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## Main Question or Discussion Point

I'm referring to question #26 in chapter 3 of Pugh's Real Mathematical Analysis.

For those without the book, here's the question:

Let X be a set with a transitive relation # (Note: #is just an abstract relation). It satisfies the condition that for all x1,x2,x3 in X, we have

x1 # x1

and

if x1 # x2 # x3 then x1 # x3.

A function f:X -> R (R is the reals)

(y # x) implies |f(x) - L|< E. We write lim f = L to indicate this convergence.

Prove that limits are unique: if lim f = L1 and lim f = L2, then L1 = L2.

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Ok now this seems simple enough, but I'm not sure if it's true (even though the book asks us to prove it). Is this a counterexample?

Take X to be the set of natural numbers with the transitive relation = (equality). Let f be a function from the naturals to the reals such that f(1) = 1 and f(2) = 2. Then 1 is a limit of f (take y=1) and 2 is also a limit of f (take y=2).

That would mean that lim f isn't unique. I'm just wondering where this counterexample goes wrong or if the question really is flawed.

For those without the book, here's the question:

Let X be a set with a transitive relation # (Note: #is just an abstract relation). It satisfies the condition that for all x1,x2,x3 in X, we have

x1 # x1

and

if x1 # x2 # x3 then x1 # x3.

A function f:X -> R (R is the reals)

**converges to a limit L**with respect to X if, given any E>0, there is a y in X such that, for all x in X,(y # x) implies |f(x) - L|< E. We write lim f = L to indicate this convergence.

Prove that limits are unique: if lim f = L1 and lim f = L2, then L1 = L2.

------------------------

Ok now this seems simple enough, but I'm not sure if it's true (even though the book asks us to prove it). Is this a counterexample?

Take X to be the set of natural numbers with the transitive relation = (equality). Let f be a function from the naturals to the reals such that f(1) = 1 and f(2) = 2. Then 1 is a limit of f (take y=1) and 2 is also a limit of f (take y=2).

That would mean that lim f isn't unique. I'm just wondering where this counterexample goes wrong or if the question really is flawed.

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