Talk:Projection from Product Topology is Open

I don't understand. What further justification is needed? The product topology is defined such that if $U_1$ and $U_2$ are open in each of $T_1$ and $T_2$, then $U_1 \times U_2$ is open in $T_1 \times T_2$ by definition. Therefore if $U = U_1 \times U_2$ is open then the first projection of $U$ is open. Okay, so I lost the unnecessary and confusing "$= U_1$" in the statement, but what else am I missing? I need help here. --prime mover (talk) 15:26, 7 October 2012 (UTC)


 * Well, not all open sets of $T$ are of the form $U_1 \times U_2$ where $U_1 \in \vartheta_1$ and $U_2 \in \vartheta_2$. Adding the necessary justification is not difficult; I just have other important things to do right now. --abcxyz (talk) 15:33, 7 October 2012 (UTC)


 * Afraid I still don't understand. A set in $T$ is defined as open iff it is of the form $U_1 \times U_2$ where $U_1 \in \vartheta_1$ and $U_2 \in \vartheta_2$. Any set which is not such a product is not open. Again, there's something I'm missing here. Feel free to wait until you have done whatever it is that's important, no hurry. --prime mover (talk) 15:36, 7 October 2012 (UTC)


 * The sets of the form $U_1 \times U_2$ form a basis for the topology on $T$. You can just check the definition page; I think you'll see what I mean. --abcxyz (talk) 15:40, 7 October 2012 (UTC)


 * As I suggested on another talk page, I'm out of my depth. I'm going to have to take up a different hobby. --prime mover (talk) 18:58, 7 October 2012 (UTC)


 * Well, let's see how you feel tomorrow. --abcxyz (talk) 20:54, 7 October 2012 (UTC)


 * Quick response: consider a circle without its boundary, or for that matter just an L-shaped region without its boundary. These would be open under the product topology (they are the union of open sets), but are not themselves representable in the form $U_1 \times U_2$.  --Alec  (talk) 04:07, 8 October 2012 (UTC)


 * I suppose you mean an open disk in $\R \times \R$? --abcxyz (talk) 04:24, 8 October 2012 (UTC)