Accumulation point of rational numbers in (0,1)

PirateFan308 · Nov 12, 2011

1. The problem statement, all variables and given/known data
Find all accumulation points of the following sequence:
an enumeration of all rational numbers in (0,1)

2. Relevant equations
Boltzano-Weirstrass Theorem:
Every bounded sequence has a convergent sequence (hence, an accumulation point)

3. The attempt at a solution
Because the enumeration of all rational numbers in (0,1) is bounded, it must have at least one convergent sequence. But if there is an accumulation point for the rational numbers in (0,1) there must also be an accumulation point for the rational numbers in (0,0.5), and the logic continues so there must be infinitely many accumulation points in (0,1). So are the accumulation points every rational number in (0,1)?

This seems odd to me that any rational number can be an accumulation point

blinktx411 · Nov 12, 2011

Do you know the fact that for every real number y there is a rational r such that |r-y| < epsilon for any epsilon > 0? Or, in a similar fashion, for any real number y there is a sequence of rationals converging to it? Perhaps these facts will help.

HallsofIvy · Nov 12, 2011

In fact, the real numbers can be defined as equivalence classes of sequences of rational numbers, the equivalence being "[itex]\{a_n\}[/itex] is equivalent to [itex]\{b_m\}[/itex] if and only if [itex]\{a_n- b_n\}[/itex] converges to 0".

Damidami · Nov 12, 2011

PirateFan308 said: ↑

Boltzano-Weirstrass Theorem:
Every bounded sequence has a convergent sequence (hence, an accumulation point)

I think there is a mistake here, that a bounded sequence has a convergent subsequence doesn't mean that it as an accumulation point. For example the constant bounded sequence [tex]a_n = 1[/tex] is such that every subsequence converges to 1, but 1 is not an accumulation point of the set [tex]\{ a_n : n \in \mathbb{N} \}[/tex].

Am I right? The definition of accumulation point is that for every epsilon there exists infinitely different elements of the sequence distant less that epsilon from that number.

PirateFan308 · Nov 13, 2011

blinktx411 said: ↑

Do you know the fact that for every real number y there is a rational r such that |r-y| < epsilon for any epsilon > 0? Or, in a similar fashion, for any real number y there is a sequence of rationals converging to it? Perhaps these facts will help.

Thanks! That helped a lot. I get it now. :)

PirateFan308 · Nov 13, 2011

Damidami said: ↑

The definition of accumulation point is that for every epsilon there exists infinitely different elements of the sequence distant less that epsilon from that number.

I believe the definition of an accumulation point is just that there exists infinite elements of the sequence that converges to c. So from this definition, the subsequence can consist of the same element repeated.

Our professor used the example of (x_n) = (-1)ⁿ
(x_n) does not converge but it has the convergent subsequences (-1)²ⁿ and the subsequence (-1)²ⁿ⁺¹. Hence, -1 and 1 are accumulation points.