22-08-2024

Problem: Consider the standard topology on $\mathbb{R} ^2$. Prove that $\mathbb{R} ^2 \setminus \{ (a,b) \} $ is path-connected and hence, connected.
Solution: We need to show that the space $\mathbb{R} ^2 - \{ (a,b) \} $ is path connected. Let us denote $P = (a,b)$. Take two points $A_1 = \left( x_1, y_1 \right) $ and $A_2 = \left( x_2, y_2 \right) $. We will show that there is a path joining $A_1$ and $A_2$ in $\mathbb{R} ^2 - \{ P \} $. Consider the line $L$ joining $A_1$ and $P$. There could be two possibilities, first the point $A_2$ does not lie on the line $L$ and second it lies on the line $L$. In both the cases we will create pats joining $A_1$ and $A_2$ which does not pass through $P$.
  • The point $A_2$ does not lie on the line $L$. Then join $A_1$ and $A_2$ by any line segment (see figure below).
    solutions of the differential equation

  • If $A_2$ lies on the line $L$, then take any point $Q$ which does not lie on the line $L$. Consider the broken line segment $A_1 Q$ and $QA_2$, which is a path in $\mathbb{R} ^2 \setminus \{ P \} $ joining $A_1$ and $A_2$.
    solutions of the differential equation
Therefore, combining the two cases, we proved that the space $\mathbb{R} ^2 \setminus \{ P \} $ is path connected and hence connected.