Product topology, basis

1. Mathematical Analysis, Apostol, 2nd Ed, pp57, Theorem 3.27
   Pick $δ∈ℚ∩(0,ε/2)$. Since $ℚ^n$ is dense in $ℝ^n,∃𝐲∈ℚ^n$ such that $𝐱∈B(𝐲, δ)$.
   $∀𝐳∈B(𝐲, δ):{‖𝐱-𝐳‖}_∞≤{‖𝐱-𝐲‖}_∞+{‖𝐲-𝐳‖}_∞<2δ<ε$. So $B(𝐲, δ)⊂B(𝐱, ε)$.
2. For any open set $B$ in $ℝ^n$, $∀𝐱∈B,∃ε:B(𝐱,ε)⊂B$.
   By (1), $∃𝐲∈ℚ^n$ and $δ∈ℚ^+$ such that $𝐱∈B(𝐲, δ)⊂B(𝐱, ε)$, therefore $B$ is the union of all $B(𝐲, δ)⊂B$.
   So $\{B(𝐲, δ): 𝐲∈ℚ^n$ and $δ∈ℚ, δ>0\}$ is a basis for the Euclidean topology on $ℝ^n$.
   It is countable because $B(𝐲,δ)↦(𝐲,δ)$ is a bijection to $ℚ^n×ℚ^+$.

Let $(x,y)∈(X×Y)∖G$. Since $y$ and $f(x)$ are distinct points in $Y$, and $Y$ is Hausdorff, ∃ disjoint open sets $U∋y$ and $V∋f(x)$ in $Y$. Since $f$ is continuous, $W=f^{-1}(V)$ is an open neighborhood of $x$.
$W×U$ is an open neighborhood of $(x,y)$ in the product topology. We will show it is disjoint from $G$.
Let $(z,f(z))$ be any point of $G$. If $z∉W$, then clearly $(z,f(z))∉W×U$. If $z∈W$, then $f(z)∈V$, so $f(z)∉U$, and therefore $(z,f(z))∉W×U$. Thus, in all cases $(z,f(z))∉W×U$, and it follows that $(W×U)∩G=∅$. Thus, each point of $(X×Y)∖G$ has an open neighborhood disjoint from $G$, and $G$ is therefore closed.

Connectedness

1. Suppose $A∪B$ is disconnected, then $A∩B=∅$ and ∃ disjoint open sets $Y,Z$ such that $A∪B⊂Y∪Z$ and $(A∪B)∩Y∩Z=∅$ and $(A∪B)∩Y,(A∪B)∩Z≠∅$.
   Since $(A∪B)∩Y∩Z=∅$, we have $A∩Y∩Z=∅$, also $A⊂Y∪Z$, but $A$ is connected, say $A⊂Y$, then $A∩Z=∅$ and $B∩Z=(A∪B)∩Z≠∅$, but $B$ is connected, we have $B⊂Z$.
   Since $B⊂X∖Y$ and $X∖Y$ is closed, we have $\bar B⊂X∖Y$, but $A⊂Y$, so $A∩\bar{B}=∅$, contradiction.
2. • $B((1,0),1),B((-1,0),1)$ are disjoint open sets, so $B((1,0),1)∪B((-1,0),1)$ is disconnected.
   • $\overline{B((1,0),1)}∩\overline{B((-1,0),1)}≠∅$, by (1) $\overline{B((1,0),1)}∪B((-1,0),1)$ is connected.
   • Denote the set by $A$. For any $(x,y)∈A$, wlog let $x$ be rational. Consider the polyline Γ from $(0,0)$ to $(x,0)$ to $(x,y)$. Any point on Γ has a rational coordinate, so Γ$⊂A$, so $A$ is path connected, so $A$ is connected.

1. Consider all connected subsets of $X$ containing $a$. Since they intersect in $a$, the union is connected, so it is the largest connected subset of $X$ containing $a$.
2. $C_a$ is connected, by Theorem 1.85, $\overline{C_a}$ is a connected subset of $X$ containing $a$, so $\overline{C_a}⊂C_a$, so $C_a$ is closed.
3. If $y∈C_x$, then $C_x$ is a connected subset of $X$ containing $y$, by definition $C_x⊂C_y$, so $x∈C_y$.
   Now $C_y$ is a connected subset of $X$ containing $x$, by definition $C_y⊂C_x$, so $C_x=C_y$.
   $x∼y⇔C_x=C_y$ is clearly an equivalence relation.
4. By (3), connected components are equivalence classes, so they are disjoint.
5. $X=X_1∪X_2$ where $X_1=\set{(x, y)∈ℝ^2: x<y},X_2=\set{(x, y)∈ℝ^2: x>y}$ are disjoint, open and connected. So they are the connected components of $X$.

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   $X=\set{(z, w)∈ℂ^2: z≠w}=ℂ^2∖\set{(z, z):z∈ℂ}$ is the complement of a 2D space in a 4D space, we guess $X$ is connected. Let $(z_1,w_1),(z_2,w_2)∈X$, we have $z_1≠w_1$ and $z_2≠w_2$.
   Case 1. $z_2≠w_1$
   In $ℂ∖\{w_1\}$ there exist path $γ_1$ from $z_1$ to $z_2$, then $Γ_1(t)=(γ_1(t),w_1)$ is a path from $(z_1,w_1)$ to $(z_2,w_1)$ in $X$.
   In $ℂ∖\{z_2\}$ there exist path $γ_2$ from $w_1$ to $w_2$, then $Γ_2(t)=(z_2,γ_2(t))$ is a path from $(z_2,w_1)$ to $(z_2,w_2)$ in $X$.
   Hence $Γ_1∗Γ_2$ is a path from $(z_1,w_1)$ to $(z_2,w_2)$ in $X$.
   Case 2. $z_2=w_1$
   Take $w∈ℂ∖\{z_1,w_1\}$.
   In $ℂ∖\{z_1\}$ there exist path $γ_1$ from $w_1$ to $w$, then $Γ_1(t)=(z_1,γ_1(t))$ is a path from $(z_1,w_1)$ to $(z_1,w)$ in $X$.
   By Case 1, there exist path $Γ_2$ from $(z_1,w)$ to $(z_2,w_2)$ in $X$.
   Hence $Γ_1∗Γ_2$ is a path from $(z_1,w_1)$ to $(z_2,w_2)$ in $X$.
   $ℂ^2$去掉对角线的连通分支
   有一种做法, 不需要分情况, 容易推广到证明“$ℝ^n(n≥4)$去掉2维子空间仍连通”:
   $Y=\{(a,b,a,b):a,b∈ℝ\}$ is a linear subspace of $ℝ^4$, we have $Y^⟂=\{(a,b,-a,-b):a,b∈ℝ\}$.
   For $(a_1,b_1,c_1,d_1),(a_2,b_2,c_2,d_2)∈X=ℝ^4∖Y$, let $y_1=\left(-\frac{a_1+c_1}2,-\frac{b_1+d_1}2,-\frac{a_1+c_1}2,-\frac{b_1+d_1}2\right)$, $y_2=\left(-\frac{a_1+a_2}2,-\frac{b_2+d_2}2,-\frac{a_2+c_2}2,-\frac{b_2+d_2}2\right)$, $x_1=(a_1,b_1,c_1,d_1)+y_1=\left(\frac{a_1-c_1}2,\frac{b_1-d_1}2,\frac{c_1-a_1}2,\frac{d_1-b_1}2\right)$ and $x_2=(a_2,b_2,c_2,d_2)+y_2=\left(\frac{a_2-c_2}2,\frac{b_2-d_2}2,\frac{c_2-a_2}2,\frac{d_2-b_2}2\right)$, then $x_1,x_2∈Y^⟂$, and $(a_1,b_1,c_1,d_1)+ty_1,t∈[0,1]$ is a path in $X$ from $(a_1,b_1,c_1,d_1)$ to $x_1$, $(a_2,b_2,c_2,d_2)+ty_2,t∈[0,1]$ is a path in $X$ from $(a_2,b_2,c_2,d_2)$ to $x_2$. Finally, since $Y∩Y^⟂=\{0\}$ and $Y^⟂≅ℝ^2$ and $ℝ^2∖\{0\}$ is connected, there is a path connecting $x_1,x_2$ in $Y^⟂∖\{0\}⊂X$.
6. Let $X$ be any connected component of $A$. Since $X$ is connected in ℝ, it is an interval.
   Since $X⊂A,\mathring A=∅$, we have $\mathring X=∅$, so $X$ is a singleton.
   In particular, $\mathring ℚ=∅$, so any connected component of ℚ is singleton.

The set of path connected components of $X$ is $π_0 (X) = [S^0, X]_+$
[ ] up to homotopy
$π_1 (X) = [S^1, X]_+$ is the fundamental group.
$π_n (X, X_0) = [S^n, X]_+$ are Abelian groups, called higher homotopy groups. Remark. intermediate value property ⇔ preserve connectedness.

1. Any 2 points in $T$, $(x_1,y_1),(x_2,y_2),x_1<y_1,x_2<y_2$ can be connected by a line $t(x_1,y_1)+(1-t)(x_2,y_2),t∈[0,1]$, since $tx_1+(1-t)x_1<ty_2+(1-t)y_2$. So $T$ is path connected. So $T$ is connected.
2. Let $(x,y)∈T$. By the Mean-Value Theorem, $∃c∈(x,y)⊂I$ such that $$ g(x,y) = \frac{f(x) - f(y)}{x - y} = f'(c)∈f'(I). $$ As $(x,y)∈T$ is arbitrary, $g(T)⊂f'(I)$.
   Next, since $I$ is open, $∀x∈I,∃$ a sequence $(y_n)_{n∈ℕ}$ in $I$ such that $x<y_n∀n∈ℕ$ and $\lim_{n→∞}y_n=x$. $$ f'(x) = \lim_{n→∞} \frac{f(x) - f(y_n)}{x - y_n} = \lim_{n→∞} g(x,y_n)∈\overline{g(T)} $$As $x∈I$ is arbitrary, $f'(I)⊂\overline{g(T)}$.
3. By (1), $T$ is connected, since $g$ is continuous on $T$, $g(T)$ is connected. By (2), $g(T)⊂f'(I)⊂\overline{g(T)}$, by Theorem 1.85, $f'(I)$ is connected.

Compactness

1. For any finite $J⊆ℕ$, $⋂_{n∈J}V_n=V_{\max J}≠∅$. Since $X$ is compact, by Proposition 2.4. $⋂_{n=1}^∞ V_n≠∅$
2. Let $x∈X$, since $f^{∘n}(x)∈X_n$, $X_n≠∅$.
   $f(X)⊂X⇒X_{n+1}=f^{∘n}(f(X))⊂f^{∘n}(X)=X_n$.
   By continuity of $f$, $X_n$ closed ⇒ $X_{n+1}=f(X_n)$ closed.
   $\left(X_n\right)_{n∈ℕ}$ is a nested sequence of non-empty closed subsets of $X$, by (1), $A≠∅$.
3. $A$ is an intersection of closed sets, so it is closed. By (2), $A≠∅$.
   To prove $f(A)⊂A:A=\bigcap_{n≥1} X_n = \bigcap_{n≥0}f(X_n)⊃f(\bigcap_{n≥0} X_n) = f(A)$
   To prove $A⊂f(A)$: Let $a∈A,V_n≔X_n∩f^{-1}(\{a\})$.
   By Hausdorff, $\{a\}$ is closed, by continuity $f^{-1}(\{a\})$ is closed, so $V_n$ is closed.
   $∀n:a∈X_{n+1}⇒∃y:y∈X_n,a=f(y)⇒y∈V_n⇒V_n≠∅$.
   $\left(V_n\right)_{n∈ℕ}$ is a nested sequence of non-empty closed set, by (1), $∃p∈\bigcap_{n∈ℕ} V_n$
   $f(p)=a,p∈A⇒a∈f(A)⇒A⊂f(A).$

Summary: any (continuous) self-map of a compact Hausdorff space has fixed subset.
“have fixed subset” doesn't imply “have fixed point” e.g. rotation of a circle
Note: rotation of a sphere has fixed points (2 poles)
Fact: Any map $f:𝔻^n→𝔻^n$ has a fixed point. [Brouwer's fixed point theorem] For any $b∈B$, Applying Proposition 2.13. to $A$ and $b$, ∃ disjoint open sets $U_b,V_b$ such that $A⊂U_b,b∈V_b$
$(V_b)_{b∈B}$ is an open cover of $B$, so ∃ finite subcover $V_{b_1},…,V_{b_n}$, so $V≔V_{b_1}∪⋯∪V_{b_n}$ is open and $B⊂V$.
$U≔U_{b_1}∩⋯∩U_{b_n}$ is open and $A⊂U$ since $∀b:A⊂U_b$.
For $i=1,…,n$ we have $U⊂U_{b_i},U_{b_i}∩V_{b_i}=∅⇒U∩V_{b_i}=∅⇒U∩V=∅$.