# Connected Sets in this Topology of $\mathbb{R}^2?$

Consider the topology on $\mathbb{R}^2$ given by $$\mathscr{T} = \{A \subset \mathbb{R}^2 \, ; \, \pi_y(A) = A_y \, \, \text{is open} \, \}$$ for the projection map $\pi_y: \mathbb{R}^2 \to \mathbb{R}$ in the $y$-coordinate.

How do we characterize open connected subsets of $\mathbb{R}^2$ in this topology? Clearly a set like $\{1\} \times (0,1)$ is connected, while a set $(0,1) \times (0,1)$ or $\{0,1\} \times (0,1)$ is not connected.

So, my guess is that, since $\pi_y^{-1} (A_y) = \mathbb{R} \times A_y$, given a set $X$ with more than one $y$-value, such that $\pi_y(X)$ is open, the union $$X \subset \dot\bigcup_{x \in \mathbb{R}} M_x$$ of open sets of the form $M_x = \{x\} \times A_y$ with $x \in X$ are disjoint and contain $X$. Thus, I believe that open connected sets in this topology are those sets that have only one $x$ coordinate and $\pi_y(X)$ is connected.

Is this correct?

Your collection of sets is not a topology. The sets $$\{(|y|,y)\mid -1<y<1\}$$ and $$\{(-|y|,y)\mid -1<y<1\}$$ would be open, but their intersection $\{(0,0)\}$ would not be open.
In general, if $f:X\to Y$ is a function from a set to a topological space, then $$\{A\subseteq X\mid f(A)\text{ is open in }Y\}$$ does not form a topology.
• Perhaps I messed up then, because the original condition was that $\mathscr{T} =\{ A\subset \mathbb{R}^2; \ \pi_y(A\cap (\{x\} \times \mathbb{R})) \ \mbox{is an open subset of} \ \mathbb{R}\}$ Commented Mar 1, 2015 at 22:24
• What is the $x$ in your definition? Some fixed value? Commented Mar 1, 2015 at 22:25
• An element of $\mathbb{R}$ Commented Mar 1, 2015 at 22:26
• is $\pi_y (A \cap (\{x\} \times \mathbb{R})) \neq A_y?$ Commented Mar 1, 2015 at 22:27
• That set could also be written as $$\{A\subseteq\Bbb R^2\mid A\cap(\{x\}\times\Bbb R)\text{ is open in }\{x\}\times\Bbb R\}$$ Then it is a special case of a final topology. It's the final topology for the map $\Bbb R\to \Bbb R^2,\ y\mapsto (x,y)$. Commented Mar 1, 2015 at 22:28