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?

  • 2
    $\begingroup$ 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. $\endgroup$ – Cheerful Parsnip Oct 12 '13 at 2:11
  • $\begingroup$ Yes, indeed. So the argument would be the space is disconnected, so is not path-connected, right? $\endgroup$ – user100106 Oct 12 '13 at 2:15
  • $\begingroup$ Yes, that is correct. $\endgroup$ – dfeuer Oct 12 '13 at 2:16
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.