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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?

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  • $\begingroup$ HINT: Try to draw a picture where $X,Y$ are real intervals, and $A,B$ are real subintervals. What do you see? $\endgroup$ – Crostul Nov 12 '16 at 17:43
  • $\begingroup$ @user160738: $A$ is a proper subset of $X$ and $B$ is a proper subset of $Y$. $\endgroup$ – user170039 Nov 12 '16 at 17:48
  • $\begingroup$ @Crostul: I am still lost. Can you elaborate? $\endgroup$ – user170039 Nov 13 '16 at 7:15
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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.

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  • $\begingroup$ I understand that if $\{x_0\}\times Y\subseteq (X\times Y)\setminus (A\times B)$ then (since $\{x_0\}\times Y$ is path connected) there exists a path from $(x_0,y_0)$ to $(x_0,y_1)$. But I don't understand how to get a path from $(x_0,y_1)$ to $(x_1,y_1)$. Can you elaborate? $\endgroup$ – user170039 Nov 13 '16 at 4:31
  • $\begingroup$ @user170039: I have repaired my answer, thanks for catching my error. $\endgroup$ – copper.hat Nov 13 '16 at 18:09
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enter image description here

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.

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