How to prove the following

Let $I\neq\emptyset$ and $\{X_\alpha:\alpha\in I\}$ a family of nonempty sets. For each $\alpha\in I,$ let $A_\alpha\subset X_\alpha$ and let $F=\{\alpha\in I: A_\alpha\neq X_\alpha\}.$ Show that $\displaystyle\prod_{\alpha\in I}A_\alpha=\bigcap_{\alpha\in F} \pi _\alpha^{-1}(A_\alpha).$

I have no idea how to prove it, could anyone guide me please?

Maybe a hint? or maybe 2 hints?

Note I am not asking for the proof since I know this site doesn't work like that

  • $\begingroup$ What is $\pi_{\alpha}$? You never define it $\endgroup$ Oct 23 '17 at 4:03
  • $\begingroup$ A single term of the intersection is A_alpha × the product of all X's except X_alpha. $\endgroup$ Oct 23 '17 at 4:52
  • $\begingroup$ F is misleading you. $\endgroup$ Oct 23 '17 at 8:39

$$ \cap_{j\in I} \ pi_j^{-1}(A_\alpha) = \cap_{j\in I} \{ A_j × \prod_{k\in I-\{j\}} X_k \} = \prod_{j\in I} A_j $$


Let $f \in \prod_{\alpha \in I} A_\alpha$. This means (by definition) that $f: I \to \bigcup_{\alpha \in I} A_\alpha$ and $f(\alpha) \in A_\alpha$ for all $\alpha \in I$.

$\pi_\alpha(f) = f(\alpha)$ (by definition), so for each $\alpha$ we have that $\pi_\alpha(f) \in A_\alpha$< or equivalently $f \in \pi_\alpha^{-1}[A_\alpha]$, and as this holds for each $\alpha$, so a fortiori for $\alpha \in F$, we have that $f \in \bigcap_{\alpha \in F} \pi_\alpha^{-1}[A_\alpha]$. This shows one inclusion.

The reverse direction: let $f \in \bigcap_{\alpha \in F} \pi_\alpha^{-1}[A_\alpha]$. Consider $\alpha \in I$: If $\alpha \in F$ we we know that $f \in \pi_\alpha^{-1}[A_\alpha]$, so $f(\alpha) = \pi_\alpha(f) \in A_\alpha$; if $\alpha \notin F$ ,$A_\alpha = X_\alpha$ by definition of $F$, so it's trivial in that case (as $f \in \prod_{\alpha \in I} X_\alpha$) that indeed $f(\alpha) \in X_\alpha = A_\alpha$ also in that case. So $f \in \prod_{\alpha \in I} A_\alpha$.

If you unpack the definitions it's basically a tautology.

  • $\begingroup$ Just one little detail in the original exercise it was F here $\bigcap_{\alpha \in F} \pi_\alpha^{-1}[A_\alpha]$ instead of $I$, but It doesn't really matter right? The proof is the same right? $\endgroup$
    – user486983
    Oct 23 '17 at 21:47
  • $\begingroup$ William Elliot modified the title of my question and changed F by I $\endgroup$
    – user486983
    Oct 23 '17 at 21:50
  • $\begingroup$ why $f \in \prod_{\alpha \in I} X_\alpha$ ? $\endgroup$
    – user486983
    Oct 23 '17 at 22:34
  • $\begingroup$ Because both sides are subsets of it. It’s the total space, and we’re proving equality of two of its subsets.@Isabella $\endgroup$ Oct 24 '17 at 4:15
  • 1
    $\begingroup$ @Isabella indeed a typo. Usually I use $\alpha\in A$ or $i \in I$ as index sets and elements of it. Hence the confusion. $\endgroup$ Nov 5 '17 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy