My Math Forum limit of a function in a metric space

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 May 6th, 2007, 12:35 AM #1 Member   Joined: Mar 2007 Posts: 57 Thanks: 0 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?
 May 6th, 2007, 01:45 AM #2 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 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)) h. This means that you can find an index N for which n>=N => d2(f(v_n),j)d2(j,h)/3, which contradicts your hypothesis.
 May 6th, 2007, 01:59 AM #3 Member   Joined: Mar 2007 Posts: 57 Thanks: 0 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
 May 6th, 2007, 05:16 AM #4 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 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.
 May 6th, 2007, 05:45 PM #5 Member   Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 ??? so why is that x useful sometimes? Isn't that true for any x around y and z?
 May 6th, 2007, 08:10 PM #6 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 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 0x0 condition: for all epsilon>0, there exists delta>0, for every x such that d2(x,x0)
 May 7th, 2007, 12:21 AM #7 Member   Joined: Mar 2007 Posts: 57 Thanks: 0 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 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!
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.

 May 7th, 2007, 05:02 AM #9 Member   Joined: Mar 2007 Posts: 57 Thanks: 0 That's cool! Now I understand it. Thanks a lot!

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