Toda coleção de abertos não-vazios, dois a dois disjuntos é enumerável.
Seja a coleção de abertos disjuntos
X
=
(
A
λ
)
λ
∈
Λ
{\displaystyle X=(A_{\lambda })_{\lambda \in \Lambda }}
x
λ
∈
A
λ
{\displaystyle x_{\lambda }\in A_{\lambda }}
x
λ
∈
I
λ
{\displaystyle x_{\lambda }\in I_{\lambda }}
I
λ
;
x
λ
∈
I
λ
⊂
A
λ
{\displaystyle I_{\lambda };x_{\lambda }\in I_{\lambda }\subset A_{\lambda }}
Q
{\displaystyle \mathbb {Q} }
R
{\displaystyle \mathbb {R} }
x
λ
∈
I
λ
⊂
A
λ
⇒
∃
q
λ
∈
Q
;
q
λ
∈
I
λ
{\displaystyle x_{\lambda }\in I_{\lambda }\subset A_{\lambda }\Rightarrow \exists \;q_{\lambda }\in \mathbb {Q} ;q_{\lambda }\in I_{\lambda }}
(
A
λ
)
λ
∈
Λ
{\displaystyle (A_{\lambda })_{\lambda \in \Lambda }}
q
λ
∈
I
λ
⊂
A
λ
⇒
q
λ
∉
A
μ
{\displaystyle q_{\lambda }\in I_{\lambda }\subset A_{\lambda }\Rightarrow q_{\lambda }\not \in A_{\mu }}
μ
≠
λ
{\displaystyle \mu \not =\lambda }
f
:
X
→
Q
{\displaystyle f:X\rightarrow \mathbb {Q} }
A
λ
{\displaystyle A_{\lambda }}
q
λ
{\displaystyle q_{\lambda }}
Q
{\displaystyle \mathbb {Q} \;}
O conjunto dos valores de aderência de uma sequência é um conjunto fechado.
Seja
X
=
(
x
n
)
n
∈
N
=
{
x
1
,
x
2
,
.
.
.
,
x
n
,
.
.
.
}
{\displaystyle X=(x_{n})_{n\in \mathbb {N} }=\{x_{1},x_{2},...,x_{n},...\}\;}
lim
x
n
=
a
,
a
≠
x
n
,
∀
n
∈
N
{\displaystyle \lim x_{n}=a,a\neq x_{n},\forall \;n\in \mathbb {N} }
X
∪
{
a
}
=
X
∪
X
′
=
X
¯
{\displaystyle X\cup \{a\}=X\cup X'={\overline {X}}}
l
i
m
x
n
=
a
,
X
=
{
x
1
,
x
2
,
.
.
.
,
x
n
,
.
.
.
}
⇒
X
¯
=
X
∪
{
a
}
{\displaystyle limx_{n}=a,X=\{x_{1},x_{2},...,x_{n},...\}\Rightarrow {\overline {X}}=X\cup \{a\}\;}
a é o único ponto de acumulação de X
⇒
X
′
=
{
a
}
,
{\displaystyle \Rightarrow X'=\{a\},}
X
∪
{
a
}
=
X
∪
X
′
=
X
¯
{\displaystyle X\cup \{a\}=X\cup X'={\overline {X}}\;}
O número
1
4
{\displaystyle {1 \over 4}\;}
editar
Leve em consideração a Falhou a verificação gramatical (Erro de conversão. O servidor ("https://wikimedia.org/api/rest_") devolveu: "Cannot get mml. upstream connect error or disconnect/reset before headers. reset reason: connection termination"): {\displaystyle {sigma_{n=1}_{\infty }}<!--<math>\;}
{\displaystyle }
⇒
⇔
⇐
→
∃
∀
{
a
b
{
}
a
3
X
¯
R
↦
∖
⟨
⟩
‖
‖
∈
⊂
⊃
x
¯
∩
∪
e
∅
{\displaystyle \;\Rightarrow \;\Leftrightarrow \;\Leftarrow \;\rightarrow \;\exists \;\forall \;{\begin{cases}\;\\\;\end{cases}}\;{a \over b}\;{\bigg \{}{\bigg \}}\;{\sqrt[{3}]{a}}\;{\overline {X}}\;\mathbb {R} \;\mapsto \;\setminus \left\langle \right\rangle \|\|\;\in \;\subset \;\supset \;{\bar {x}}\;\cap \;\cup \;{\mbox{e}}\;\emptyset }
Sejam F, G conjuntos fechados disjuntos tais que
F
∪
G
{\displaystyle F\cup G\;}
F
=
∅
{\displaystyle F=\varnothing }
G
=
∅
{\displaystyle G=\varnothing \;}
Pela Hipótese
F
∩
G
=
∅
{\displaystyle F\cap G=\varnothing \;}
Suponhamos que F,G sejam infinitos, como
F
∪
G
{\displaystyle F\cup G\;}
∃
a
,
b
∈
F
∪
G
;
a
∈
F
,
b
∉
F
{\displaystyle \exists \;a,b\in F\cup G;a\in F,b\not \in F\;}
F
∩
G
=
∅
⇒
a
∉
G
,
b
∈
G
{\displaystyle F\cap G=\varnothing \Rightarrow a\not \in G,b\in G}
{
∃
h
∉
F
,
h
<
b
⇒
(
h
,
b
)
∩
F
=
∅
,
se
a
<
b
∃
h
∉
F
,
b
<
h
⇒
(
b
,
h
)
∩
F
=
∅
,
se
b
<
a
{\displaystyle {\begin{cases}\exists h\not \in F,h<b\Rightarrow (h,b)\cap F=\varnothing ,{\mbox{ se }}a<b\\\exists h\not \in F,b<h\Rightarrow (b,h)\cap F=\varnothing ,{\mbox{ se }}b<a\end{cases}}}
t
=
{
inf
{
h
∈
G
;
[
h
,
b
]
⊂
G
}
,
se
a
<
h
<
b
sup
{
h
∈
G
;
[
b
,
h
]
⊂
G
}
,
se
b
<
h
<
a
{\displaystyle t={\begin{cases}\inf\{h\in G;[h,b]\subset G\},{\mbox{ se }}a<h<b\\\sup\{h\in G;[b,h]\subset G\},{\mbox{ se }}b<h<a\end{cases}}}
F
∩
G
=
∅
⇒
t
∉
F
{\displaystyle F\cap G=\varnothing \Rightarrow t\not \in F}
Como
F
∩
G
=
∅
⇒
{\displaystyle F\cap G=\varnothing \Rightarrow }
[
a
,
t
)
⊂
F
,
se
a
<
b
{\displaystyle [a,t)\subset F,{\mbox{ se }}a<b}
(
t
,
b
]
⊂
F
,
se
b
<
a
{\displaystyle (t,b]\subset F,{\mbox{ se }}b<a}
Logo
∄
a
,
b
∈
F
∪
G
;
a
∈
F
,
b
∉
F
{\displaystyle \not \exists \;a,b\in F\cup G;a\in F,b\not \in F\;}
a
,
b
∈
F
,
G
=
∅
ou
a
,
b
∈
G
,
F
=
∅
{\displaystyle a,b\in F,G=\varnothing {\mbox{ ou }}a,b\in G,F=\varnothing }
Suponhamos que G seja finito, tome
t
∈
G
{\displaystyle t\in G\;}
F
∩
G
=
∅
⇒
t
∉
F
,
{\displaystyle F\cap G=\varnothing \Rightarrow t\not \in F,}
∀
ϵ
>
0
,
(
t
−
ϵ
,
t
+
ϵ
)
∩
F
≠
∅
⇒
t
∈
F
′
⊂
F
{\displaystyle \forall \;\epsilon >0,(t-\epsilon ,t+\epsilon )\cap F\neq \varnothing \Rightarrow t\in F'\subset F}
G
=
∅
{\displaystyle G=\varnothing }
Analogamente se supormos que F é finito será um absurdo, portanto
F
=
∅
{\displaystyle F=\varnothing }
Pela Hipótese
F
∩
G
=
∅
{\displaystyle F\cap G=\varnothing \;}
Como
F
∪
G
{\displaystyle F\cup G\;}
F
∪
G
=
[
a
,
b
]
{\displaystyle F\cup G=[a,b]\;}
Tomemos F e G infinitos e
c
∈
(
a
,
b
)
{\displaystyle c\in (a,b)\;}
c
∈
F
{\displaystyle c\in F\;}
F
∩
G
=
∅
⇒
c
∉
G
{\displaystyle F\cap G=\varnothing \Rightarrow c\not \in G}
Suponha que
a
∉
F
{\displaystyle a\not \in F}
F
∪
G
=
[
a
,
b
]
,
F
∩
G
=
∅
⇒
a
∈
G
{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow a\in G}
∃
t
∈
[
a
,
c
)
;
[
a
,
t
]
⊂
G
⇒
t
∉
F
,
∃
b
−
a
>
ϵ
>
0
;
(
a
,
a
+
ϵ
)
⊄
G
{\displaystyle \exists \;t\in [a,c);[a,t]\subset G\Rightarrow t\not \in F,\exists b-a>\epsilon >0;(a,a+\epsilon )\not \subset G}
F
∪
G
=
[
a
,
b
]
,
F
∩
G
=
∅
⇒
(
a
,
a
+
ϵ
)
⊂
F
{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow (a,a+\epsilon )\subset F}
a
∈
F
{\displaystyle a\in F}
Analogamente, se supormos que
b
∉
F
{\displaystyle b\not \in F}
b
∈
F
{\displaystyle b\in F}
Suponha que
G
≠
∅
{\displaystyle G\neq \varnothing }
c
∈
F
∩
(
a
,
b
)
{\displaystyle c\in F\cap (a,b)}
∃
I
p
⊂
G
∩
(
(
a
,
c
)
∪
(
c
,
b
)
)
{\displaystyle \exists \;I_{p}\subset G\cap ((a,c)\cup (c,b))}
I
p
{\displaystyle I_{p}\;}
Assim
I
p
=
[
u
,
v
]
⊂
G
∩
(
(
a
,
c
)
∪
(
c
,
b
)
)
{\displaystyle I_{p}=[u,v]\subset G\cap ((a,c)\cup (c,b))\;}
u
,
v
∉
F
,
∃
inf
{
u
−
a
,
b
−
v
}
>
ϵ
>
0
;
(
v
,
v
+
ϵ
)
,
(
u
−
ϵ
,
u
)
⊄
G
{\displaystyle u,v\not \in F,\exists \inf\{u-a,b-v\}>\epsilon >0;(v,v+\epsilon ),(u-\epsilon ,u)\not \subset G}
F
∪
G
=
[
a
,
b
]
,
F
∩
G
=
∅
⇒
(
v
,
v
+
ϵ
)
,
(
u
−
ϵ
,
u
)
⊂
F
{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow (v,v+\epsilon ),(u-\epsilon ,u)\subset F}
Suponhamos que G seja finito, tome
t
∈
G
∩
[
a
,
b
]
{\displaystyle t\in G\cap [a,b]\;}
F
∪
G
=
[
a
,
b
]
,
F
∩
G
=
∅
⇒
{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow }
⇒
t
∉
F
,
∃
inf
{
t
−
a
,
b
−
t
}
>
ϵ
>
0
;
(
t
−
ϵ
,
t
−
ϵ
)
−
t
⊄
G
{\displaystyle \Rightarrow t\not \in F,\exists \inf\{t-a,b-t\}>\epsilon >0;(t-\epsilon ,t-\epsilon )-{t}\not \subset G}
F
∪
G
=
[
a
,
b
]
,
F
∩
G
=
∅
⇒
(
t
−
ϵ
,
t
−
ϵ
)
−
t
⊂
F
{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow (t-\epsilon ,t-\epsilon )-{t}\subset F}
Portanto
G
=
∅
{\displaystyle G=\varnothing }
c
∈
[
a
,
b
]
∩
G
F
=
∅
{\displaystyle c\in [a,b]\cap GF=\varnothing }
Tome
a
∈
F
∪
G
{\displaystyle a\in F\cup G}
(
a
−
ϵ
,
a
+
ϵ
)
∩
F
,
G
≠
∅
⇒
a
∈
F
′
,
G
′
⇒
a
∈
F
∩
G
{\displaystyle (a-\epsilon ,a+\epsilon )\cap F,G\neq \varnothing \Rightarrow a\in F',G'\Rightarrow a\in F\cap G}
Seja E enumerável. Consiga uma sequência cujo conjunto dos valores de aderência é
E
¯
{\displaystyle {\overline {E}}\;}
F
⊂
R
{\displaystyle F\subset \mathbb {R} }
editar
Por ser E enumerável,
Se E for finito,
E
=
{
x
1
,
x
2
,
.
.
.
,
x
n
}
{\displaystyle E=\{x_{1},x_{2},...,x_{n}\}\;}
∀
x
i
∈
E
,
ϵ
>
0
,
(
x
i
−
ϵ
,
x
i
+
ϵ
)
∩
E
=
{
x
i
}
⇒
E
′
=
∅
⇒
E
¯
=
{
x
1
,
x
2
,
.
.
.
,
x
n
}
.
{\displaystyle \forall \;x_{i}\in E,\epsilon >0,(x_{i}-\epsilon ,x_{i}+\epsilon )\cap E=\{x_{i}\}\Rightarrow E'=\varnothing \Rightarrow {\overline {E}}=\{x_{1},x_{2},...,x_{n}\}.}
Tomando E um fechado finito qualquer, temos que
E
=
{
x
1
,
x
2
,
.
.
.
,
x
n
}
⇒
E
=
{
x
i
;
x
i
∈
(
x
n
)
}
{\displaystyle E=\{x_{1},x_{2},...,x_{n}\}\Rightarrow E=\{x_{i};x_{i}\in (x_{n})\}}
(
x
i
−
ϵ
,
x
i
+
ϵ
)
∩
E
=
{
x
i
}
,
{\displaystyle (x_{i}-\epsilon ,x_{i}+\epsilon )\cap E=\{x_{i}\},}
Se E é um conjunto enumerável, que não é finito, assim dado
E
=
{
x
1
,
x
2
,
.
.
.
,
x
n
,
.
.
.
}
,
∃
ϕ
:
N
→
E
{\displaystyle E=\{x_{1},x_{2},...,x_{n},...\},\exists \;\phi :\mathbb {N} \rightarrow E}
ϕ
:
i
↦
ϕ
(
i
)
=
x
i
∈
E
,
i
∈
N
{\displaystyle \phi :i\mapsto \phi (i)=x_{i}\in E,i\in \mathbb {N} }
E
0
=
{
y
n
∈
E
;
y
i
<
y
i
+
1
,
∀
i
∈
N
}
com
y
i
=
m
i
n
{
E
−
{
y
1
,
y
2
,
.
.
.
,
y
i
−
1
}
}
.
{\displaystyle E_{0}=\{y_{n}\in E;y_{i}<y_{i+1},\forall \;i\in \mathbb {N} \}{\mbox{ com }}y_{i}=min\{E-\{y_{1},y_{2},...,y_{i-1}\}\}.}
E
=
E
0
{\displaystyle E=E_{0}}
y
p
,
y
p
+
1
,
{\displaystyle y_{p},y_{p+1},}
(
y
p
,
y
p
+
1
)
∩
E
=
∅
⇒
{\displaystyle (y_{p},y_{p+1})\cap E=\varnothing \Rightarrow }
∀
y
i
∈
E
,
ϵ
>
0
,
(
y
i
−
ϵ
,
y
i
+
ϵ
)
∩
E
=
{
y
i
}
⇒
E
′
=
∅
⇒
E
¯
=
E
.
{\displaystyle \forall \;y_{i}\in E,\epsilon >0,(y_{i}-\epsilon ,y_{i}+\epsilon )\cap E=\{y_{i}\}\Rightarrow E'=\varnothing \Rightarrow {\overline {E}}=E.}
editar
![{\displaystyle \;\Rightarrow \;\Leftrightarrow \;\Leftarrow \;\rightarrow \;\exists \;\forall \;{\begin{cases}\;\\\;\end{cases}}\;{a \over b}\;{\bigg \{}{\bigg \}}\;{\sqrt[{3}]{a}}\;{\overline {X}}\;\mathbb {R} \;\mapsto \;\setminus \left\langle \right\rangle \|\|\;\in \;\subset \;\supset \;{\bar {x}}\;\cap \;\cup \;{\mbox{e}}\;\emptyset }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b46bf2b2eb8febb3150f7620eecae9461a8be4f6)
![{\displaystyle t={\begin{cases}\inf\{h\in G;[h,b]\subset G\},{\mbox{ se }}a<h<b\\\sup\{h\in G;[b,h]\subset G\},{\mbox{ se }}b<h<a\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92421260a5a6ec745f02caba237f950c0c74abb8)
![{\displaystyle (t,b]\subset F,{\mbox{ se }}b<a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/afc90552f37dab7ddf9086c6965811134b615182)
![{\displaystyle F\cup G=[a,b]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a85a44dcfc924252cb676bd2f5fc9d9256e58fd9)
![{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow a\in G}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd95f52ef5d30000b931e437c75f48c06ea71626)
![{\displaystyle \exists \;t\in [a,c);[a,t]\subset G\Rightarrow t\not \in F,\exists b-a>\epsilon >0;(a,a+\epsilon )\not \subset G}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c1b59b1fcff74689e75292ebf58a22ccdcfc8cf)
![{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow (a,a+\epsilon )\subset F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/037e15a4917955341adbba65745324d53211c99c)
![{\displaystyle I_{p}=[u,v]\subset G\cap ((a,c)\cup (c,b))\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81de62bd4ec98e705add15454d5b4b86facf418c)
![{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow (v,v+\epsilon ),(u-\epsilon ,u)\subset F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1e562c32a06876991c6e56a587ee0a6c25cef6e)
![{\displaystyle t\in G\cap [a,b]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84af65db29907948505cb5cabeb3ee5ed8f698dc)
![{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow }](https://wikimedia.org/api/rest_v1/media/math/render/svg/44856af941de5dcdb4264f8f972997a25381f424)
![{\displaystyle F\cup G=[a,b],F\cap G=\varnothing \Rightarrow (t-\epsilon ,t-\epsilon )-{t}\subset F}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd6c98ed6f422da8c29806fd2ea016176fcb7b93)
![{\displaystyle c\in [a,b]\cap GF=\varnothing }](https://wikimedia.org/api/rest_v1/media/math/render/svg/38eb50fea666226a354d24b86fcf7376649322e4)
