If $X$ $\times$ $Y$ is path connected so is $X$ $\times$ $Y$ - $(x_0,y_0)$ for some $(x_0,y_0)$

Question

I want to prove the following

Let $X$ $\times$ $Y$ be path-connected show that $X$ $\times$ $Y$ $\setminus$ $(x_0,y_0)$ is still path connected for some point $(x_0,y_0)$

My attempt was by considering the infinite tangents that can be drawn for each point so that it is possible to construct a path without intersecting $(x_0,y_0)$ (by drawing perpendicular if $(x_0,y_0)$ is directly in the way of the two points) This seems to work for $\mathbb{R}^2$ but I don't know about arbitrary product spaces

But this does not sound rigorous enough is this idea/sketch even correct, is there a more rigorous way to solve this?

Any ideas or hints are appreciated

Thanks!

Answer 1

Let $X$ and $Y$ be path-connected, and suppose that $X$ and $Y$ each have at least two elements. Show that $Z=X\times Y-\{(x,y)\}$ is path-connected.

My guess is that this is the actual question.

The answer is easy:

$\\$

Notice that every set of the form $X\times \{y'\}$ or $\{x'\}\times Y$ is path-connected, for $x'\neq x$ and $y' \neq y$.

Let's say you want to connect $(b,c),(d,e)\in Z$ by a path. I will deal with one of the cases and you can verify the rest.

If $b,d\neq x$ and $c,e\neq y$, then connect $(b,c)$ to $(d,c)$ and then connect $(d,c)$ to $(d,e)$.