A question on proving the chain rule

[mathwonk]

you lost me. i think i have given a complete proof, with ample detail.

added Later:

@ Fredrik: Sorry to be so slow. Did you want something like this?

Lemma: If (1) f(x)-->L as x-->x0, and (2) x(t)-->x0 as t-->t0, then f (x(t))-->L as t-->t0.
proof: given e>0 choose d1>0 so that |x-x0|<d1 implies | f(x)-L | < e. (ok by (1)).

Then choose d>0 so that |t-t0| < d implies |x(t)-x0| < d1. (ok by (2)).

Then |t-t0| < d implies |f(x(t) - L| < e. QED.

(I admit I tend to take for granted that someone can fill in such details on his own. But i do recall being puzzled by exactly such matters, hmmm, maybe some 50 years ago.)

 

[Fredrik]

Yes, that's what I had in mind. That completes the proof.

 

[mathwonk]

here is another nice way to make logical sense of vague statements like "t-->t0 implies x-->x0".

think of t as a sequence {tn}, and tn-->t0 as the statement "the sequence {tn} converges to t0 (as n goes to infinity)".

then the statement " tn-->t0 implies x(tn)-->x0" is an implication. and since also "xn-->x0 implies f(xn)-->L" is an implication,

one can compose them and get that "tn-->t0 implies f(x(tn))-->L".

 

[Svein]

think of t as a sequence {tn}, and tn-->t0 as the statement "the sequence {tn} converges to t0 (as n goes to infinity)".

Mathematically, this is not equivalent with an ε - δ argument. A sequence contains a countable set of points, the ε - δ argument talks about all points (which is usually not countable). The argument would make sense in ℚ, but then the limit might not be in ℚ.

 

[mathwonk]

As a matter of fact the sequence approach is equivalent to the epsilon delta approach, for convergence in the reals, as you can show as an easy exercise. i.e.

lemma: the following are equivalent:
1) for every e>0 there is a d>0 such that |x-x0| < d implies |f(x)-L| < e.

2) for every sequence {xn} converging to x0, the sequence {f(xn)} converges to L.

the point is that there is a countable sequence of rationals, {1/n} if you like, converging to 0, and if (1) fails, these can be used (as d's) to construct a sequence for which (2) fails. the proof that if (1) is true then (2) is true is even easier.

sequential convergence is equivalent to more general convergence in any "first countable space", such as any metric space. here is a little article on them from wikipedia:

http://en.wikipedia.org/wiki/First-countable_space

 

[Svein]

for every sequence {xn} converging to x0, the sequence {f(xn)} converges to L.

Yes, now I can agree.

 

[mathwonk]

one often omits a universal quantifier. e.g. you did not object to "for all x" being omitted in statement (1) of the lemma above.

so since you point it out, let me balance both statements. i.e. they might reasonably read:

either:
lemma: the following are equivalent: (the universe of all variables is the real numbers, except for f which denotes a function);
1) for every e>0 there is a d>0 such that for all x, |x-x0| < d implies |f(x)-L| < e.

2) for every sequence {xn} converging to x0, the sequence {f(xn)} converges to L.

or: sloppier:
lemma: the following are equivalent: (omitting both universal quantifiers on the x's)
1) for every e>0 there is a d>0 such that |x-x0| < d implies |f(x)-L| < e.

2) {xn} --> x0, implies {f(xn)} --> L.