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


*

*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.

*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.
A: 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.
A: 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).
