Can Convergent Sequences with Different Limits Have Infinite Intersections?

[bedi]

Let x_n and y_n be two convergent sequences with different limits. Show that the set {x_n : n€N} n {y_n : n€N} is finite.

Attempt: by definition, for each £>0 there exists an N such that |x_n - x|<£ and similarly |y_n - y|<£ holds for every n with n>N. Take £=(x-y)/3 and assume that x_n and y_n are equal for a while. Call N_1 the number which satisfies |x_n - x|<(x-y)/3 and call N_2 which satisfies |y_n - y|<(x-y)/3. Put N=max(N_1,N_2). So after that N, the distance between x_n and y_n is minimum (x-y)/3. Hence there are only N many elements of the set. Is this correct?

 

[bedi]

I think I should have assumed the contrapositive, which is actually equivalent to what I did, right?

 

[haruspex]

Let x_n and y_n be two convergent sequences with different limits. Show that the set {x_n : n€N} n {y_n : n€N} is finite.

Attempt: by definition, for each £>0 there exists an N such that |x_n - x|<£ and similarly |y_n - y|<£ holds for every n with n>N. Take £=(x-y)/3 and assume that x_n and y_n are equal for a while. Call N_1 the number which satisfies |x_n - x|<(x-y)/3 and call N_2 which satisfies |y_n - y|<(x-y)/3. Put N=max(N_1,N_2). So after that N, the distance between x_n and y_n is minimum (x-y)/3. Hence there are only N many elements of the set. Is this correct?

That's basically right but there are a couple of inaccuracies along the way.

ε=(x-y)/3​

ε needs to be guaranteed > 0.

Call N_1 the number which​

N_1 does not appear in the expression which follows. Need a 'for all' in there.