Examples of net convergence
1. In R', let £ =
(f e R*
f(x) 0 or 1, and
fix} -- 0 only finitely often) and let p be the function in R^ which is identically 0. Then, in the product topology on R*, p e ñ (refer to Example 10.6\ Find a net
t f ) in £ which converges to p.
10.6 Exampies. a) Consider N = R^ with the product topology. Let
E —— f e R° ( /(x) = 0 or 1 and /(x) = 0 only finitely often},
and let q e R‘ be the function which is 0 everywhere. Then if
UQ) is a basic nhood of q, we have
UQ) —— (/i e R" |/i(y) —
g(y)| for some finite set I R and some r > 0. But such a nhood U(p) meets £ in the
function h which is 0 on elements of
F arid 1 elsewhere. Hensce, q CU £. On
the other hand, if (/„) is a sequence in £, with each /, being 0 on the finite set
A„then any function which is a limit of the sequence
f j can be rem at most on the
countable set _
A . Since p does not meet this requirement, no sequence
in
E can converge to p,
Since sequences cannot describe the topology of R‘, the criterion (or con- tinuity given in Theorem 10.5 for first-countable spaces probably fails here. In Exercise l0B, you are asked to find a noncontinuous function
F , R^ —• R with the property that whenever
f —• f in R‘, then
F(f ) —+
F(/).b) Recall that n denotes the set of ordinals co„ the first uncountable
ordinal, and f2, = n — ;‹»,}. Put the order topology (6D) on f2, for which a
subbase consists of all sets [1, e) — {y | 1 y
(e) is a nhood of « in this topology, while if e is a limit ordinal, the nhoods (§, «], Q
Now note that m
l e f2o i n this topology. But if (e,) were a sequence in f2
0
with limit sequences
=l. we would have ni = sup (z,{ contradicting Theorem 1.20. Thus
rail to describe the topology on f2.
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