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c University of Leeds, School of Mathematics
MATH 3224 Topology: Some useful results and formulae (2)
J.R. Partington
November 2000
Please let me know of any errors, omissions or obscurities!
Completeness. A sequence in a metric space is Cauchy if ∀ǫ > 0, ∃n0 s.t. d(xn , xm ) < ǫ
for all n, m ≥ n0 . Any convergent sequence is Cauchy. A metric space is complete if all
Cauchy sequences in X have limits in X. Example: Rn . Complete subsets are always
closed; closed subsets of complete spaces are complete.
Two metrics d1 and d2 are equivalent if they define the same topology, i.e., ι : (X, d1 ) →
(X, d2 ) is a homeomorphism. A function f : X → Y between metric spaces is an
isometry if dY (f (x1 ), f (x2 )) = dX (x1 , x2 ) for all x1 , x2 ∈ X.
Completion. For any metric space (X, d) there is a completion, i.e., a metric space
˜ and a mapping i : X → X̃, such that X̃ is complete, i is an isometry, and i(X)
(X̃, d),
is dense in X̃. Construct X̃ by taking equivalence classes of Cauchy sequences in X,
where (xn ) ∼ (yn ) if and only if d(xn , yn ) → 0. From Q we get R, from (0, 1) we get
[0, 1].
Contraction mappings. T : (X, d) → (X, d) (metric) is a contraction mapping if for some
k with 0 < k < 1, d(T x, T y) ≤ kd(x, y) for all x, y ∈ X. Every contraction mapping has
a unique fixed point x such that T x = x. Proof: uniqueness—easy (suppose T x = x,
T y = y, show d(x, y) = 0); existence—pick any x0 and let xn = T n x0 , which form a
Cauchy sequence whose limit is the fixed point of T .
Completeness of C[a, b] with d(f, g) = max |f (t) − g(t)|. Identify the limit f (pointwise
values fn (t) are Cauchy in R for each t, and hence define f (t) = lim fn (t)). Now f is
continuous since fn → f uniformly.
Picard’s theorem. Let F : R2 → R be continuous and satisfy the Lipschitz condition
|F (x1 , y) − F (x2 , y)| ≤ |x1 − x2 | ∀y. Then the differential equation dx/dt = F (x, t),
x(0) = a, has a unique solution on (−1, 1). Proof: work with C[−k, k], 0 < k < 1,
Rt
and show that (T f )(t) = a + 0 F (f (s), s) ds defines a contraction mapping on the
closed subspace of all functions f with f (0) = a. Its fixed point x is the solution to the
differential equation.
Implicit function theorem. Let f : [a, b] → R be continuous in the first variable,
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differentiable in the second, with 0 < m ≤ fy (x, y) ≤ M ∀x, y. Then there is a unique
ψ ∈ C[a, b] with f (x, ψ(x)) = 0. Proof: (T φ)(x) = φ(x) − (1/M )f (x, φ(x)) defines a
contraction on C[a, b]. Its fixed point ψ solves the equation.
Hausdorff spaces and normal spaces. (X, τ ) is Hausdorff if we can separate points by
open sets, i.e., ∀x, y ∈ X with x 6= y, ∃U , V ∈ τ , such that U ∩ V = ∅, and x ∈ U ,
y ∈ V . (X, τ ) is normal if we can separate closed sets by open sets, i.e., ∀F , G closed
with F ∩ G = ∅, ∃U , V ∈ τ , such that U ∩ V = ∅, and F ⊆ U , G ⊆ V . In Hausdorff
spaces, one point sets are closed. All metric spaces are Hausdorff and normal. In a
Hausdorff space, if xn → x and xn → y, then x = y (sequences have unique limits). A
non-Hausdorff example—the indiscrete topology on R.
Compactness. Let K ⊆ X, where (X, τ ) is a topological space. Then K is compact if
S
every open cover of K has a finite subcover, which means that whenever K ⊆ λ∈Λ Uλ ,
with each Uλ open, then K ⊆ Uλ1 ∪ . . . ∪ Uλn for some λ1 , . . . , λn ∈ Λ. Continuous
(real) functions on compact sets are bounded and attain their bounds. In a Hausdorff
space any compact set is closed. Any compact subset of a metric space is bounded.
Closed subsets of compact sets are compact. Let K be compact and f continuous; then
f (K) is compact. So if f : X → Y is a continuous bijection, where X is compact and
Y Hausdorff, then f is a homeomorphism.
Heine–Borel. Any closed bounded real interval [a, b] is compact. Hence compactness
in R is equivalent to being closed and bounded.
One point compactification. Given (X, τ ), add a new point, ∞, say, to get (X∞ , τ∞ )
where X∞ = X ∪ {∞} and U ∈ τ∞ ⇐⇒ U ∈ τ or X∞ \ U is a closed compact subset
of X. Then (X∞ , τ∞ ) is compact and τ = τ∞ ∩ X. Examples: R2 gives the sphere by
stereographic projection; R and (0, 1) give a circle.
Sequential compactness. K is sequentially compact if every sequence in K has a convergent subsequence with limit in K.
Precompact sets. A subset K of a metric space is precompact or totally bounded if
for each ǫ > 0 it can be covered with finitely many balls B(xk , ǫ). Any compact
or sequentially compact set is precompact. Note that the closed unit ball in C[0, 1],
B(0, 1) = {f ∈ C[0, 1] : max |f (t)| ≤ 1} is not totally bounded, hence not compact
despite being closed and bounded.
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