What is {empty set} x something with the product topology?

Question

The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies. From what I understand, then $\{\emptyset\}\times]a,b[$ is an open set of $\mathbb{R}^2$ with the product topology, right? But what is this geometrically? I can't see what this is.

Answer 1

By definition, the elements of $A\times B$ (for sets $A$ and $B$) are the ordered pairs $\langle a,b\rangle$ such that $a\in A$ and $b\in B.$ Since $a\in\emptyset$ is impossible, then $\emptyset\times B$ has no elements, meaning $\emptyset\times B=\emptyset,$ regardless of the set $B.$

The open sets of $\Bbb R^2$ in the product topology can be obtained from the basis $\{A\times B:A,B\text{ are open subsets of }\Bbb R\}.$ This will include elements of the form $\emptyset\times B=\emptyset,$ as well as $A\times\emptyset=\emptyset.$ However, it will not have any elements of the form $\{\emptyset\}\times B,$ since $\emptyset$ isn't an element of $\Bbb R.$

Answer 2

Instead of $\{\emptyset\}$, I'll consider $\{*\}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that $\{*\}$ is a singleton as a set.

Let $X$ be a topological space. Then, basis for topology of $\{*\}\times X$ is given by $\{ \{*\}\times U\mid U\subseteq X\ \text{open}\}$.

Define $f\colon X\to \{*\}\times X$ by $f(x) = (*,x)$ and $g\colon \{*\}\times X\to X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $U\subseteq X$ open we have

$$f^{-1}(\{*\}\times U) = U,\ g^{-1}(U) = \{*\}\times U.$$

Thus, $X$ and $\{*\}\times X$ are homeomorphic.

If $X = \mathbb R$, instead of $*$ choose some real number $a$. Then $\{a\}\times\mathbb R$ is naturally a line embedded in $\mathbb R^2$. Here are examples for some different $a$'s: