I apologize if this seems trivial, would still appreciate help.


• $X=\underset{\alpha\in I}{\prod}X_{\alpha}$ be the cartesian product of sets $\left\{ X_{\alpha}\right\} _{\alpha\in I}$ .

• $U=\underset{\alpha\in I}{\prod}U_{\alpha}$ s.t $\forall\alpha\in I$ : $U_{\alpha}\subseteq X_{\alpha}$ .

• $\pi_{\beta}:X\to X_{\beta} , \left(x_{\alpha}\right)_{\alpha\in I}\mapsto x_{\beta}$ be projection from $X$ to $X_{\beta}$ for $\beta\in I$ .

I want to show that $U=\underset{\alpha\in I}{\bigcap}\pi_{\alpha}^{-1}\left(\pi_{\alpha}\left(U\right)\right)$ :

• $\subseteq$: Let $u\in U$ and $\alpha\in I$ . By image definition $\pi_{\alpha}\left(u\right)\in\pi_{\alpha}\left(U\right)$. $u$ is an input of $\pi_{\alpha}\left(u\right)$ , therefore by preimage definition $u\in\pi_{\alpha}^{-1}\left(\pi_{\alpha}\left(U\right)\right)$ .(I hope this direction is fine)

• $\supseteq$: This is the part I have a problem with. Would appreciate direction. Thanks!


Your proof of $U \subseteq \bigcap_\alpha \pi_\alpha^{-1}\bigl[\pi_\alpha[U]\bigr]$ is fine.

For the other direction: Suppose that $x \in \bigcap_\alpha \pi_\alpha^{-1}\bigl[\pi_\alpha[U]\bigr]$, then for each $\alpha$ we have $x \in \pi_\alpha^{-1}\bigl[\pi_\alpha[U]\bigr]$, that is, by definition of the preimage, $\pi_\alpha(x) \in \pi_\alpha[U]$. Now, by definition of $U$, we have $\pi_\alpha[U] = U_\alpha$, hence $x_\alpha \in U_\alpha$ for every $\alpha$. Therefore, $x \in U$.

  • $\begingroup$ Appreciate the help. Thanks a lot! $\endgroup$ – Ungoliant Nov 30 '15 at 5:14

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