# Accumulation point of rational numbers in (0,1)

• PirateFan308
In summary, the sequence of all rational numbers in the interval (0,1) has infinitely many accumulation points, as every real number in that interval can be defined as the limit of a sequence of rational numbers. This is supported by the Boltzano-Weirstrass Theorem, which states that every bounded sequence has at least one convergent subsequence, and the fact that for every real number there exists a sequence of rational numbers converging to it. Additionally, the definition of an accumulation point allows for the same element to be repeated in a subsequence, as exemplified by the sequence (-1)n with the accumulation points of -1 and 1.

## Homework Statement

Find all accumulation points of the following sequence:
an enumeration of all rational numbers in (0,1)

## Homework Equations

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

## 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

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.

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

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 $$a_n = 1$$ is such that every subsequence converges to 1, but 1 is not an accumulation point of the set $$\{ a_n : n \in \mathbb{N} \}$$.

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.

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. :)

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 (xn) = (-1)n
(xn) does not converge but it has the convergent subsequences (-1)2n and the subsequence (-1)2n+1. Hence, -1 and 1 are accumulation points.