Limit definition gives a contradiction

[torquerotates]

Limit definition gives a contradiction!

say we are given sequences a(n), b(n) such that, a(n)->a, b(n)->b

that means for epsilon>0,

a-epsilon<a(n)<a+epsilon when n>N1

b-epsilon<b(n)<b+epsilon when n>N2

set N=max(N1,N2)

when n>N,
we can subtract the two inequalities

(b-a) +0<b(n)-a(n)<(b-a)+0

the epsilons cancel

b-a<b(n)-a(n)<b-a this is a contradiction. How can a number be bigger and smaller than the same number at the same time? There must be something wrong with my logic.

 

[chiro]

For one your epsilons are not going to be the same for both sequences unless they are equal.

 

[torquerotates]

But in the case that they are equal, we would get a contradiction.

 

[IsometricPion]

While inequalities exhibit transitivity it is not as complete as that of equalities. a>b, b>c implies a>c however, a>c, b>c does not necessarily imply a=b, in fact, it provides no information on the relation between the values of a and b. Similarly, [tex] |a(n)-a|<\epsilon [/tex] and [tex] |b(n)-b|<\epsilon [/tex] gives no information about the relationship between the values of the quantities inside the absolute signs. The inequalities are preserved under operations that are monotonic on their arguements, e.g., adding the same number to both sides, exponentiating both sides. So, it is still possible to bound [tex] |a(n)-a|-|b(n)-b| [/tex], but the bound is not zero, and is not equivalent to the contradiction you obtained.

 

[Citan Uzuki]

The problem with your logic is that it is not in valid to subtract inequalities from each other. As a simple counterexample, consider that if this were valid, we could subtract 0<1 from 0<1 to obtain 0<0, no limit concept required.

Now, it is valid to add inequalities. Suppose that we tried to perform a similar operation by first negating one of the inequalities and then adding them. Negation reverses the inequality, so we get -a+ε > -a(n) > -a-ε, which we can also write as -a-ε < -a(n) < -a+ε. Adding this to the other equality yields b-a-2ε < b(n)-a(n) < b-a + 2ε, which is not contradictory, and is in fact useful in demonstrating that b(n)-a(n) → b-a.