The ordered square is connected but not path connected.

Question

consider the following proof , from Munkres . My question is why I define $U_x$ using $x\times (0,1)$ and for example not other open set such as $x\times (0,\frac{1}{2})$ or $x\times (\frac{1}{3},\frac{1}{2})$ ?

Answer 1

It looks like any open subset $V$ of $I$ could take the place of $(0,1)$ and the construction would still work.

• $f^{-1}(x \times V)$ is nonempty by the intermediate value theorem
• $f^{-1}(x \times V)$ is open because $f$ is continuous
• $f^{-1}(x \times V) \cap f^{-1}(y \times V) = \emptyset$ whenever $x \neq y$.
• $f^{-1}(x \times V)$ contains a rational number because the rational elements of $I$ are dense in $I$.

So it's a matter of style. The question becomes, "what's your favorite open subset of $I$, if you had to pick one?" One good candidate would be the interior of $I$, or the largest open subset of $I$, which is $(0,1)$. If the author had chosen $(0,\tfrac{1}{2})$, for example, the reader might have wondered if it was necessary to exclude certain points.