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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.

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

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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).

The "fact" is in fact false. Of course, if two sets do intersect they will intersect in all factors, by definition. So if they are disjoint in some factor, they are in the product. It's quite easy to find examples in the plane of disjoint sets, whose two projections do intersect on both axes. So that's not to way to go, see before.

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