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We know that $\mathbb{Q}$ is countable so let $\{q_n\}_{n \in \mathbb{N}}$ a sequence of all rational numbers.

My Question is:

Is any subsequence $\{q_{n_k}\}_{k \in \mathbb{N}}$ dense in $\mathbb{R}$

Thanks.

Matey Math
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2 Answers2

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"Any". Obviously. The sequence itself is a subsequence. And if you meant proper sequence just remove one term. Or take the sequence of those with even denominators, every neighborhood of a real contain a rational with an odd denominator.

"All". Obviously not. Take the $q_k$ that are integers.

fleablood
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  • +1 I was about to write a comment (as if the existing comments were not enough . . .), but after reading your answer, I think it both answers the question and addresses the ambiguity issues. It's fairly well known (or so I thought) that "any" is often ambiguous and one should mostly avoid its use in formal mathematical statements, and this question gives another example of the kinds of things discussed In proofs, are “for each” and “for any” synonyms? – Dave L. Renfro Jun 16 '18 at 18:01
  • FWIW I take "any" literally. Is any subsequence dense? Well, obviously. It is it's own subsequence and it is dense... – fleablood Jun 16 '18 at 19:35
  • The problem is that "any" can sometimes mean "there exists", and "any" can sometimes mean "for all". This is discussed in the links I gave in the question I linked to in my previous comment. See Be careful with your use of "any" here and see #13 here. – Dave L. Renfro Jun 16 '18 at 20:45
  • Ah, I see what you mean. Yes that should be avoided. – fleablood Jun 16 '18 at 23:53
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This is equivalent to asking is any subset of $\mathbb Q$ dense?.

Some subsets are dense some are not e.g. $\mathbb N$

zwim
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