3

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.

Martin Sleziak
  • 53,687
user73957
  • 311

2 Answers2

2

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.
Berci
  • 90,745
  • but doesn't $(\mathbb{R}\times \mathbb{R})\setminus ([0,1]\times[0,1])=\mathbb{R}\setminus[0,1]\times \mathbb{R}\setminus[0,1]$? – user73957 Apr 24 '13 at 12:49
  • 1
    No, this is a plane with an infinite thickened 'cross' removed. So, in particular it has 4 path components. – Dan Rust Apr 24 '13 at 13:52
  • @DanielRust Ah, yes sorry I was getting confused thanks very much. – user73957 Apr 24 '13 at 14:25
  • @DanielRust do we need the extra assumption that $Y$ be compact? otherwise could we not consider the closed set $Z={(x,y)|\frac{y}{x}=1}$then $P(Z)=(-\infty,0)\cup (0,\infty)$ which is not closed? – user73957 Apr 24 '13 at 14:35
  • But the projection is not closed, it maps the graph of $x\mapsto 1/x$ to the positive real numbers. It is closed, however, if $Y$ is compact. – Stefan Hamcke Apr 24 '13 at 15:11
  • @StefanH. yeah that is what I have said in my comment – user73957 Apr 24 '13 at 15:16
  • @user73957: Oh, sorry, I didn't read all the comments. You are right, $Y$ should be compact. However, your $Z$ is just the diagonal minus the origin, which is not closed. Maybe just a typo, as you just have to switch $1$ and $y$ in the formula. By the way, there is sort of a converse of the statement: $Y$ is compact iff for all spaces $X$ the projection $Y\times X\to X$ is closed. – Stefan Hamcke Apr 24 '13 at 15:24
1

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 $\mathbb R$.

enter image description here

(Sorry the picture i draw is not good.)

Paul
  • 20,553