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This is a basic question and I know this is typically not how this is proven but I was wondering if the following is a valid proof of showing that given two disjoint sets, say V and W in the co-domain, their pre-images are also disjoint:

If we assume that the pre-image($V$) and pre-image($W$) had an intersection, the points in the intersection would be in the pre-image of $V$, and the same points would be in the pre-image of $W$ and so from the definition of pre-image $f$(these points) would map to both $V$ and $W$ at the same time since $V$ and $W$ are disjoint, but this can’t be because $f$ is a function. Thus a contradiction and so our assumption that pre-image($V$) and pre-image($W$) had an intersection is wrong and so pre-image($V$) and pre-image($W$) are disjoint.

M47145
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bryan
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    Looks good. Of course it's also true that for any family (indexed by a set of any cardinality) of sets ${U_\alpha}{\alpha \in A}$, we have $f^{-1}(\bigcap{\alpha \in A}U_\alpha) = \bigcap_{\alpha \in A}f^{-1}(U_\alpha)$. Pre-images are very nice. – Noah Olander Jun 11 '15 at 04:39
  • The logic seems fine. Using some symbols would clarify your argument, though. Also, how is this not the typical way of proving the statement? It seems pretty standard to me... – 5xum Jun 11 '15 at 04:39

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