# path connected subspaces of $\mathbb R^2$

I am trying to prove two statements that visually I think they are "obvious", but I am totally lost when it comes to do a formal proof. The statement of the exercise is: Decide whether $\mathbb R^2 \setminus \{(0,0)\}$ and $\mathbb R^2 \setminus \{(x,0)\mid x \in \mathbb R\}$ are path connected.

Visually, this is what I see : removing a point from the plane doesn't alterate much. I mean, the only problem I would have is with the points $x$,$y$ that live in a line that also contains the point $(0,0)$. That being the case, I could define a curve joining those two points without passing through $(0,0)$.

If I remove a line from the plane, then I am in trouble. This line divides the plane into two regions. If I take a point $x$ that lives in one part and another point $y$ that is contained in the other side, then there is no continuous function that can join me those two points. I have no idea how to prove it without the "hand-waving".

Also, I have a doubt when it comes to generalizing this idea to $\mathbb R^n$. I suppose that removing a hyperplane is what affects path-connectedness, but I'm not so sure, I am just generalizing from the case of the real line and the plane. Is there any good topology textbook that treats this topic of metric spaces?

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To show that removing a line disconnects the plane, it's probably easiest just to argue that there is a partition of the resulting space into two open sets. –  Grumpy Parsnip Oct 12 '13 at 2:11
Yes, indeed. So the argument would be the space is disconnected, so is not path-connected, right? –  user100106 Oct 12 '13 at 2:15
Yes, that is correct. –  dfeuer Oct 12 '13 at 2:16
From people's help, you can see that a path-connected space must be connected at least. Can you prove that? A hint would be using continuity of the map of your path. –  Secret Math Oct 12 '13 at 2:21

Any path connected space is connected. Show that the two sets $\{ (x, y) : y \gt 0 \}$ and $\{(x,y) : y \lt 0 \}$, are disjoint open sets whose union is the whole space. That is the definition of a disconnected space. Since it's not connected, it can't be path connected, since that would lead to a contradiction.
To show formally that any line dividing $\mathbb{R}^2$ results in a disconnected space. Consider the homeomorphism $z = x + yi \mapsto e^{i\theta}z + z_0$. It maps lines to any other line given the angle to the new line and the proper offset. Since $\mathbb{R}^2$ is homeomorphic to $\mathbb{C}$ you have a sequence of homeomorphisms from your cut space to a cut space with an arbitrary line, so any line cutting $\mathbb{R}^2$ results in a disconnected space.
The fact that $\mathbb R^2 \setminus \{(x,0)\mid x \in \mathbb R\}$ is not path connected follows trivially from the intermediate value theorem.
Furthermore, we can generalize this to an arbitrary line by re-aligning the line. For a given line $\mathbb L$ partitioning $\mathbb R^2$ construct a homeomorphism which sends $\mathbb L$ to $\{(x,0)\mid x \in \mathbb R\}$. The result follows from the fact that homeomorphisms preserve connectedness.