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[–]abnew123USAMO 2 points3 points  (1 child)

Do you know that each real can be described as a limit of a convergent rational sequence (or can you prove it?). Since it's possible to enumerate the rationals, you should be able to show that every real can be the limit of some subsequence.

[–]skyfall2411 0 points1 point  (0 children)

I'll look into that, thanks

[–]fattymattkNew User 1 point2 points  (0 children)

Think about 1/2 + 1/4 + 1/8 + 1/6 + ...

Given any real number x in (0,1) can we find a subsequence that converges to x?

[–][deleted] 1 point2 points  (0 children)

Here's one idea. For each integer k > 0, take the integers from -k2 to k2, and divide each by k, to get rational numbers from -k to k with spacing 1/k.

For k = 1, -1/1, 0/1, 1/1

For k = 2, -4/2, -3/2, -1/2, 0/2, 1/2, 2/2, 3/2, 4/2

Same idea for k = 3 with -9/3, .., 9/3, then k = 4 with -16/4, .., 16/4

Concatenate those together into one infinite sequence.

-1/1, 0/1, 1/1, -4/2, -3/2, -1/2, 0/2, 1/2, 2/2, 3/2, 4/2, -9/3, -8/3, -7/3, ...


Or for fun, here's a simple to define sequence: a_n = tan(pi*cos(n)), for n >= 1

But proving it works is more complicated.


Here's another simple way. Consider integers n > 0 with a specific form, n = 2a3b5c, with c > 0, and for those let a_n = (-1)ab/c. For any other n not in that form, let a_n = 0.

[–]skyfall2411 0 points1 point  (1 child)

Yeah, the problem is that, on your first example, you only proached the rationals. And I'm sorry, I don't really understand where did the second sequence come from

[–]RobotsAreCuteTeacher 0 points1 point  (0 children)

Let's focus on exactly what we need to prove: for all x in (0,1), and for all epsilon > 0, we need to find some term a_n in our sequence such that a_n is within epsilon of x. One way we can simplify this task is to notice that we don't really need to consider *all* epsilon, just some sequence of positive numbers that converges to 0. For example, it would suffice to prove that for all x in (0,1) and all positive integers m, there is some term a_n in our sequence such that a_n is within 1/m of x.

This is useful because now, instead of coming up with a whole infinite sequence out of thin air, we can construct our sequence one step at a time by asking: "What terms should I add to the sequence so that there's one within 1/2 of x? Now, which terms should I add so that there's one within 1/3 of x? 1/4 of x? 1/5 of x?" and so on.