[위상수학 솔루션]J.r Munkres topology

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[위상수학 솔루션]J.r Munkres topology에 대한 보고서 자료입니다.

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Solutions to Topology Homework #3, due Week 6.
Problems: Munkres Homework:
Section 13: 2, 3, 7
Sections 14-16: 2, 3, 10
13.2 Consider the nine topologies on the set X = {a, b, c} indicated in Example
1 of Section 12. Compare them; that is, for each pair of topologies,
determine whether they are comparable, and if so, which is the finer.
I will number the topologies from the upper left as 1,2,3 (top row), 4,5,6
(middle row) and 7,8,9 (bottom row). It is clear that 1 is the coarsest topology
of them all, being the trivial topology, and it is comparable to all the
rest. Furthermore, 9 is the finest of them all, being the discrete topology,
and is also comparable to all the rest. Other comparables are as follows:
Topology 2 finer than 7 and coarser than 8.
Topology 3 is finer than 4 and 7, and coarser than 6.
Topology 4 is coarser than 3, 6, and 8.
Topology 6 is finer than 3, 4 and 7.
Topology 7 is coarser than 2, 3, 6, and 8.
Topology 8 is finer than 2, 4, and 7.
13.3 Show that the collection Tc given in Example 4 of Section 12 is a topology
on the set X.
Is the collection T∞ = {U|X − U is infinite or empty or all of X} a
topology on X?
Proof: From the example we have Tc is the collection of all subsets U of
X such that X − U either is countable or is all of X. We check the three
properties of a topology as follows:
(i) ∅ ∈ Tc since X −∅ is all of X. Furthermore, X ∈ Tc since X − X = ∅
is countable.
(ii) If A is a collection of elements of Tc then X − A∈A A ⊆ X − A
for each A ∈ A. If all A are empty then clearly X − A∈A A = X ∈ Tc.
Otherwise, X −A∈A ⊆ X −A for some nonempty A ⊆ Tc. But since X −A
is countable, so is X − A∈A A, so this set too is in Tc.
(iii) Now consider finite intersections of elements of Tc. Since finite unions
of countable sets are countable, so using DeMorgan’s Law we get the result
that X −n
i=1 Ai = (X − Ai) which is countable. Therefore, n
i=1 Ai ∈ Tc.
It follows from these three properties that Tc is a topology. The same
is not necessarily true in the case of T∞. The reason is that intersection of
infinite sets may be finite. For example, let X = R, and U1 = Q − {0} and
U2 = I. Then U1, U2 ∈ T∞ but U1 U2 = R − {0}, with complement {0}
which is neither infinite nor empty nor all of X. Therefore U1U2 is not in
T∞, which is consequently not a topology.

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