Formal logic and functions (I am struggling with writing my proof)

Question

I wish to show that $f\left(\bigcup_aX_a\right)=\bigcup_af(X_a)$

My attempt:
$$y\in f\left(\bigcup_aX_a\right)\implies\exists x\in\bigcup_aX_a:y\in f(\{x\})$$ which I didn't like, so I re-wrote it as:
$$\forall y\in f\left(\bigcup_aX_a\right)\exists x\in\bigcup_aX_a:y\in f(\{x\})$$

I have to introduce an $x$, I need to put an exists somewhere (unlike proving De Morgan's laws, where you can just say $x\in$left hand side $\implies\in$RHS and then the other way)

Now I can write the "if $x$ is in the union of some sets it is in at least one of them" nicely as:

$$\forall x\in\bigcup_aX_a\exists b:x\in X_b$$ - given an $x$ I can get a $b$.

How do I use this above?

$$\forall y\in f\left(\bigcup_aX_a\right)\exists x\exists b:x\in X_b\implies y\in f(\{x\})$$

I am saying for all y there exists an x (based on that y) there exists a b based on that x and y with $x\in X_b$.... surely I could also say "there exists a b where there exists an x such that x in $X_b$ implies...."

I will want to go to:

$\forall y\in f\left(\bigcup_aX_a\right)\exists b:y\in f(X_b)$ which makes me want to swap round the order of exists.

Then from this we go (not sure how to write it with equal rigor) that y would be in the union of $f(X_a)$ over a.

Completing the proof

My first attempt:

$y\in f\left(\bigcup_aX_a\right)\implies f^{-1}(\{y\})\subset \bigcup_aX_a$ so $\exists x\in f^{-1}(\{y\})$ with $x\in\bigcup_aX_a$ which doesn't feel as concrete as I'd like.

Answer 1

This is not a real answer, but a long comment to jdc's answer.

To be totally "formal" is an hard task, due to the huge amont of defined symbols used in set language [see as example : Yuri Manin, A Course in Mathematical Logic for Mathematicians (2010), Chapter 1.3 : Beginenrs' Course in Translation, page 9-on].

Let us try with :

$∃x \in \bigcup_a X_a (f(x)=y)$.

In order to simplify it, abbreviate the formula $(f(x)=y)$ as $\varphi(x,y)$; thus, we have (more "formally") :

$(\exists x) (x \in \bigcup_a X_a \land \varphi(x,y))$.

Now, if we have the simpler case of $X_1 \cup X_2$, our formula will be :

$(\exists x) [(x \in (X_1 \cup X_2)) \land \varphi(x,y)]$.

This is "quite" first-order logic ...

The condition $x \in (X_1 \cup X_2)$, by definition of union of sets, is equivalent to:

$(x \in X_1) \lor (x \in X_2)$.

Now, with the "$\bigcup$-case", we are not allowed to write a "potentially" infinite disjunction; thus, we have to use $\exists$.

We have that $x \in \bigcup_a X_a$ is equivalent to :

$(\exists a) [(a \in A) \land (x \in X_a)]$,

where $A$ is the (left implicit) index-set.

We try now to "assemble" all together :

$(\exists x) ( \quad (\exists a) [(a \in A) \land (x \in X_a)] \land \varphi(x,y) \quad )$.

Now we need a rule for quantifiers :

$\exists x P(x) \land Q \leftrightarrow \exists x (P(x) \land Q), \quad$ provided that $x$ is not free in $Q$.

In our formula above, $i$ is not free in $\varphi(x,y)$; thus we apply the rule to get :

$(\exists x) (\exists a) ( [(a \in A) \land (x \in X_a)] \land \varphi(x,y) )$.

Now we need a second rule for quantifiers :

$\exists x \exists y P(x,y) \leftrightarrow \exists y \exists x P(x,y)$.

Applying it we have :

$(\exists a) (\exists x) ( [(a \in A) \land (x \in X_a)] \land \varphi(x,y) )$.

An this is (again "omitting" the index-set) :

$(\exists a) (\exists x) [(x \in X_a) \land f(x)=y]$

that, with a little "sloppiness", we write it as :

$∃a ∃x \in X_a (f(x)=y)$.

An then we can try to go on ...

Recalling that a function in set theory is a set of ordered couples, we have that $f(x)=y$ is :

$(x,y) \in f$

where in turn $f \subseteq X_a \times f[X_a]$.

Thus : $x \in X_a$ and $y \in f[X_a]$.

In conclusion, from : $∃a ∃x \in X_a (f(x)=y)$

we have that :

$\exists a (y \in f[X_a])$.

The last step is easy ... and I hope it can help.

Answer 2

I'm not sure what you mean by "concrete," but the following is correct.

$y \in f[\bigcup_\alpha X_\alpha]$ if and only if there exists $x \in \bigcup_\alpha X_\alpha$ such that $f(x) = y$. This happens if and only if there exists an $\alpha$ such that $x \in X_\alpha$ and $f(x) = y$, which in turn means that there is an $\alpha$ such that $y \in f[X_\alpha]$. But that means $y \in \bigcup_\alpha f[X_\alpha]$.

Given the number of things you've said above you don't like, this may or may not be anything like what you're looking for.

Edit: In the vague hope that you might like something more formal (and don't mind chaining "iff" signs in a manner that some mathematicians consider abusive), I offer the following.

\begin{align*} y \in f\Big[\bigcup_\alpha X_\alpha\Big] &\iff \exists x \in \bigcup_\alpha X_\alpha \ \big(f(x) = y\big)\\ &\iff \exists \alpha\ \exists x \in X_\alpha \big(f(x) = y\big)\\ &\iff \exists \alpha\ \big(y \in f[X_\alpha]\big) \\ &\iff y \in \bigcup_\alpha f[X_\alpha]. \end{align*}

Answer 3

Using slightly different notations which I find more convenient, the statement to be proved is $$ \tag{0} f[\langle \cup a :: X_a \rangle] \;=\; \langle \cup a :: f[X_a] \rangle $$ where the definitions are \begin{align} \tag{1} & y \in f[X] \;\equiv\; \langle \exists x : f(x) = y : x \in X \rangle \\ \tag{2} & z \in \langle \cup a :: V(a) \rangle \;\equiv\; \langle \exists a :: z \in V(a) \rangle \\ \end{align}

For me, the simplest way to prove $(0)$ is to go to the element level using these definitions, and use the laws of logic: investigate which elements are part of the left hand side, and then transform into the right hand side.

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In other words, for any $\;y\;$ we calculate as follows:

\begin{align} & y \in f[\langle \cup a :: X_a \rangle] \\ \equiv & \qquad \text{"definition (1) for $\;\cdot[\cdot]\;$"} \\ & \langle \exists x : f(x) = y : x \in \langle \cup a :: X_a \rangle \rangle \\ \equiv & \qquad \text{"definition (2) for $\;\langle \cup \ldots \rangle\;$"} \\ & \langle \exists x : f(x) = y : \langle \exists a :: x \in X_a \rangle \rangle \\ (*) \;\;\; \equiv & \qquad \text{"logic: exchange $\;\exists\;$ quantifications"} \\ & \langle \exists a :: \langle \exists x : f(x) = y : x \in X_a \rangle \rangle \\ \equiv & \qquad \text{"definition (1) for $\;\cdot[\cdot]\;$"} \\ & \langle \exists a :: y \in f[X_a] \rangle \\ \equiv & \qquad \text{"definition (2) for $\;\langle \cup \ldots \rangle\;$"} \\ & y \in \langle \cup a :: f[X_a] \rangle \\ \end{align}

which proves $(0)$ by set extensionality.

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This proof clearly shows that the key step $(*)$ is that two nested $\;\exists\;$ quantifications (or two nested unions) may be exchanged.