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Say I have a countably infinite initial set $A$ with $A \subset [0,1]$. My set satisfies the property that $\inf_{x\in A} |x-y| = 0,\,\forall{y\in [0,1]}$ (For example $A$ can be the set of rational numbers). I have a "construction" that works in the following manner:

Step 1: Remove an arbitrary $x_1\in A$ from $A$ to come up with the set $A_1 = A - \{x_1\}$,

and for $n=2,\ldots,N$,

Step n: Remove an arbitrary $x_n \in A_{n-1}$ from $A_{n-1}$ to come up with the set $A_n = A- \{x_1, \ldots, x_n\}$.

It can be shown that for any $n$, we have $\inf_{x\in A_n}|x-y| = 0,\,\forall{y\in [0,1]}$ as well. That is, removing finitely many terms from $A$, we still get $0$. I cannot guarantee though that repeating the above process infinitely many times I still get $0$, as I may have removed all the elements of $A$. My first question is why this is happening? Or to better put it, how can I avoid making mistakes when going from a finite number of steps to a countably infinite number of steps?

Now, consider another "construction"; at step $n$ I just remove an arbitrary element $x_n$ from $A$ to come up with the set $A_n = A - \{x_n\}$. Similar to the previous case we have $\inf_{x\in A_n} |x-y| = 0,\,\forall{y\in [0,1]}$ for any $n$. But now, I guess, even if I repeat the steps of this new construction infinitely many times, I would still get $0$ as my final answer. How can I prove this? In general, I have a construction in which I repeat some well-defined steps infinitely many times; how can I verify the correctness of my final answer? (You may suggest that I need to find a set-limit $\lim A_n$ for this case, but it appears to me that I need not, as I am "sure" that the final answer should be $0$ in any case).

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Mostly depends on whether one decides whether the inf of an empty set is defined. I would say not. – André Nicolas Sep 24 '12 at 0:14
If you want to avoid removing all (or too many of) the elements of $A$, you have to ensure somehow (when setting up your construction) that there are certain elements of $A$ that are never removed. – Robert Israel Sep 24 '12 at 0:17

As Robert Israel said in the comments, you have to carry out the construction in a way that ensures that it (provably!) does what you want. I’ll illustrate by starting with a countably infinite set $A\subseteq(0,1)$ whose infimum is $0$ and removing infinitely many elements of $A$ in such a way that the set that remains and the set that’s removed both have $0$ as their infima.

Since $0=\inf A\notin A$, $(0,r)\cap A\ne\varnothing$ for any $r>0$. Recursively pick distinct points $a_n,b_n$ such that for each $n\in\Bbb N$,

$$a_n,b_n\in\Big(\big(0,2^{-n}\big)\cap A\Big)\setminus\Big(\{a_k:k<n\}\cup\{b_k:k<n\}\Big)\;.$$

Then all the points $a_n,b_n$ are in $A$, and the sequences $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$ both converge to $0$. Now let $A_0=A\setminus\{b_n:n\in\Bbb N\}$; then $A_0\supseteq\{a_n:n\in\Bbb N\}$, so $\inf A_0=0$, and we’ve removed the infinitely many points of the set $\{b_n:n\in\Bbb N\}$, whose infimum is also $0$.

When you want to make sure that you leave at least as many as you take away, the trick of picking two points at a time is very handy.

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