# Show that the enumeration diverges

## Homework Statement

Let A = (0,1)$\cap$Q
Let (Xn) be an enumeration of A. Show that (Xn) diverges

## The Attempt at a Solution

I am not quite sure of what 'diverges' means. Does this simply mean that a tail of (Xn) diverges to +∞ or -∞? If this is what the definition is, I cannot see how (Xn) diverges. I would say instead that it converges to 1.

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Ray Vickson
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## Homework Statement

Let A = (0,1)$\cap$Q
Let (Xn) be an enumeration of A. Show that (Xn) diverges

## The Attempt at a Solution

I am not quite sure of what 'diverges' means. Does this simply mean that a tail of (Xn) diverges to +∞ or -∞? If this is what the definition is, I cannot see how (Xn) diverges. I would say instead that it converges to 1.
The question claims (truly) that if you have *any* list containing all the rationals in (0,1) [which is possible because this is a countable set], the list is not a convergent sequence; in other words, x_n does not have a well-defined limit as n --> infinity. Of course, 0 < x_n < 1 for all n, so there is no question x_n going to +- infinity as you seem to think.

RGV