Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Im unsure of an example Give an example of a sequence of real numbers with subsequences converging to every real number

share|improve this question
add comment

3 Answers 3

HINT: The rationals are a countable set, so they can be enumerated, and every real is the limit of a sequence of rationals.

share|improve this answer
add comment

It's a bit of work, but you may show that taking the sequence $$a_n = n \bmod{2\pi}$$ has a limit point at every point in the interval $[-\pi, \pi]$, where the endpoints points are never actually attained attained (otherwise $\pi$ would be rational). So you may continuously map the interval $(-\pi,\pi)$ into $\mathbb{R}$ in you favorite way. Take, for instance, the map $f: (-\pi,\pi) \to \mathbb{R}$ sending $x \mapsto \tan(x)$. So now you have a sequence $a_n$ converging to every point $a \in (-\pi,\pi)$, then since tangent is continuous, $\tan(a_n) \to \tan(a)$, and $\tan$ is surjective onto $\mathbb{R}$, so you are done.

I realize that the first part, taking integers modulo $2\pi$ is not entirely necessary for the proof, since $\tan(x) = \tan(x + 2k\pi)$, but I think it's easier to break it into smaller problems.

share|improve this answer
    
Or based on the opposite idea, $a_n = f(n\pi\mod 1)$ where $f(x) = \frac1x\sin\left(\frac1x\right)$. The modulo operation is necessary in this case, though. –  user1551 Dec 12 '12 at 4:53
    
I think what you're referring to when you say $n\pi \bmod{1}$ is actually $n \pi - \lfloor n\pi \rfloor$ (the "floor" function). The way you have written is not well defined. i.e. $f(x)$ must be equal to $f(x + k)$ for any $k \in \mathbb{N}$, however $f(1) \neq f(2)$. –  andybenji Dec 12 '12 at 5:02
add comment

A related question that you can try:

Let $(a_k)_{k\in\mathbb{N}}$ be a real sequence such that $\lim_k a_k=0$, and set $s_n=\sum_{k=1}^na_k$. Then the set of subsequential limits of the sequence $(s_n)_{n\in\mathbb{N}}$ is connected.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.