第 2 回レポート問題 解答
幾何学 AII/ 幾何学 I ( 担当 : 新國 )
2013
年12
月18
日(水)
出題問題
.
以下の大問1 , 2 , 3 , 4
の全てに解答せよ.1
標準的な位相によるn
次元Euclid
空間( R n , O ( R n ))
において, 単位(n − 1)
次 元球面S n − 1 = {
(x 1 , x 2 , . . . , x n ) ∈ R n
∑ n
i=1
x 2 i = 1 }
は
R n
の有界な閉集合であることを示せ.2 (X, O )
を位相空間とし,A, B
はX
の部分集合でA ⊃ B
であるとする. いま,B
がA
のコンパクト部分集合であるならば,B
はX
のコンパクト部分集合で もあることを示せ.3 (X, O )
を位相空間とし,X ′
を集合とする. また,f : X → X ′
を写像とし,f
によりO
から誘導されるX ′
の位相をO ′
で表す.1
このとき,f
が全射で, か つ(X, O )
がコンパクトであれば, 位相空間(X ′ , O ′ )
もコンパクトであること を示せ.4 (X, O )
を位相空間とし,U
をその閉集合系とする. このとき,以下の条件(T 4 ),
(T 4 ) ′
は互いに必要十分条件であることを示せ.(T 4 ) A 1 ∩ A 2 = ∅
なるA 1 , A 2 ∈ U
に対し, あるO 1 , O 2 ∈ O
が存在して,A 1 ⊂ O 1 , A 2 ⊂ O 2 , O 1 ∩ O 2 = ∅
が成り立つ.(T 4 ) ′ A ⊂ O
なるA ∈ U , O ∈ O
に対し, あるO 1 ∈ O
が存在して,A ⊂ O 1
か つO ¯ 1 ⊂ O
が成り立つ.以上
1 § 4.3
定義4.3.2
を参照せよ.位相空間(X, O ),
及び写像f : X → X ′
に対し,X ′
の部分集合族O ′
をO ′ = {
O ′ ⊂ X ′ | f − 1 ( O ′ )
∈ O }
で定義すると,これは
X ′
の位相となる(定理 4.3.1).
このO ′
を,f
によりO
から誘導されるX ′
の位相と呼ぶのであっ た.解答 .
1 S n − 1
は閉球体B ∗ (0; 1)
に含まれるので有界である. 以下,S n − 1
がR n
の閉集合 であることを示す. そのためには,R n − S n − 1
がR n
の開集合であること, 即ち, 任 意のx ∈ R n − S n − 1
に対し, あるε > 0
が存在して,x
を中心とする半径ε
の開球 体B (x; ε)
がR n − S n − 1
に含まれることを示せば良い. 以下,x, y ∈ R n
に対し,x
とy
の距離をd (x, y)
で表すとき,2
任意のx ∈ R n − S n−1
はd (x, 0) ̸ = 1
をみたす.即ち
d (x, 0) > 1
またはd (x, 0) < 1
が成り立つので,この2
通りに場合を分ける.(1) d (x, 0) > 1
の場合: 正数ε > 0
をε = d (x, 0) − 1 (i)
で定義しよう. いま,
a ∈ B (x; ε)
に対し, 三角不等式(命題 1.1.3 (4))
からd (x, 0) ≤ d (x, a) + d (a, 0) (ii)
が成り立つ. そこで(i), (ii)
よりd (a, 0) ≥ d (x, 0) − d (x, a) > (ε + 1) − ε = 1
となるので,
d (a, 0) > 1,
即ちa ∈ R n − S n − 1
である. 従ってB (x; ε) ⊂ R n − S n − 1
となる.(2) d (x, 0) < 1
の場合: 正数ε > 0
をε = 1 − d (x, 0) (iii)
で定義しよう. いま,
a ∈ B (x; ε)
に対し, 三角不等式及び(iii)
からd (a, 0) ≤ d (a, x) + d (x, 0) < ε + (1 − ε) = 1
が成り立つ. 故に
d (a, 0) < 1,
即ちa ∈ R n − S n − 1
である. 従ってB (x; ε) ⊂ R n − S n − 1
となる.以上により,
R n − S n − 1
はR n
の開集合であることが示された(図 1
は, 特にn = 2
の場合に, 上記議論を図示したものである). 従ってS n − 1
はR n
の閉集合である.注意.
1
の結果と定理7.3.4
を合わせて, 単位(n − 1)
次元球面S n−1
はコンパクト な位相空間であることがわかる( § 7.3
例1 (3)).
2
定義1.1.1
を参照.x = (x 1 , x 2 , . . . , x n ) , y = (y 1 , y 2 , . . . , y n ) ∈ R n
に対し,d(x,y) = v u u t ∑ n
i=1
(x i − y i ) 2
である.
x
0
a ε
1 x
0 a
1
ε B ( ; ) x ε
B ( ; ) x ε
(1) (2)
図
1: B (x; ε) ⊂ R n − S n − 1
なるx
の開近傍B (x; ε)
の存在(n = 2)
2 C = { O λ } λ ∈ Λ
をB
のX
における開被覆としよう. 即ち,B ⊂ ∪
λ ∈ Λ
O λ , O λ ∈ O (iv)
である
(図 2
の左図参照. 各O λ
はA
に含まれているとは限らないことに注意). そ こで各λ ∈ Λ
に対し,A
の部分集合O λ ′
をO λ ′ = O λ ∩ A (v)
で定義すると,
A
におけるO
の相対位相O A
の定義から,O ′ λ
は部分位相空間(A, O A )
の開集合であり, (iv), (v)及びB ⊂ A
から,B ⊂ ( ∪
λ ∈ Λ
O λ )
∩ A = ∪
λ ∈ Λ
(O λ ∩ A) = ∪
λ ∈ Λ
O λ ′ , O ′ λ ∈ O A
となる. 即ち,
C ′ = { O ′ λ } λ ∈ Λ
はB
のA
における開被覆である(図2
の右図参照). そこ で仮定よりB
はA
のコンパクト部分集合であるから,ある有限個のλ 1 , λ 2 , . . . , λ k ∈ Λ
が存在して,B ⊂ O ′ λ 1 ∪ O ′ λ 2 ∪ · · · ∪ O λ ′
k (vi)
が成り立つ. このとき
(vi), (v)
からB ⊂
∪ k
i=1
O ′ λ i =
∪ k
i=1
(O λ ∩ A) = ( k
∪
i=1
O λ i )
∩ A
となり, 故にB ⊂
∪ k
i=1
O λ i = O λ 1 ∪ O λ 2 ∪ · · · ∪ O λ k
である. これは
B
がX
のコンパクト部分集合であることを意味する.X
A B
X
A B O λ
'
O λ
''
O λ
O' λ
'
O' λ
''
O' λ
図
2:
左図:B
のX
における開被覆C = { O λ } λ ∈ Λ ,
右図:B
のA
における開被覆C ′ = { O λ ′ } λ ∈ Λ
3 C ′ = { O ′ λ } λ ∈ Λ
をX ′
の開被覆としよう. 即ち,X ′ = ∪
λ ∈ Λ
O λ ′ , O λ ′ ∈ O ′ (vii)
である. このとき, (vii)及び写像の演算から,
X = f − 1 (X ′ ) = f − 1
( ∪
λ ∈ Λ
O ′ λ )
= ∪
λ ∈ Λ
f − 1 (O λ ′ ) (viii)
となり, 更に位相O ′
の定義から, 各λ ∈ Λ
に対し,f − 1 (O λ ′ )
は(X, O )
の開集合で ある. 従って(viii)
より,C = { f − 1 (O ′ λ ) } λ ∈ Λ
はX
の開被覆である. そこで仮定より(X, O )
はコンパクトであるから, ある有限個のλ 1 , λ 2 , . . . , λ k ∈ Λ
が存在して,X = f − 1 ( O ′ λ 1 )
∪ f − 1 ( O ′ λ 2 )
∪ · · · ∪ f − 1 ( O λ ′
k
) (ix)
が成り立つ. 更にまた仮定より
f
は全射であるから,X ′ = f(X) (x)
である. このとき
(x), (ix)
及び写像の演算(f
は全射であることに注意)から,X ′ = f ( f − 1 (
O λ ′ 1 )
∪ f − 1 ( O ′ λ 2 )
∪ · · · ∪ f − 1 ( O ′ λ
k
))
= f ( f −1 (
O λ ′ 1 ))
∪ f ( f −1 (
O ′ λ 2 ))
∪ · · · ∪ f ( f −1 (
O ′ λ
k
))
= O ′ λ 1 ∪ O λ ′ 2 ∪ · · · ∪ O λ ′ k
となる. これは
(X ′ , O ′ )
がコンパクトであることを意味する.注意. 位相空間
(X, O )
及びX
における同値関係∼
において,X/ ∼
をX
の∼
によ る商集合とし,π : X → X/ ∼
を自然な全射とする. このとき,π
によりO
から誘導 されるX/ ∼
の位相O ′
を商位相といい, 位相空間(X/ ∼ , O ′ )
を, (X,O )
の( ∼
によ る)商空間というのであった(定義 4.3.3).
このとき, 3の結果から,もし(X, O )
が コンパクトならば, 商空間(X/ ∼ , O ′ )
もコンパクトである. 例えば,M¨ obius
の帯( § 4.3
例1 (2))
がコンパクトであることは, このことからわかる.4 (T 4 ) ⇒ (T 4 ) ′ : A ⊂ O
なるA ∈ U , O ∈ O
に対し,A ′ = O c
とおくと,A ′ ∈ U
か つA ∩ A ′ = ∅
である. 故に(T 4 )
から, あるO 1 , O 2 ∈ O
が存在して,A ⊂ O 1 , A ′ ⊂ O 2 , O 1 ∩ O 2 = ∅
が成り立つ
(図 3
参照). このとき,O 1 ⊂ O c 2
かつO 2 c ∈ U
から, 補題2.2.6 (2)
よりO ¯ 1 ⊂ O ¯ 2 c = O 2 c (xi)
となり, 一方,A ′ ⊂ O 2
より,O 2 c ⊂ A ′ c = O (xii)
となる. (xi), (xii)から
O ¯ 1 ⊂ O
である. 即ち, (T4 ) ′
が成り立つことが示された.(T 4 ) ′ ⇒ (T 4 ): A 1 ∩ A 2 = ∅
なるA 1 , A 2 ∈ U
に対し,O = A c 2
とおくと,O ∈ O
かつA 1 ⊂ O
である. 故に(T 4 ) ′
から, あるO 1 ∈ O
が存在して,A 1 ⊂ O 1 , O ¯ 1 ⊂ O
が成り立つ. そこでO 2 = ( O ¯ 1 ) c
とおくと,
O 2 ∈ O
かつO 1 ∩ O 2 = ∅
で,A 2 = O c ⊂ ( O ¯ 1 ) c
= O 2
となる
(図 4
参照). 即ち, (T4 )
が成り立つことが示された.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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O
X
A'
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A
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X
A'
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A X
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A
O 1 O 2
図
3: A ⊂ O 1
かつO ¯ 1 ⊂ O
なるO 1 ∈ O
の存在X X
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A 1 A 2
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X
1
O 2
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A 1
O
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A 2
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O
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A 1
O 1 O 1
図
4: A 1 ⊂ O 1 , A 2 ⊂ O 2 , O 1 ∩ O 2 = ∅
なるO 1 , O 2 ∈ O
の存在