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Find a sequence $\{x_n\}$ such that for any $y \in \Bbb R$, there exists a subsequence $\{x_{n_i} \}$ converging to $y$.

I actually have almost zero clue where to start? Thanks to anyone who can help!

user860374
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chloemillz
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4 Answers4

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An enumeration of $\mathbb Q$.

Surb
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As pointed out by Surd, an enumeration of $\Bbb Q$ (that means, a listing of the elements of $\Bbb Q$ - i.e. a sequence in $\Bbb Q$ ).

The reason for this being as follows:

Since $\Bbb Q$ is dense in $\Bbb R$, we know that for any $y \in \Bbb R$ there exists a sequence $\{q_n\}\in \Bbb Q$ such that $q_n \to y$. Since the sequence $\{q_n\}$ converges to $y\in \Bbb R$, we then (clearly) must have that every subsequence $\{q_{n_k}\}$ of $\{q_n \} \in \Bbb Q$ converges to the same $y \in \Bbb R$.

user860374
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  • That helps a lot! If I were going to explicitly define one, how would I go about it? – chloemillz Nov 04 '15 at 22:58
  • This can be quite a daunting task as there this usually involves playing around with some kind of numerical analysis, I will try and find you an example now and post it :). The purpose of this question (as done in Real Analysis), is usually simply to help you understand (theoretically) why this is possible :). – user860374 Nov 04 '15 at 23:00
  • See this question as an example :). In this one, we first prove that there is a sequence in $\Bbb Q$ that converges to $\sqrt{3}$ (which is in $\Bbb R$). This then directly implies that every subsequence must also converge to $\sqrt{3}$. The second part shows different way you can go about finding such a sequence.

    http://math.stackexchange.com/questions/1226922/rational-number-that-approximates-sqrt3

     – user860374 Nov 04 '15 at 23:04
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Think about the fact that every real number can be arbitrarily approximated by a finite decimal expansion (i.e., a rational number)...

In particular, the density of the rational in the reals is a fundamental, elementary notion from real analysis. Also that the rationals are countable...

This question was similar Every real number is the limit of some subsequence of a given sequence in the sequence of all rationals

Community
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charlestoncrabb
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Although it's not necessary, it's easier if every rational occurs in the sequence infinitely often. There is a bijection $$f\colon \mathbb{N} \to \mathbb{N} \times \mathbb{Q} \text{.}$$ So let the sequence be $$q_n = \text{2nd coordinate of }f(n)\text{.}$$ Given a real $x$, let its decimal expansion be $\Sigma_{i=0}^{\infty}a_i 10^{-i}$. Then each partial sum $x_n = \Sigma_{i=0}^n a_i 10^{-i}$ occurs infinitely often in the sequence $(q_n)$, and we can define the following subsequence: $$ \begin{align} n_0 &= \text{least $n$ where $q_n = x_0$,} \\ n_{k+1} &= \text{least $n > n_k$ where $q_n = x_{k+1}$.} \end{align} $$ Clearly, $(q_{n_k})_{k\ge 0}$ is a sequence of the partial sums, in order, and $\lim_{k\to \infty} q_{n_k} = x$.

BrianO
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