Show that the enumeration diverges

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SUMMARY

The enumeration of the set A = (0,1) ∩ Q diverges, meaning that the sequence (Xn) does not converge to a specific limit as n approaches infinity. This is established by the fact that any list containing all rational numbers in the interval (0,1) is countable and does not have a well-defined limit. The values of x_n remain bounded between 0 and 1, thus ruling out divergence to ±∞.

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Homework Statement


Let A = (0,1)[itex]\cap[/itex]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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PirateFan308 said:

Homework Statement


Let A = (0,1)[itex]\cap[/itex]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
 

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