dec 02 '13 at 20:17
  • @Omnomnomnom,corrct.thnx – Haha Dec 02 '13 at 20:20
  • @Dimitris it is, however, sufficient to require that $A$ be the continuous image of some $1$-dimensional measure-zero set, as explained here – Ben Grossmann Dec 02 '13 at 20:21
  • @Omnomnomnom,somehow this is what i wanted to say.i haven't studied yet measure theory. i will start this semester:P – Haha Dec 02 '13 at 20:23
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    Half of your maps $f^r$ have exactly the same image: the straight line segment connecting $x$ and $y$. Others do no make sense: for example, when $r<0$. Others are in fact constant (when you deal appropriately with the fact that $0^0$ is not defined...) You need to make sure that things you write do make sense! In particular, youe argument —while well-oriented— is quite not sound. – Mariano Suárez-Álvarez Dec 02 '13 at 20:35
  • (You need to take $x$ and $y$ distinct for any chance of your claims to be true. Even in that case, it is impossible to have two distinct paths from $x$ to $y$ have distinct images, for the two points will allways be in the intersection) – Mariano Suárez-Álvarez Dec 02 '13 at 20:43
  • @MarianoSuárez-Alvarez I see now that in my notebook i only really considered $r \in R_{+}$ thanks for pointing that out! What i meant implicitly by distinct path is distinct with regard to points other than x and y. – Saal Hardali Dec 02 '13 at 20:49
  • If you restrict to $r>0$ all the maps have exactly the same image. (It is never a good idea to «mean something implicitly»: be explicit.) – Mariano Suárez-Álvarez Dec 02 '13 at 20:49
  • @MarianoSuárez-Alvarez There must be something fundamental I’m not seeing here because $f^{1/2}(t)$ can't possibly be a straight line. Could you please write a more detailed answer explaining this? – Saal Hardali Dec 02 '13 at 20:59
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    I don't know why it cannot possibly be a straight line, since it is :-) I suggest you make pictures of the images of your functions $f^r$ for various values of $r$ (fix $x=(0,0)$ and $y=(1,0)$, for example) – Mariano Suárez-Álvarez Dec 02 '13 at 21:01
  • @MarianoSuárez-Alvarez I see now! thanks for making me feel retarded :-) – Saal Hardali Dec 02 '13 at 21:04

1 Answers1

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This proof fails since as Mariano in the comments pointed out:

For $x=(0,0)$ and $y=(0,1)$, $f^r(t) = (0,t^r)$ which is a straight line (and obviously intersects the other paths).

A great sound proof of this statement can be found here.

ADDED: Mariano found a way to fix the argument:

instead of step (2): Let $v$ be a non zero vector orthogonal to $y-x$ define $f^r$ as follows: $$f^r(t)=x+t(y-x)+rt(1-t)v \space \space \space \space \space r \in (0,1)$$

It follows that if $f^{r_1}(t)=f^{r_2}(t)$ for some $x,y \in \mathbb{R}^2$ and some $t,r_1,r_2 \in(0,1)$ then:

$$x+t(y-x)+r_1t(1-t)v = x+t(y-x)+r_2t(1-t)v$$

$$\implies (y-x)+r_1(1-t)v = (y-x)+r_2(1-t)v$$

By orthogonality of $v$ and $y-x$ $\implies r_1(1-t)=r_2(1-t)$

$$\implies r_1=r_2$$

Namely we've shown that if two paths intersect at one point (other than $t=0,1$) then they correspond to the same parameter $r$.

instead of step (4): show that $f^r$ is bijective with (0,1) and hence uncountable.

change in step (6): $r \in (0,1)$

Community
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Saal Hardali
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    You can fix your argument! For example: let $v$ be a non-zero vector orthogonal to $y-x$, and consider the functions $f^r:t\in[0,1]\mapsto x+t(y-x)+r t(1-t)v\in\mathbb R^2$, for each $r\in(0,1)$. – Mariano Suárez-Álvarez Dec 02 '13 at 21:27
  • @MarianoSuárez-Alvarez Fixed! – Saal Hardali Dec 02 '13 at 22:01
  • @SaalHardali : I don't understand one thing , from $x+t(y-x) + r_1t(1-t)v=x+t(y-x)+r_2t(1-t)v$ , by cancellation of addition of vectors , doesn't it follow that $r_1t(1-t)v=r_2t(1-t)v$ and $v$ is non-null , so $r_1t(1-t)=r_2t(1-t)$ , and if $t \in (0,1)$ then $t(1-t) \ne 0$ , so $r_1=r_2$ ? Am I missing something ? Why the orthogonality of $\theta \ne v$ with $y-x$ is needed to deduce this $r_1=r_2$ ? –  May 09 '15 at 16:35
  • Orthogonality isn't really needed only linear dependence. – Saal Hardali May 09 '15 at 17:28
  • @SaalHardali : I can feel that linear independence is needed , we need a vector $v$ that is not a scalar multiple of $x-y$ (otherwise $\mathbb R \setminus A$ also would be path connected ) , but where is it needed in the proof , that I cannot understand .... –  May 10 '15 at 08:01
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    @SaalHardali : Ah huh ! that shouldn't both be $t$ on both sides of your equation , that should be $x+t_1(y-x)+r_1t_1(1-t_1)v=x+t_2(y-x)+r_2t_2(1-t_2)v$ for some $t_1,t_2 \in (0,1)$from which $(t_1-t_2)(y-x)+(r_1t_1(1-t_1) - r_2t_2(1-t_2) ) v=0$ now since $v , y-x$ are linearly independent , so $t_1=t_2$ and $r_1t_1(1-t_1)=r_2t_2(1-t_2)=r_2t_1(1-t_1)$ , so $r_1=r_2$ –  May 12 '15 at 04:08