Thus, for any sequence (Sn,tn) decreasing to (s,t), the sequence
{f(s n,n ) N is Cauchy in E complete; hence lim f(s ,t ) exists. We
n n ncN n n
n
show now that we get the same limit for any sequence decreasing to
(s,t).
Consider two sequences (S ,t ) and (s',t'), both decreasing to
n n n n
(s,t). We construct a new sequence (p n,rn) as follows.
We set (P1,r ) = (sl,t ), (p2,r ) = the first term of (s3,t')
smaller than (p1,r ), (p ,r ) = the first term of (sn,t ) after
(s ,t ) smaller than (p2,r2), etc. The sequence (pn,r ) then
decreases and converges to (s,t) since both the even- and odd-
numbered terms do. Then L = lim f(p ,r ) exists from above. Looking
n
at the odd-numbered terms, we have L = lim f(p 2k+,r 2k1). But the
k
odd-numbered terms form a subsequence of (sn,tn), and we know
f(s ,t ) converges, so L = lim f(s ,t ) as well. Similarly, since
n n n n
n
the even-numbered terms form a subsequence of (s',tn), we obtain
n n
L = lim f(s',t') as well. The limit is then independent of the
n n
particular sequence, so lim f(s',t') exists, and it is
(s',t')++(s,t)
this limit we denote by f(s ,t ).
The proofs of (2)-(4) are similar, and we will omit some
computational details where they are identical to the ones for (1).
Proof of (2). We consider a sequence (s ,t ) (s,t), with s n+s,
n n n
tn++t, and show that the sequence {f(s ,t )}nN is Cauchy in E, which
we again do by denial.