I recalled this interesting proposition from my real analysis course, for which the answer is true but I forgot the construction of such a sequence. I remember that's a point to get to the answer:
theorem: every real number has a strictly increasing infinite sequence of rational numbers whose limit is that real number. i.e. $\{1, 1.4, 1.414, \ldots\} \to \sqrt{2}$ (could be generalized to $\mathbb{Q}'$), $\{1,1.5,1.75,1.875,\ldots\}\to 2$ (could be generalized to $\mathbb{Q}$).
I know there is a way to include all rational numbers into a sequence using the numerator/denominator table construction. Although this sequence contains all distinct rational numbers, they are not in an increasing order so I can't utilize the theorem above. I'm stuck here then, do you have any hint to it?
Best Answer
The 'theorem above' (that 'every real number has a strictly increasing infinite sequence of rational numbers tending to it') offers a little help with a question given in the title: Is there a sequence that has all real numbers as cluster points?
The set $\Bbb Q$ is countably infinite, so there is a sequence $S$ containing all members of $\Bbb Q$. The set is dense in $\Bbb R$ — every open interval contains infinitely many rational numbers, so for every real number $y$ and arbitrarily small positive $\varepsilon$ there is infinitely many terms of $S$ in a neighborhood $(y-\varepsilon, y+\varepsilon)$. So every real $y$ is a cluster point of $S$,
Q.E.D.
Of course, for any $y$ you can find an infinite monotonic subsequence of $S$, which has a limit $y$ – but you don't need a 'subsequence', let alone 'monotonic'. Of course having 'infinitely many terms' in a neghborhood implies some infinite subsequence, but that subsequence is not necessary, 'infinitely many terms' is enough.