Projection of a closed subspace

Question

If we have two topologies $(X,\mathcal{T})$ and $(Y,\mathcal{U})$, then we may take the product topology. We define the projection map $\prod_x$ in the usual way.

If $A\subset X\times Y$ is closed, is $\prod_x(A)$ closed in $X$? I think that it is, as:

$A=(X\times Y)\setminus (\cup_\lambda T_\lambda \times U_\lambda=(X\setminus \cup_\lambda T_\lambda \times Y\setminus \cup_\lambda U_\lambda)$ so that:

$\prod_x(A)=\prod_x((X\setminus \cup_\lambda T_\lambda \times Y\setminus \cup_\lambda U_\lambda)))=X\setminus \cup_\lambda T_\lambda$ which is closed in $X$. Is this correct?

Cheers.

Answer 1

No, we don't have $A=(X\times Y)\setminus(\bigcup_\lambda T_\lambda\times U_\lambda)=R\times S$ in general for any sets $R,S$: think about e.g. $A=(\Bbb R\times\Bbb R)\setminus (0,1)\times (0,1)\ \subseteq \Bbb R\times\Bbb R$.

However,

1. $\Pi_X$ is an open mapping (sending open sets to open sets),
2. and it is surjective, and therefore it sends closed sets to closed sets.

Answer 2

It may be helpful for you, however I'm not sure.

The example shows that the projections are not closed:

The projection $p: \mathbb R^2 \rightarrow \mathbb R$ of the plane $ \mathbb R^2 $ onto the $x$-axis is not closed. Indeed, the set $F=\{(x,y)\in \mathbb R^2 : xy=1\}$ is closed in $\mathbb R^2 $ and yet its image $p(F)= \mathbb R \setminus \{0\}$ is not closed in $\mathbb R$.

(Sorry the picture i draw is not good.)