# Pugh, exercise in err?

1. Mar 24, 2008

### SpaceTag

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) 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.

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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.

Last edited: Mar 24, 2008
2. Mar 25, 2008

### Crosson

The statement is false as written here. The relation # as defined is called a pre-order, satisfying for all x1,x2,x3 in X:

x1 # x1 (reflexive property)

( (x1 # x2) and (x2 # x3) ) implies (x1 # x3) (transitive property)

I think the statement might still be false even if we specify that the relation # is antisymmetric (which would make it a partial ordering relation instead of a pre-ordering relation):

( (x1 # x2) and (x2 # x1) ) implies (x1 = x2) (antisymmetric property)

Maybe someone else will see a mistake in the counterexample that I don't see, but otherwise it looks sound to me.

3. Mar 25, 2008

### chronon

I think that for the proof you also need the condition for all x1, x2 either x1#x2 or x2#x1