Let X be a normed vector space and let $Y\subset X$ be a closed linear subspace. Suppose we have a sequence $(x_k)_{k\in \mathbb{N}}$ such that $$\inf_{y\in Y} ||x_k - x_{k+1} + y||_X < \frac{1}{2^k}, \quad \quad \text{for all} \ k\in \mathbb{N}.$$ Then by the Axiom of Countable Choice we can find a sequence $(\eta_k)_{k\in\mathbb{N}}$ in $Y$ such that $$||x_k - x_{k+1} + \eta_k||_X < \frac{1}{2^k}, \quad \quad \text{for all} \ k\in \mathbb{N}.$$

Why did we need to invoke the Axiom of Countable Choice here, surely it is obvious that we can just take the $y$'s such that the first expression above is satisfied, to be our $(\eta_k)$'s?

If not, what is the technical reason behind how the Axiom of Countable Choice works..how does it enable us to choose the sequence $(\eta_k)$?

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    $\begingroup$ There may be uncountably infinitely many values of $y$ that satisfy the first inequality. You would use the axiom of choice to get countably many arranged into a sequence. Even if there were only countably many such $y$ values, picking which one goes first, second, etc. is a matter of arbitrary choice. $\endgroup$ – Aurel Aug 8 '17 at 11:42
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    $\begingroup$ We don't need the AoCC to pick out single elements $\eta_k$. What the axiom is needed for is to guarante that we can pick out a whole infinite sequence. $\endgroup$ – Arthur Aug 8 '17 at 11:59
  • $\begingroup$ So do we just say 'by the AoCC we can obtain a sequence that that satisfies {some property}...' or is there a more rigorous that we can obtain such a sequence. How could we formally state the existence of the $(\eta_k)$ sequence above for example? $\endgroup$ – eurocoder Aug 8 '17 at 12:18
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    $\begingroup$ @eurocoder If you want to formally state the existence of a sequence when you need to choose the elements, practically one-by-one, then you need the AoCC. If you can find some function or other way to "automate" the choosing process, then you can get by without the axiom; the axiom guarantees a choosing function in all cases, if you already have one there is no need to invoke it. $\endgroup$ – Arthur Aug 8 '17 at 12:27

Let's look at this from another perspective for a moment. Let's fix $x=x_k-x_{k+1}$, and ask what are the possible values of $y$ which satisfy $\|x-y\|<\frac1{2^k}$.

For a concrete example, consider $\Bbb R^2$ as $X$ and $\{(r,0)\mid r\in\Bbb R\}$ as $Y$. Now let $x=(0,0.01)$, and tell me how many possible values of $y$ we have to witness that $\inf_{y\in Y}\|x-y\|<\frac12$, for example. There are uncountably many possible values. You simply have no canonical choice of a vector.

So now you need to choose for every $k$ some point $\eta_k$ which witnesses that $x_k-x_{k+1}$ is sufficiently close to $Y$. And since there is no distinguished vector that just the job, there is no way of choosing these $\eta_k$'s uniformly. There is where the axiom of choice kicks in.

Your mistake is that when you write $\inf_{y\in Y}$, then $y$ is a quantified variable, and you can't treat it as a single valued notion. It ranges over all the possible values.

  • $\begingroup$ Ok that gives me a better conceptual idea of what is talking place, thanks! I'm still not sure though how this corresponds to the definition of the AoCC, which is (from Wikipedia): Given a function $A$ with domain $\mathbb{N}$ such that $A(n)$ is a non-empty set for every $n\in\mathbb{N}$, then there exists a function $f$ with domain $N$ such that $f(n) \in A(n)$ for every $n \in N$. I don't see how to relate this to the particular sequence in my post? $\endgroup$ – eurocoder Aug 9 '17 at 10:58
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    $\begingroup$ For every $k$, the set of suitable candidates $E_k=\{y\mid \|x_k-x_{k+1}-y\|<\frac1{2^k}\}$ is non-empty. So by ACC there is a choice of $\eta_k\in E_k$. $\endgroup$ – Asaf Karagila Aug 9 '17 at 11:23
  • $\begingroup$ Excellent, I fully understand it now! $\endgroup$ – eurocoder Aug 9 '17 at 12:25

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