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May 6th, 2007, 12:35 AM   #1
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limit of a function in a metric space

How does the statement

"for all epsilon > 0 there exists delta > 0 such that
( ( d1(x,y) < delta ) AND ( d1(x,z) < delta ) ) IMPLIES ( d2( f(y), f(z) ) < epsilon )"
imply that f : X1 -> X2 has a limit at x in X1

where (X1, d1) and (X2, d2) are metric spaces, and X2 is complete?

Does that statement imply the usual epsilon-delta definition of the limit of functions in metric spaces (which I couldn't get through to) or does that imply some other statement which implies f has a limit at x?
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May 6th, 2007, 01:45 AM   #2
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This is not what you mean. If you set the problem like this, this is obvious because taking x=y in your definition, you basically say that for every epsilon>0, there is delta>0, such that d2(f(x),f(z))<epsilon for every z such that d1(x,z)<delta. Well, this implies that f has a limit at point x, and this limit is f(x) (note that this is equivalent to the definition of limit as we use it in France and in certain arab countries, but not to yours, because yours does not assume anything about the behaviour at point x). So basically this says that f is continuous at point x, and does not use the completeness of your metric space.

What you want to assume is that for every epsilon, there is delta such that for every x,y,z satisfying 0<d1(x,y)<delta and 0<d1(x,z)<delta, we have that d2(f(y),f(z))<epsilon.

Now a proof. Rather than using the topological definition of limit, we will use its sequential characterization, which is equivalent in metric spaces but not in more general topological spaces; that is to say, a function f admits a limit h at point x if and only if for every sequence u_n of points converging to x - while being all distinct from x -, the sequence f(u_n) converges to h.

Consider such a sequence (u_n) in your metric space X1. Clearly the sequence (f(u_n)) is Cauchy, thence convergent to a limit h since the metric space X2 is complete (this limit h is unique because we are working in a metric space, which is separated; this fact may not hold anymore in certain non-separated topological spaces). Now consider another such sequence (v_n). If we can prove that the limit of (f(v_n)) is still h, then we are done. Assume the contrary, i.e that (f(v_n)) converges to a point j <> h. This means that you can find an index N for which n>=N => d2(f(v_n),j)<d2(j,h)/3 and d1(f(u_n),h)<d2(j,h)/3. This means that d2(f(v_n),f(u_n))>d2(j,h)/3, which contradicts your hypothesis.
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May 6th, 2007, 01:59 AM   #3
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Oh yes, whoops, I missed typing out the "y and z not equal to x in X1" sorry, and thanks. Haha...

The usual definition of limit we use is:
f : X1 -> X2 has a limit at x in X1 if
for all epsilon > 0 there exists delta(depending on epsilon) > 0 such that
d1(y,x) < delta implies d2( f(y), f(x) ) < epsilon, for all y not equal to x in X1
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May 6th, 2007, 05:16 AM   #4
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And this is precisely this hypothesis x<>y which is removed in our definition of limit (the one set by the group Bourbaki); this may be very confusing sometimes.
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May 6th, 2007, 05:45 PM   #5
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??? so why is that x useful sometimes? Isn't that true for any x around y and z?
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May 6th, 2007, 08:10 PM   #6
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It's the contrary in this problem. Things are true for y and z around x. Actually, it would have been more convenient to call x "x0", to show that it is fixed. I was just underlining that while the English definition of a limit (be it in R, a normed vector space, a metric space, a uniform space, or topological space), the point at which the function is studied is not taken into account. In the setting of metric spaces:

f admits a limit l at point x0 if and oly if:

for all epsilon>0, there exists delta>0, for every x such that 0<d2(x,x0)<delta, then d2(f(x),l)<epsilon.

The definition given by Bourbaki relaxes the x<>x0 condition:

for all epsilon>0, there exists delta>0, for every x such that d2(x,x0)<delta, then d2(f(x),f(x0))<epsilon.

Meaning to Bourbaki's followers, a function which, for instance, has a limit at point x0 in R and which is defined at this point will automatically be continuous. This definition has the advantage of making the formulation of a lot of theorems simpler (especially those involving the composition operator).
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May 7th, 2007, 12:21 AM   #7
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Could you explain a bit more how you got from the step


there exists N for which n>=N => d2(f(v_n),j)<d2(j,h)/3 and d2(f(u_n),h)<d2(j,h)/3


to the step


d2(f(v_n),f(u_n))>d2(j,h)/3


and why this contradicts the fact that f(v_n) converges to j in X2?




Oh, and also, why or how is it that

"clearly" the sequence (f(u_n)) is cauchy?

Isn't there something else needed in between before we can claim that it is cauchy? (I've been told not to use the words "clearly" and "obviously" in analysis proofs so I'm just making sure if in this case we can claim (f(u_n)) is cauchy straight away at that step)



Thanks!
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May 7th, 2007, 01:40 AM   #8
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Quote:
Originally Posted by aptx4869
to the step d2(f(v_n),f(u_n))>d2(j,h)/3
Take two distinct points in the plane, say A and B, such that d(A,B)=1. Draw circles with center A and B and radii 1/3 around them. Pick two points, one in each of these two circles. The distance separating these two points will be greater than 1/3. What's going on above is the same, just a little more abstract. In our problem, A and B are represented by j and h, while v_n and u_n represent the two points. Around j and h are considered the two open balls with radius d2(j,h)/3.

So basically, in the context of metric spaces, you want to prove that:

Given two distinct points A,B and two other points a,b such that d(a,A)<1/3*d(a,b) and d(b,B)<1/3*d(a,b), show that d(a,b)>1/3. Proof: by the triangle inequality, d(a,b)>|d(A,B)-d(a,A)-d(b,B)|=1/3.

Quote:
Originally Posted by aptx4869
and why this contradicts the fact that f(v_n) converges to j in X2?
Because we assumed that it does and it led to "d2(f(v_n),f(u_n))>d2(j,h)/3" for all n, which contradicts the fact that we can render d2(f(z),f(y)) as small as desired when z and y are close enough to x (this is merely your hypothesis).

Quote:
Originally Posted by aptx4869
"clearly" the sequence (f(u_n)) is cauchy?
Well, we have that d2(f(u_n),f(u_m)) ->0 by hypothesis (because for any epsilon >0, there is N such that n,m>=N => d1(u_n,x)<delta and d1(u_m,y)<delta, which in turns implies that d2(f(u_n),f(u_m))<epsilon by your hypothesis (note that our delta depends on epsilon). This just says that limit d2(f(u_n),f(u_m)) goes to 0 as n,m->+infinity, which is precisely the definition for a limit to be Cauchy.
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May 7th, 2007, 05:02 AM   #9
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That's cool! Now I understand it. Thanks a lot!
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