Proof that the interval and the plane is not homeomorphic

I tried to prove this by contradiction. I assumed that there is a homeomorphism $f: (0,1) \longrightarrow \mathbb{R^2}$. A submap $ab(f): (0,1)\setminus\{\frac{1}{2}\} \longrightarrow \mathbb{R^2}\setminus\{f(\frac{1}{2})\}$ (it's also homeomorphism). It is obvious that the first space is not connected space, but I very confused about $\mathbb{R^2}\setminus\{f(\frac{1}{2})\}$. As far as I understand, it's something like a plane with a hole, but I can't prove that this space is connected.

I would be pleased if you help me with proof of this fact or if you give me some hints about my original question.

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Do you mean $\setminus$ (\setminus) or $/$ (/)? –  Asaf Karagila Feb 14 '13 at 14:33
Yes, I mean (\setminus), I will correct it. –  Oiale Feb 14 '13 at 14:35
What is harder, though, is showing that $2$-space and $3$-space are not homeomorphic. May require some algebraic topology, or at least the Jordan curve theorem. –  GEdgar Feb 14 '13 at 14:39
This is just one point you take away, try to show that the remaining space is pathconencted –  Stefan Feb 14 '13 at 14:54
If it helps, since there is a self-homeomorphism of $\mathbb{R}^2$ taking one point to any other, you can assume wlog $f\left(\frac{1}{2}\right) = (0,0)$. –  Jason DeVito Feb 14 '13 at 15:17

You should know that path-connectedness implies connectedness. Here's a sketch of the proof that there is always a path between two points in $\mathbb{R}^2 \setminus \{0\}$. I choose to remove $0$ for ease of notation, but this argument runs the exact same way if we remove $f(1/2)$. Of course, you could have removed $f^{-1}(0)$ from $(0,1)$, and then you do end up with $\mathbb{R}^2 \setminus \{0\}$. Regardless, let's get to the proof.

Take any two points $x,y \in \mathbb{R}^2 \setminus \{0\}$. Using polar coordinates, we can write $x = (r_1, \theta_1)$ and $y = (r_2,\theta_2)$, and without loss of generality $r_{2} \geq r_{1}$. First consider the path

$$\gamma_{1}(t) = (r_{1}, \theta_{2}t + (1-t)\theta_{1})$$

Since $r_{1} \neq 0$, this path never passes through the origin. Note that this path starts at $x = (r_{1},\theta_{1})$ and ends at $(r_{1},\theta_2)$. Then let

$$\gamma_{2}(t) = (r_{2}t + (1-t)r_{1},\theta_{2})$$

Since $r_{2} \geq r_{1}$, $r_{2}t + (1-t)r_{1}\geq r_{1} > 0$, so this path never passes through the origin. Moreover, it starts at $(r_{1},\theta_{2})$ and ends at $(r_{2},\theta_{2}) = y$.

If you concatenate $\gamma_{2}$ and $\gamma_{1}$, that is if you run along $\gamma_{1}$ at twice the speed and then run along $\gamma_{2}$ at twice the speed, you get a new path $\gamma$ that goes from $x$ to $y$ without touching the origin.

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Well you can show, that there is always a path connecting two points.

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Here are some non-traditional arguments, which hopefully will enlighten you why path-connectedness is so much more elementary to use in a situation like this.

Eventhough it could seem easy to show that $\mathbb{R}^{2}\setminus\{f(\frac{1}{2})\}$ is connected by using the fact that $\mathbb{R}^{2}$ is connected, a straightforward attack might fail as one somehow needs to use the dimension of $\mathbb{R}^{2}$. And this might be difficult. Too generic arguments do not lead anywhere as for example $\mathbb{R}^{1}\setminus\{x\}$ is not connected eventhough $\mathbb{R}^{1}$ is. One could proceed e.g. by any of the following ways.

1. There is a result (e.g. in Hurewicz, Wallman - Dimension Theory, 1948, Theorem IV 4.) which states that $\mathbb{R}^{n}$ (or in fact any $n$-dimensional manifold) cannot be disconnected by a subset of topological dimension $\leq n-2$. The singleton $\{f(\frac{1}{2})\}$ is zero dimensional.

2. Another result in the same book states that if $U\subseteq \mathbb{R}^{n}$ is non-empty, open and not dense (i.e. its complement contains an open set), then the topological dimension of the boundary of $U$ is precisely $n-1$ (Corollary 2 of Theorem IV 3). Now if $\mathbb{R}^{2}\setminus\{f(\frac{1}{2})\}$ was disconnected, then since it is open in $\mathbb{R}^{2}$, there exists non-empty disjoint open sets $U_{1},U_{2}\subseteq \mathbb{R}^{2}$ with $U_{1}\cup U_{2}=\mathbb{R}^{2}\setminus\{f(\frac{1}{2})\}$. Show that the boundary of each $U_{i}$ is either empty or equals the singleton $\{f(\frac{1}{2})\}$ and derive a contradiction with the above result.

These are just to show what one could do with plain connectedness arguments. But ofcourse, using path-connectedness is much more easier in this case, as any two points in $\mathbb{R}^{2}\setminus\{f(\frac{1}{2})\}$ can be connected with a continuous path that is a proper composition of lines going through origin and a rotation. Use the fact that path connectedness implies connectedness.

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Let $I$ be an interval. Then the only simple closed curves in $I$ are trivial (since a one-to-one continuous function from $[0,1]$ into $I$ is necessarily monotonic) whereas there are many such curves on the plane (eg. a circle).