# Subspace convergence question

1. Jun 30, 2009

### Tokipin

1. The problem statement, all variables and given/known data

From Introduction to Topology by Bert Mendelson, Chapter 2.7, Exercise 8:

Consider the subspace (Q, d_Q) (the rational numbers) of (R, d). Let a1, a2, ... be a sequence of rational numbers such that $\lim_{n} a_n = \sqrt{2}$. Does the sequence converge when considered to be a sequence of points of (Q, d_Q)?

2. The attempt at a solution

I think the answer is that it doesn't converge in Q because the value is outside the space and the distance function can't tell how far away the point is. At the same time though, the sequence would seem to converge, though maybe to no particular value. Confusing. @_@

2. Jun 30, 2009

### Dick

That's about right. It doesn't converge in Q, because there is no point in Q to converge to. You would agree that for any point q in Q, the sequence doesn't converge to q, right? The sequence is Cauchy (which is think what you mean by 'seem to converge', if you haven't covered that yet, look it up). The confusing part is that Q isn't 'complete'. Look that up too.

Last edited: Jul 1, 2009
3. Jul 2, 2009

### HallsofIvy

Staff Emeritus
And the part I bolded is where you are confused. As Dick said, a sequence converges to a particular value. There is no such thing as a sequence that "converges but to no particular value". The Definition of "converges" is that there exist some number L, "the number the sequence converges to, such that for every $\epsilon> 0$ there exist N such that if n> N, then |a_n- L|< $\epsilon$". Without "L" that definition makes no sense!

It is true that the numbers in the sequence are getting closer and closer together. That is what Dick referred to as a "Cauchy sequence" and it is an important property of real numbers that every Cauchy sequence converges- but that is NOT true of the rational numbers. Another example is the sequence of rational numbers 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ... where each number includes one more decimal place in the decimal expansion of $\pi$. Those are all rational numbers because they are terminating decimals. The sequence is a "Cauchy sequence" because past the nth number all numbers are equal to the nth decimal place and so the differences go to 0. Thought of as a sequence of real numbers, it converges to $\pi$ but in the rational numbers it does not converge.

In fact, any sequence of rational numbers that, thought of as a sequence of real numbers, converges to an irrational number, cannot also converge to a rational number- because then it would converge to two different real numbers, which cannot happen. And therefore, as a sequence in the rational number system, does not converge.

4. Jul 20, 2009

### Tokipin

Thanks Dick, Cauchy + Completeness clarified it. I noticed that the problems in this book reference things from all over the place. And many thanks Ivy for taking the time to set everything straight. Very appreciated!