Re: Calculus XOR Probability

Tony Orlow wrote:
cbrown@xxxxxxxxxxxxxxxxx said:
Tony Orlow wrote:
cbrown@xxxxxxxxxxxxxxxxx said:


Fine; since it is irrelevant to providing you with a counter-example to
"infinite induction", continue to call them "sets of pairs in R^2".

(1) Do you think I can't define a sequence of staircases {C_n} as a
sequence of sets of pairs in R^2? I have already done this for you at
least once: see, for example, the end of:

http://groups.google.com/group/sci.math/msg/8b80a8687cead0c6?&hl=en&q=+risers+treads

That seems to point to this message. :(

Apaologies; that should be the same link as I gave elsewhere:

http://groups.google.com/group/sci.math/msg/8b80a8687cead0c6?dmode=source

For convenience, here is the section I was reffering to in that post,
modified to be in keeping with terms I am currently using:

----
Call the kth staircase C_k. Then the set C_k of pairs in R^2 is the
union of two sets: R_k (the kth set of risers) and T_k (the kth set of
treads).

R_k = ((x,y) : x = j/k, j in N and 0< j <= k; x - 1/k < = y <= x)

T_k = ((x,y) : y = j/k, j in N and 0 < j <= k; y - 1/k <= x <= y)

C_k = R_k union T_k.

To repeat the earlier argument:

Define D = {(a,b) : b = 1 - a, a,b >= 0}

Let p = (p_a, p_b) be any pair in D.

Define "the line passing through p with slope 1" as the set L_p =
{(a,b): x is a real number, a = x + p_a, b = x + p_b}.

Let p_k be the intersection of L_p and the set C_k.

It is tedious to prove, but it should be fairly obvious that the
sequence (p_1, p_2, p_3, ...) is well-defined (i.e., there is a single,
unique pair p_k for all k in N); and that furthermore this sequence has
limit p using the usual delta-epsilon proof; since the (Euclidean)
distance between p and p_k is always <= sqrt(2)/(2*k).

Therefore, every pair p in D is a limit point of the sequence C = (C_1,
C_2, ..., C_n, ...); so D is at least a subset of the limit of C.

Conversely, suppose p is not in D; then let d be the minimum of the
distances from p to each point q in D. Then there is some n such that
for all m > n, the distance from the closest pair in C_m to p is
greater than 0; so it cannot be the case that there is a sequence of
pairs {p_k} where p_k in C_k such that the sequence has limit p;
therefore p is not a limit point of C if it is not a pair in D.

Therefore every pair p in D is a limit point of {C_k}; and no pair p
not in D can be a limit point of {C_k}; so therefore lim n->oo {C_k} =
D.

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

Cheers - Chas

.

Relevant Pages

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