Prove Xn->x is Integer if Xn is Sequence of Real Numbers

[sleventh]

if Xn is a sequence of integers and Xn--->x as n----> infinity and x is an element of the reals. show that x must be an integer.

i know that since the sequence is convergent it will be bounded. i don't however see how i can prove the above. thank you very much for any help.

 

[Office_Shredder]

A convergent sequence also has to be Cauchy. See what you can get out of that

 

[sleventh]

im not exactly sure what you mean by ''bounded away'', is this the same as will not converge? it seems to me that even if x were a real value that |Xn-x|<e given that e>0. is this wrong?

 

[HallsofIvy]

If x_n\to x, then, given any \epsilon, there exist N such that if n> N, |x- x_n|&lt; \epsilon. Take \epsilon to be less than 1 and remember that x_n is an integer.

 

[sleventh]

using what you said in that a convergent sequence has a cauchy sequence, and from hallsofly, could the argument go as follows:
the sequence Xn converges and therefore is a cauchy sequence, by definition for any value n,m>N where N is a natural number |Xn-Xm|<e, given e>0 must be true. each element of Xn,m however is an integer and by the commutative ring of integers Xn-Xm will also produce an integer. if x were a real number it would then be between to integer values. The nearest value Xn can assume are these two integers, therefore taking e<1 |Xn-Xm|<e will not hold

 

[HallsofIvy]

No, that's not true and not my point. The difference of two integers can be 0!

 

[sleventh]

it seems like i need to show that x must be an integer in order for the |Xn-x| to equal zero, but i do not see why |Xn-x|<e would not hold if Xn was an integer and x Real. is it because there is a set distance between a integer and a real value, and if epsilon was set to be this value, |Xn-x|could not be less then epsilon?

 

[Mark44]

it seems like i need to show that x must be an integer in order for the |Xn-x| to equal zero, but i do not see why |Xn-x|<e would not hold if Xn was an integer and x Real. is it because there is a set distance between a integer and a real value, and if epsilon was set to be this value, |Xn-x|could not be less then epsilon?

Integers are real numbers! The converse is not necessarily true, though.

If you have a sequence of integers, it either doesn't converge or converges to an integer. Examples of each kind:
{1, 2, 3, 4, ..., n, ...}
{4, 3, 2, 1, 0, 0, 0, ..., 0, ...}

 

[sleventh]

assuming x is a value between two successive integers, there will be a finite distance E between x and the nearest integer. if e<E then |Xn-x|<e will not hold. therefore by contradiction x must be an integer. This also shows that |Xn-x|=0, and therefore Xn will eventually be a constant value.
How does that sound?