Is the Axiom of Choice needed in this kind of reasoning?

Question

I do not understand if the Axiom of Choice is needed in a certain kind of reasoning.

Example 1. Let $X,Y$ be two sets and $f:X\to Y$ be a function. Let $A,B$ be two subsets of $X$ such that $A\subset B$. Statement: $f(A)\subset f(B)$. Proof: consider arbitrary element $y\in f(A)$. Choose $x\in A$ such that $y=f(x)$. Since $A\subset B$, we conclude that $x\in B$. Then $y=f(x)\in f(B)$. Thus every element of the set $f(A)$ belongs to the set $f(B)$ and consequently $f(A)\subset f(B)$.

Question 1: do we need in these arguments a function $f(A)\ni y\mapsto x\in A$ such that $y=f(x)$?

If such a function is needed here, then it turns out that these innocent arguments require the Axiom of Choice in full generality...

Example 2. Let $V$ and $W$ be vector spaces and $A:V\to W$ be a linear operator. Statement: the range of $A$, $Ran(A)$, is a subspace of $W$. Proof: first we will show that for every $y_1$ and $y_2$ from $Ran(A)$ their sum $y_1+y_2$ also belongs to $Ran(A)$. Consider arbitrary $y_1,y_2\in Ran(A)$. Choose $x_1\in V$ such that $y_1=Ax_1$. Choose $x_2\in V$ such that $y_2=Ax_2$. Then $y_1+y_2=Ax_1+Ax_2=A(x_1+x_2)\in Ran(A)$. Now we will show that for every scalar $\alpha$ and every $y\in Ran(A)$ the element $\alpha y$ also belongs to $Ran(A)$. Consider arbitrary scalar $\alpha$ and arbitrary $y\in Ran(A)$. Choose $x\in V$ such that $y=Ax$. Then $\alpha y=\alpha Ax=A(\alpha x)\in Ran(A)$. Therefore $Ran(A)$ is a subspace of $W$.

Question 2: Do we need in these arguments a function $Ran(A)\ni y\mapsto x\in V$ such that $y=Ax$?

If such a function is needed here, then it turns out that these innocent arguments require the Axiom of Choice in full generality...

Example 3. In one of my papers I have to prove that a certain set $S$ in a Hilbert space $H$ is closed. My proof: we will show that every limit point of the set $S$ in $H$ belongs to $S$. Consider arbitrary limit point of the set $S$ in $H$, denote it by $x_0$. Using the Axiom of Countable Choice we can choose a sequence $x_k\in S$, $k\geqslant 1$, such that $x_k\to x_0$ as $k\to\infty$. After this I use some arguments (without the Axiom of Choice) and show that $x_0\in S$. Therefore every limit point of the set $S$ in $H$ belongs to $S$ and consequently $S$ is closed in $H$.

Question 3: Do we need in these arguments a function $Lim(S)\ni x_0\mapsto \{x_k\,|\,k\geqslant 1\}$ such that $x_k\in S$, $k\geqslant 1$, and $x_k\to x_0$ as $k\to\infty$?

If such a function is needed here, then it turns out that these innocent arguments require the Axiom of Choice in full generality...

Please help me to understand if a choice function (as above) is needed in the arguments of this kind, or we simply need a few existential instantiations. Please do not write just "yes" or "no", but provide reasons for your answer.

Answer 1

There is no need for the full axiom of choice in any of these arguments. These are all existential instantiations.

Note that I'm trying to provide formal justification for what you're doing. Your informal arguments are perfectly rigorous and acceptable, since it is clear that they can be translated to a formal argument using the methods I am about to describe.

The key rule here is that the statements

$$\forall x . (P(x) \to Q)$$

and $$(\exists x . P(x)) \to Q$$

are logically equivalent. This is just a basic part of first-order logic.

Now let's consider example 1. Here, we know that $A \subseteq B \subseteq X$ and that $f : X \to Y$. We wish to prove that $Im(A) \subseteq Im(B)$. So formally, we wish to prove that $\forall y \in Im(A) . (y \in Im(B))$. To prove this, we suppose that we have some $y$, and we wish to prove that $y \in Im(A) \to y \in Im(B)$.

Now $y \in Im(C)$ is, by the definition of $Im$, equivalent to $\exists x \in C . f(x) = y$. So what we wish to prove is $(\exists x \in A . f(x) = y) \to (\exists x \in B . f(x) = y)$.

Now by using the above equivalence, we can instead prove that $\forall x \in A . (f(x) = y \to \exists x' \in B . f(x') = y)$. We prove this by supposing we have some $x \in A$ such that $f(x) = y$. Then we see that $x \in B$ and $f(x) = y$.

Moving on to example 2, we see that we simply imply the existential instantiation rule twice. The first time, we use it to go from

$$(\exists x \in V . Ax = y_1) \to ((\exists x \in V . Ax = y_2) \to \exists x \in V . Ax = y_1 + y_2)$$

to

$$\forall x \in V . (Ax = y_1 \to ((\exists x \in V . Ax = y_2) \to \exists x \in V . Ax = y_1 + y_2))$$

We then suppose we have some $x_1 \in V$ such that $Ax_1 = y_1$, and seek to prove that $(\exists x \in V . Ax = y_2) \to \exists x \in V . Ax = y_1 + y_2$. To prove this claim, we again apply the rule to convert the claim to the statement

$$\forall x \in V . (Ax = y_2 \to \exists x \in V . Ax = y_1 + y_2)$$

We then suppose we have some $x_2 \in V$ such that $Ax_2 = y_2$, and seek to prove $\exists x \in V . Ax = y_1 + y_2$. We then write $x = x_1 + x_2$, and we verify that $Ax = y_1 + y_2$ and that $x \in V$.

For example 3, you need the axiom of countable choice to go from the statement

$$\forall n \in \mathbb{N} . \exists x \in S . ||x - x_0|| < 2^{-n}$$

to the statement

$$\exists \{x_n \in S\}_{n \in \mathbb{N}} . \forall n \in \mathbb{N} . ||x - x_n|| < 2^{-n}$$

From here, you use the rule to translate the claim

$$[\exists \{x_n \in S\}_{n \in \mathbb{N}} . \forall n \in \mathbb{N} . ||x - x_n|| < 2^{-n}] \to x \in S$$

to the equivalent claim

$$\forall \{x_n \in S\}_{n \in \mathbb{N}} . [(\forall n \in \mathbb{N} . ||x - x_n|| < 2^{-n}) \to x \in S]$$

which we then prove.

As a rule of thumb, if you're only making a finite, known number of choices, you're not using the axiom of choice.

If you are making a finite number of choices but don't know how many choices you're making, you're also not using the axiom of choice. This is because you can prove by induction that for all $n \in \mathbb{N}$, for all sets $S$ of cardinality $n$, if for all $s \in S$ there exists some $x$ such that $P(s, x)$, then there is some $\{x_s\}_{s \in S}$ such that for all $s \in S$, $P(s, x_s)$.

If you're making a countably infinite number of choices, none of which depends on the others, you will often require the axiom of countable choice (unless you can come up with some way to pick a specific thing for each $n$). Countable choice is usually all that is required for analysis.

If you're making a countably infinite number of choices and future choices depend on previous choices, you will often require the axiom of dependent choice (unless you can come up with some way to pick a specific thing for each $n$, which depends on the previously chosen things in a deterministic manner).

If you're making a choice of $x_s$ for all $s \in S$ with no idea of what $S$ is, you will usually require the full strength of the axiom of choice.