Prove or disprove that $(X\times Y)\setminus(A\times B)$ is path connected

Question

While trying to prove that $\mathbb{R}^2\setminus\{(0,0)\}$ is path connected. I was wondering whether the following proposition is true,

Let $X$ and $Y$ be two path connected spaces and let $A,B$ be two proper subsets of $X$ and $Y$ respectively then $(X\times Y)\setminus(A\times B)$ is path connected.

I know that if $X$ and $Y$ are connected then for any two proper subsets $A,B$ of $X$ and $Y$ respectively then $(X\times Y)\setminus(A\times B)$ is connected.

But I don't know how to proceed regarding this problem.

Can anyone help?

Answer 1

Pick $x_0 \in X \setminus A, y_0 \in Y \setminus B$, note that $\{x_0\} \times Y \subset (X\times Y) \setminus (A \times B)$ and $X \times \{y_0\} \subset (X\times Y) \setminus (A \times B)$.

Now pick $(x_1,y_1) \in (X\times Y) \setminus (A \times B)$.

The either $\{x_1\} \times Y \subset (X\times Y) \setminus (A \times B)$ or $X \times \{y_1\} \subset (X\times Y) \setminus (A \times B)$. (If neither is true, we must have $(x_1,y_1) \in A \times B$.)

Hence there is a path $(x_0,y_0) \to (x_0,y_1) \to (x_1,y_1)$ or $(x_0,y_0) \to (x_1,y_0) \to (x_1,y_1)$.

Since there is a path from $(x_0,y_0)$ to any $(x_1,y_1) \in (X\times Y) \setminus (A \times B)$, we see that there is a path joining any two points in $(X\times Y) \setminus (A \times B)$, hence it is path connected.

Answer 2

This picture shows how to build a path from $(x_1,y_1)$ to $(x_2,y_2)$ in the case when $x_1, x_2 \notin A$ and $y_1, y_2 \notin B$. I leave you to conclude with the details.