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Can we prove that every real number is a limit of a sequence of rational numbers without using Axiom of Choice?

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closed as off-topic by Asaf Karagila, Andres Caicedo, mookid, ml0105, John Apr 20 at 3:37

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AFAIK the standard construction of the reals as equivalence classes of sequences of rationals does not require AoC, and makes this problem trivial. –  Alex Becker Apr 12 at 6:04
Let $\langle q_n\rangle$ be a enumeration of rational numbers. For each natural number $k$, we can find the smallest natural number $n_k$ satisfy that $q_{n_k}\in (a-1/n,a+1/n)$. –  tetori Apr 12 at 6:21
This question has a trivial answer using Shoenfield's absoluteness theorem. Another question you asked today has the same trivial answer. Please make sure to check the applicability of Shoenfield absoluteness yourself before asking questions such as this. @Makoto Kato –  Carl Mummert Apr 13 at 0:14
If you prefered a concrete answer, and you knew Shoenfield's absoluteness theorem applies, you should have indicated so in your question. The appearance given by the question, to be frank, is that you did not do even minimal diligence in trying to solve it yourself before posting it here, and didn't even check whether Shoenfield's theorem applies. Also, on a separate note, I would appreciate it if you include the space in my name, because my name is not just one word - thanks. @Makoto Kato –  Carl Mummert Apr 13 at 2:14
@Carl, as this comment should illustrate, any 3+ letter string that uniquely identifies the username (among posters to one thread of comments) pings that user. However, I think it is reasonable to keep the ping in the Stackexchange-provided spaceless format, or at least not interpret that as a discourtesy, while allowing for the manual reinsertion of spaces as an extra courtesy or the automatic version of that as a feature-request for the future. –  zyx Apr 15 at 2:07
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2 Answers 2

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Let $x\in\Bbb R$. Did I use the axiom of choice in constructing this sequence?


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Well, if you define $$\lfloor x\rfloor=\begin{cases} \max\{k\in\Bbb Z\mid k\leq x\} & \sf AC\\0 &\lnot\sf AC\end{cases}$$ then you have used the axiom of choice! ;-) –  Asaf Karagila Apr 12 at 6:50
Let for example $x=\sqrt2=1.4142\cdots$ then by constructing the sequence $$\{1,1.4,1.41,1.414,1.4142,\ldots\}$$ do you think that I used the axiom of choice? –  Sami Ben Romdhane Apr 12 at 7:19
C'mon, there's a winking smiley at the end of that comment. Why do you have to take it so seriously? :-) –  Asaf Karagila Apr 12 at 7:23
It was just a question for clarification and I thought that you meant that my answer is wrong ;-)@AsafKaragila –  Sami Ben Romdhane Apr 12 at 7:31
Nah, if I'd think you're wrong you'd know it. :-) –  Asaf Karagila Apr 12 at 7:32
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Yes. Let $r$ be a real number, then $\{q\in\Bbb Q\mid q<r\}$ is a countable set. Enumerate it as $q_n$ and then by induction construct a strictly increasing sequence approaching to $r$.

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Could you explain how you enumerate the set and construct the sequence? –  Makoto Kato Apr 12 at 7:16
The rational numbers are countable, therefore can be enumerated. I'm not going to bother with constructing something that I don't have to construct. If you want to, there are constructions for enumerations of the rational numbers and you can look for them on your own. Constructing a sequence is done by picking the least index which is larger than your previously chosen rational (we're bounded below $r$), then one can show that we get arbitrarily close to $r$ by noting that every $q$ in this set has a finite index in the enumeration, so at some point we have to pick someone larger than $q$. –  Asaf Karagila Apr 12 at 7:25
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