# [SOLVED] Finally in LaTex: Pre-images and indexed set.

• Oct 11th 2008, 07:40 PM
ilikedmath
[SOLVED] Finally in LaTex: Pre-images and indexed set.
Let f be a function from X into Y.

My thinking so far (which isn't a lot because I still don't have a grasp on this concept of indexed sets):

So how I'm understanding this is that the question is asking me to prove that the pre-image (and not the inverse function!) of the intersection of a certain family of subsets of Y (the set {
$\displaystyle B_\alpha$}) is the intersection of the pre-images of $\displaystyle B_\alpha$ .

The pre-image is in the domain. In this case the domain is X. So I have to show that the intersection of the pre-images of
$\displaystyle B_\alpha$ is the same as the pre-image of the intersection of all sets $\displaystyle B_\alpha$.

Is that even close?
• Oct 12th 2008, 05:20 AM
Plato
$\displaystyle u \in \left( { \cap _\alpha f^{ - 1} (B_\alpha )} \right) \Leftrightarrow \left( {\forall \alpha \in A} \right)\left[ {u \in f^{ - 1} (B_\alpha )} \right]$
$\displaystyle \begin{gathered} \Leftrightarrow \left( {\forall \alpha \in A} \right)\left[ {f(u) \in (B_\alpha )} \right] \hfill \\ \Leftrightarrow \left[ {f(u) \in \cap _\alpha (B_\alpha )} \right] \hfill \\ \Leftrightarrow u \in f^{ - 1} \left( { \cap _\alpha (B_\alpha )} \right) \hfill \\ \end{gathered}$

That is a start. Can you do the details?
• Oct 13th 2008, 07:10 AM
ilikedmath
Quote:

Originally Posted by Plato
$\displaystyle u \in \left( { \cap _\alpha f^{ - 1} (B_\alpha )} \right) \Leftrightarrow \left( {\forall \alpha \in A} \right)\left[ {u \in f^{ - 1} (B_\alpha )} \right]$
$\displaystyle \begin{gathered} \Leftrightarrow \left( {\forall \alpha \in A} \right)\left[ {f(u) \in (B_\alpha )} \right] \hfill \\ \Leftrightarrow \left[ {f(u) \in \cap _\alpha (B_\alpha )} \right] \hfill \\ \Leftrightarrow u \in f^{ - 1} \left( { \cap _\alpha (B_\alpha )} \right) \hfill \\ \end{gathered}$

That is a start. Can you do the details?

I think I can see it now. Thanks!