What are the connected components of $\Bbb R^2 \setminus \big(\{(x,y) \in \Bbb R^2, x^2+y^2=1\} \cup \{(x,y)\in \Bbb R^2, y=x^2+1\}\big)$?

Question

What are the connected components of $\Bbb R^2 \setminus \big(\{(x,y) \in \Bbb R^2, x^2+y^2=1\} \cup \{(x,y)\in \Bbb R^2, y=x^2+1\}\big)$ ?

I made a drawing and to me the connected components are:

• $\{(x,y)\in \Bbb R^2, y>x^2+1\}$
• $\{(x,y)\in \Bbb R^2,y^2+x^2 < 1\}$

Are there any others? How to prove it formally?

Answer 1

I think that you were meaning to find the connected components, and not the connected subsets of $S=\Bbb R^2 \setminus \big(\{(x,y) \in \Bbb R^2, x^2+y^2=1\} \cup \{(x,y)\in \Bbb R^2, y=x^2+1\}\big)$.

Connected components are the maximal (for set inclusion order) connected subsets.

If that was indeed your real question, then the connected components of $S$ are $S_1 =\{(x,y) \in \mathbb R^2, x^2+y^2 <1\}$ the open unit disk, $S_2=\{(x,y)\in \Bbb R^2, y>x^2+1\}$ and $S_3=\mathbb R^2 \setminus \big(\{(x,y) \in \Bbb R^2, x^2+y^2 \le 1\} \cup \{(x,y)\in \Bbb R^2, y \ge x^2+1\}\big)$.

You can verify that those sets are maximal connected subsets of $S$, disjoint and that their union is $S$.