Prove that $A=\{(x,y)\in\mathbb{R}:(y\neq 0)\vee (x>0)\}$ is connected.

Question

I'm working on a multivariable integration problem and, at some point, I need to use Poincaré Lemma. To justify that I verify the Lemma's hypothesis conditions, I need to prove that the set $$A=\{(x,y)\in\mathbb{R}:(y\neq 0)\vee (x>0)\}$$ is connected. In fact, proving it's simply connected will work as well for my excercise, I don't know wich way is easier, and in any case how to prove it cannot be represented as the union of two or more disjoint non-empty open subsets. I will thank any help or advice.

Answer 1

Here's a proof that the space $A$ is connected. In fact we'll show it's path-connected by explicitly exhibiting a continuous path from an arbitrary point to the point $(1,0)$. To exhibit a path between two arbitrary points in $A$, simply concatenate their paths to $(1,0)$.

Consider some $(x,y) \in A$. Let $\gamma(t) = ( (1-t)x + t, (1-t)y)$ be the straight line connecting $(x,y)$ and $(1,0)$.. To see that $\gamma(t)\in A$ for all $t\in[0,1]$, first observe that $\gamma(t) = (1,0)\in A$. Now consider the following cases:

(I) If $y > 0$, then $(1-t)y > 0$ so $\gamma(t) \in A$ for all $t\in [0,1[.$

(II) If $y < 0$, then certainly $(1-t)y < 0$ for all $t\in [0,1[$.

(III) If $y = 0$ and $x > 0$, then $(1-t)y = 0$ and $(1-t)x + t > (1-t)x > 0$ for all $t\in[0,1[$.

In all cases $\gamma(t)\in A$ for all $t\in [0,1]$. Thus we have shown that any point in $A$ can be connected with $(1,0)$ by a straight line. (Such a space is called "star-shaped.")

Answer 2

The space you presented is just $\mathbb{R}^2\setminus \{ (x,0) : x\leq0 \}$. That is you are removing from the real plane the half line starting at the origin and going through the negative part of the $x$ axis.

You can easily show it is path connected, thus it is connected. It is also simply connected but from the phrasing of your questions I would think you are not entirely familiar with this concept, am I right?

Simply connected means that the first homotopy group of the space is trivial. If you do not know the definition of this group: this is equivalent to the fact that every loop in your space can be contracted to its basepoint.