Proof: Intersection, Inverse Function

**lfroehli** · March 20th 2011, 09:08 PM

For this proof, we are to show that each containment is a subset of the other.

Part 1: $f^{-1}(\bigcap_{\lambda\in\Lambda}B_{\lambda}) = \bigcap_{\lambda\in\Lambda}f^{-1}(B_{\lambda})$ .

Part 2: $f^{-1}(\bigcup_{\lambda\in\Lambda}B_{\lambda}) = \bigcup_{\lambda\in\Lambda}f^{-1}(B_{\lambda})$ .

I honestly have no idea how to get started on this...any insight is appreciated!

**FernandoRevilla** · March 20th 2011, 11:14 PM

It is almost routine knowing the definition of $f^{-1}(B)$ . For example:

$x\in f^{-1}(\bigcap_{\lambda\in\Lambda}B_{\lambda})\Rightar row f(x)\in \bigcap_{\lambda\in\Lambda}B_{\lambda}\Rightarrow f(x)\in B_{\lambda}\;\forall \lambda\in\Lambda\Rightarrow$

$x\in f^{-1}(B_{\lambda})\;\forall \lambda\in\Lambda \Rightarrow x\in \bigcap_{\lambda\in\Lambda}f^{-1}(B_{\lambda})$

etc.

**emakarov** · March 21st 2011, 12:48 AM

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