# Subsets of a Cartesian product are disjoint iff there exist projections that are disjoint

The Theorem?

Suppose $\{X_\alpha\}_{\alpha \in J}$ is a family of non-empty sets. Let $X = \prod_{\alpha \in J} X_\alpha$. For each $\alpha \in J$ define $p_\alpha : X \to X_\alpha$ to be the canonical projection from $X$ to $X_\alpha$.

Suppose $U,V$ are subsets of $X$. I want to show that $U$ and $V$ are disjoint if and only if there exists $\beta \in J$ such that $p_\beta(U) \cap p_\beta(V) = \emptyset$.

Background

I wanted to use this "theorem" to show that if the product space $X$ is $T_4$ then so is each $X_\alpha$.

My Thoughts

The $\Longleftarrow$ direction seems straightforward from the definition of the Cartesian product, although I haven't worked it out precisely since it is really $\Longrightarrow$ I care about. Likewise, it seems intuitively true that $\Longrightarrow$ is true. I've done a bit of googling and searching through textbooks, but I haven't quite found what I am looking for. Moreover, I am unsure how to proceed with a proof. Here is my problem:

I want to say that if $U$ and $V$ are disjoint subsets of $X$, then I can write $U$ is a product of subsets of $X_\alpha$ for each $\alpha \in J$, but this can't be correct. For example, $U$ could be the union of subsets of $X$ which cannot, in general, be written as a product of a union of subsets.

• I think I have a counterexample with $J={1,2}$ and $X_1=X_2=$0,1$$. Are you sure it is true? – Hoseyn Heydari Nov 6 '15 at 21:05

Simple counterexample: Let $J = \{0,1\}, X_0 = X_1 = \{0,1\}$; let $U = \{(0,0), (1,1)\}$, and let $V = \{(0,1)\}$. Then $p_0(U) = p_1(U) = \{0,1\}$, $p_0(V) = \{0\}$, and $p_1(V) = \{1\}$, but $U \cap V = \emptyset$.
To prove your theorem, it's easier to use that every $X_\alpha$ embeds as a closed subset of the product $\prod_\alpha X_\alpha$ (using that all spaces are $T_1$). Do you see how? (think of the two-dimensional case first).