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Dugunji J. Topology (1966)(T)(460s)_MDgt_.djvu |
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lates at some Yo Y, we have Y0 ( F = ( F, and therefore the
intersection is not empty...
Families of compact subsets of a space have the useful property:
1.6 Let X be Hausdorff, let o be an initial ordinal, and let {F,
be a descending family of nonempty compact subsets; that is,
/ < v F, D F...
, for each n < o there exists an an .Q and a constant cn such that f(% n) =- cn
for all c > c%; lett/ng c% = sup c, we find a0 < f) and thatfis constant on each
strip {(a, n) I a > a0}...
This
is actually weaker than normality: [0,/2] x [0, X2] - (X2, X2) is not normal, but
it is weakly normal...
Local Compactness 237
compact fiber p-(y) and, since p is a closed map, find a nbd V(y) ofy
such that p-*(V(y)) C (.J W,<,...
Since is compact and there-
fore normal, there is a continuous f: V 2 -+ I having value 1 at p and 0 on
2 - V...
Compactification 245
f: Z + --> I be the map sending E to 0 and N to 1, its extension F:/Z + -- I shows
thatffC F-(0),)C F-l(1),andsoEN = ;...
It is easy to see that this is
actually a topology and that the subspace X C 2 is homeomorphic to X...
The class of k-spaces is larger than that of the locally compact spaces;
for example, it contains also all the metric spaces, since
9.3 Every locally compact and every 1 countable space is a k-space...
Prove: A regular space Y is countably compact if and only if every point-
finite open covering has a finite subcovering...
The c-topology in yx is that having as sub-
basis all sets (A, V), where A C X is compact and V C Y is open...
Using 6., we conclude that
there is a continuous 9- = , with -+% so by XI, 1.3, -is
compact...
It is important to observe that the metric d in Z plays a dual role:
It determines the set C(Y, Z; d) and imposes a metric topology on this
set...
Since convergence in cartesian products is equivalent to coordinate
convergence, it is evident from 9.3 that convergence in Zr(p) is equivalent
to pointwise convergence...
XIII
In this chapter, we study the set of continuous real-valued functions on
a space Y; in particular, we obtain the algebraico-topological Stone-
Weierstrass theorem, which is one of the fundamental facts of modern
analysis,
I...
Now,
letting x = 1 - t, we conclude that s=(1 - t) 1 - 7 uniformly on
[0, 1], so the polynomials p,(t) = 1 - s(1 - t) fulfill the requirement
of the lemma...
For, with this additional requirement, it is easy
to see that the family { f + f +, i(f - f +) I f D} of real-valued functions
is separating; therefore it can be used to approximate separately the real
and imaginary parts of any given complex-valued function...
Now let Y be com-
pletely regular, and let O : Y -- fi(Y) be its Stone-ech compactification...
A continuous map
o: X--> Y induces a map o+: (Y)-->(X) by setting o+(g) = go
for each g (Y); note that 0+(8) = 8...
We will first discuss
complete metrics and then, in Section 9, we consider complete gauge
structures...
Proof." By 2.5, A is d-complete, so in view of 3.5, we need prove only
the general proposition: A is totally d-bounded if and only if A is totally
d-bounded...
Then d+(p + s, p) < e/3, and p + s is the function we seek, since
] p(x + h) + s(x + h) - p(x) - s(x) > s(x + h)h -s(x) I
_ p(x+ h-p(x) l '
and for each x [0,1 - 1Ci, we can evidently find an h ]0, l/n] such that the
right side is >M + n + 1 - M = n + 1...
The identity map f:/1 -/1 is uniformly
continuous, but cannot be extended to a continuous F: E --/1, since E s
connected and F cannot be constant...
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