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My book wrongly asserted that for any family $X_a$ of sets the following is true:

$$f(\cap_aX_a)=\cap_af(X_a)$$ Because I could think of a counter example (y=$x^2$ - simple but works) I decided to try and prove these as an exercise. This is self-learning. (to clarify the sets {2} and {-2} have an empty intersection, but the intersection of f of these sets is {4})

The union form of this was easy but the first time I accidentally slipped up and "proved" it. Showing the RHS $\subset$ LHS is easy (how do I do a subset going the other way in LaTeX) but going the other way isn't always true.

if $y\in\cap_af(X_a)$ then $y\in f(X_a)\ \forall a$

On my second pass I was more pedantic and looking for where I went wrong, all we know from this is:

$\forall a\ \exists x_a\in X_a\ :\ f(x_a)=y$

$f$ is injective $\implies$ $x_i=x_j$ if $f(x_i)=f(x_j)$

So if $f$ is injective all the $x_a$s are the same (how do I write this? $x_a=x_a$ is silly)

anyway if $f$ is injective then let $x=x_a$ (for all a?) and $f(x)=y$, as $x\in X_a\ \forall a$ $x\in\cap_aX_a$ thus $y\in f(\cap_aX_a)$

I have 3 questions, first how's my notation? Am I writing this out correctly? Second should I bother with there existing a set of $x_a$s that are all the same but the function not being injective (I don't see why this can't happen)?

It is am implies sign, injective means what I need to prove the identity is satisfied but it is not an if and only if.

Lastly can I have the answers to:




By answers I mean "what properties must f have for them to be true" (so I can know if I'm right)

I also have yet to use surjectivity at all.

I'm slightly worried because I didn't know I wasn't using surjectivity.

So back to the title of this question, is there a diagram I can use that will not let me accidentally neglect properties, like the first time I tried I unknowingly assumed $f$ was injective, thus "proving" it.

This is page 3 of a book in the "preliminaries" section, as it's a first edition and chapter 0 I don't hold the errors against the book (it just asserts "if $X_a$ is any family of sets then" with no constraints on $f$) but I do want to be clear on what it intends and to do this for myself.

share|cite|improve this question
$\supset$ is \supset – Henning Makholm Jan 12 '14 at 16:13
Thanks for correcting the intersects Daniel and thanks for telling me @HenningMakholm, I was hoping someone would. – Alec Teal Jan 12 '14 at 16:16

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