Limit Set of All Real Numbers Not even sure how to start this one. Does anyone know how to do this?
Prove that there exists a sequence such that its limit set is the set
of all real numbers
Limit set is the set of all subsequence limits btw
 A: Enumerate all the Rational numbers in the Real line, and you have your sequence. 
We have that the Rationals are dense in the Reals, which implies that every irrational is the limit of a sequence of Rationals. Specifically, consider any irrational $a:=a_0.a_1a_2...a_n....$, then the Rational number $b:=a_0.a_1...a_k00000...$ approximates  $a$ as well as you want by taking as many terms in the expansion as you need. Then $$|a-b|=0.000000a_{k+1}a_{k+2}..... < \frac{9}{10^{k+2}} \rightarrow 0$$ as $k \rightarrow \infty$.
A: Start the sequence with $0$. 
Then write $-1,1$.
Then write $-\frac{4}{2},-\frac{3}{2},-\frac{1}{2}, \frac{1}{2},\frac{2}{2},\frac{3}{2},\frac{4}{2}$. 
Then write $-\frac{16}{4},-\frac{15}{4}, -\frac{13}{4},\dots, -\frac{1}{4},\frac{1}{4},\dots, \frac{16}{4}$. 
Then write $-\frac{64}{8}, \frac{63}{8},\dots, \frac{63}{8},\frac{64}{8}$, skipping $\frac{0}{8}$, though it really doesn't matter.
Then write everything from $-\frac{256}{16}$ to $\frac{256}{16}$, in steps of $\frac{1}{16}$, skipping \frac{0}{16}$.  
Continue. 
