北京大学数学科学学院期末试题参考答案.pdf
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期末考试参考答案
北京大学数学科学学院期末试题参考答案
2020 – 2021 学年第 1 学期
考试科目： 拓扑学 考试时间： 2021 年 1 月 13 日
1. 设 f, g : X → Y 是连续映射，其中 Y 是 Hausdor 空间。证明
x ∈ X
f(x) = g(x)
是 X 中闭
集。
Proof. A :=
x ∈ X
f(x) = g(x)
is the compliment of B :=
x ∈ X
f(x) = g(x)
. It suces to show
that B is open. For any x ∈ B, f(x) and g(x) are distinct and hence has disjoint open neighborhoods U
and V since Y is hausdor. W = f
−1
(U) ∩ g
−1
(V ) is an open neighborhood of x. For any z ∈ W , we have
f(z) ∈ U and g(z) ∈ V , so f (z) = g(z) and z ∈ B. We have z ∈ W ⊂ B. Hence z ∈ B
o
. It follows that B
is open and A is closed.
2. 设 X 是可数的拓扑空间（即 X 作为集合是可数的）。若 X 是正则空间，证明 X 是正规空间。
Proof. Let A and B be disjoint closed subsets in X. For any x in A, the open subset B
c
contains x. Since
X is regular, there exists an open neighborhood U of x such that U ⊂ B
c
. Since A is (at most) countable,
we have countably many open subsets U
1
, U
2
, · · · such that
A ⊂
∞
[
n=1
U
n
, U
n
⊂ B
c
By same arguments, there are countably many open subsets V
1
, V
2
, · · · such that
B ⊂
∞
[
n=1
V
n
, V
n
⊂ A
c
We then have the collection of open sets
U
′
= U
n
\
n
[
k=1
V
k
and V
′
= V
n
\
n
[
k=1
U
k
Since A is disjoint from each V
k
, we have A ∩ U
′
n
= A ∩ U
n
, and therefore
A ⊂ U :=
∞
[
n=1
U
′
n
and similarly B ⊂ V :=
∞
[
n=1
V
′
n
If U ∩ V is nonempty and x ∈ U ∩ V , then x belongs to some U
′
m
and some V
′
n
. If m ⩽ n, then
x ∈ U
′
n
⊂ U
n
and x ∈ V
′
n
= V
n
\
n
[
k=1
U
k
which is clearly a contradiction; we reach a similar contradiction if n ⩽ m. Hence U and V are disjoint,
and so X is normal.
Remark. One may want to show that X is second countable, and then apply Lindelöf theorem to show the
normality of X. However, a countable topological space is not necessarily second countable. For example,
the one point compactication of rationals is not second countable, and it is even not rst countable.
1
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