Find a sequence such that for any real number $x$, there exists a subsequence which converges to $x$.

So I pick the sequence of the rational numbers ( we know this is a sequence because the rationals are countable ) so we write

$f : N \to Q$

$n \to (r_1,r_2,r_3, \cdots )$

And then I defined the gubsequence $g : N \to N$ so that $f o g : N \to Q.$

We define g(k) such that $x-1 < r(g(1)) < x+1$

$x-\tfrac12 < r(g(2)) < x+\tfrac12$

And in general,

$x-\tfrac{1}{2^{k-1}} < r(g(k)) < x+ \tfrac{1}{2^{k-1}}$

where $r(g(k))$ is the rational number corresponding to the index $g(k)$ in our list of rationals.

But then, I face a different problem.

How do I ensure that $g(k)$ is a strictly increasing function, i.e , my subsequence doesn't mess up the order the terms appear in?

  • $\begingroup$ Please use MathJax to format maths in your question to make things more readable. You can find some more information here math.stackexchange.com/help/notation $\endgroup$ – Rhys Steele Sep 27 '17 at 22:15
  • $\begingroup$ Yes, essentially you are defining a subsequence of $\{r_n\}$, $g_1,g_2,\cdots$ such that $g_k$ is the first term of $\{r_n\}$ lying in the interval $\left(x-2^{-k},x+2^{-k}\right)$. This will converge to $x$. If you want strictly increasing, pick $g_k$ in $\left(x-2^{-k},x\right)$. $\endgroup$ – John Wayland Bales Sep 27 '17 at 22:23
  • $\begingroup$ I have tried to improve your text. Next time use Mathjax... $\endgroup$ – Jean Marie Sep 27 '17 at 22:29

Note that when you choose the value of $g(k)$, there are only finitely many natural numbers smaller than $g(k)$. But there are infinitely many rationals in $(x - \frac{1}{2^{k}}, x + \frac{1}{2^{k}})$ so there are infinitely many rationals in that interval with an index higher than $g(k)$. Picking the index of such a rational to be $g(k+1)$ allows you to ensure that $g$ is strictly increasing.

  • $\begingroup$ Sorry, I don't understand how we went from infinitely many rationals in the interval to infinitely many rationals with an index higher than g(k). Suppose we want x=0.5 Then $-0.5$< $r_g(_1)$< $1.5$ And $0$< $r_g(_2)$< $1.0$ Then how do we conclude that g(2) > g(1)? $\endgroup$ – Saad Sep 27 '17 at 23:15
  • $\begingroup$ Remember we are free to choose $g(k)$ in a clever way (having chosen $g(1), \dots, g(k-1)$). Since the set $\{1, 2, 3, \dots, g(k) \}$ is finite it cannot contain the index of every one of the infinitely many rationals in $(x - \frac{1}{2^{k}}, x + \frac{1}{2^{k}})$. As a result we can pick a rational in that interval whose index is not contained in $\{1, 2, 3, \dots, g(k) \}$ to be $g(k=1)$. i.e. such that $g(k+1) \gt g(k)$ $\endgroup$ – Rhys Steele Sep 27 '17 at 23:24
  • $\begingroup$ Thank you. I think I understand now. $\endgroup$ – Saad Sep 28 '17 at 0:11

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