Source of most content: Gunhee Cho and Topology by Munkres.

Countability Axiom

Motivation. Let $X$ be a metric space and $A \subseteq X$ be open. Then,

1. If $\exists (x_n) \subseteq A$ s.t. $x_n \to x$, then $x \in \overline{A}$. Note that converse holds if $X$ is a MS.
2. $f$ is cts at $x=p$, i.e., $f(\overline{A}) \subseteq \overline{f(A)}$ iff $\forall (p_n)$ which converges to $p$, $f(p_n) \to f(p)$ holds.

Here, we might induce similar results without the assumption of a metric space:
For $x \in \overline{A}$, if $F = \{ V_x \mid V_x : \text{nbhd of }x \}$ is countable, let $\forall n\in \mathbb{N}$, $V_1 \cap \ldots V_n \in F$, $x_n \in V_1 \cap \ldots V_n\cap A$. Then, $\exists (x_n) \subseteq A$ and $x_n \to x$.

Countable TS

Def. Given T.S. $(X, T)$, $B_X (\subseteq T)$ is a local base at $x$ if $\forall$ nbhd $V$ of $x$, $\exists B \in B_X$ s.t. $x\in B \subseteq V$. And we say $X$ is 1st countable if $\forall x \in X$, $\exists B_x $ a countable local base at $x$.

NB1; If $X$ is 1st countable, then $\exists (x_n) \subseteq A$ s.t. $x_n \to x \Leftrightarrow x \in \overline{A}$.
NB2; If $X$ is 1st countable, then $f$ is cts at $x=p \Leftrightarrow \forall (p_n) \to p$, $f(p_n) \to f(p)$. Note that left arrow of NB1,2 can be shown by the def. of 1st countable.

Def. Given T.S. $(X, T)$, $X$ is 2nd countable if $\exists$ countable basis $\mathcal{B}$.

NB3; if $X$ is T3(regular) and 2nd countable, then $X$ is metrizable. (Uryshon’s metrization thm)
NB4; 2nd countable $\rightarrow$ 1st countable.

$\forall x \in X, B_x = \{ B \in \mathcal{B} : x \in B \}$ is countable since $\mathcal{B}$ is countable.

NB5; $(\mathbb{R}^w, T_{uniform})$: countable product of real space with uniform topology is 1st countable but not 2nd countable.

Propositions

\[\text{Let } X \text{ be a 2nd countable T.S.. Then, the followings hold: } \nonumber\]

1. For any open cover of $X$, $\exists$ countable subcover of $X$, i.e., $X$ is Lindelof space.
2. $\exists$ countable dense subset $A$ of $X$. Here, dense implies $\overline{A} = X$, i.e., $X$ is a separable space.

NB6; If $X$ is metrizable, TFAE: 2nd countable, Lindelof, separable.

Proof

1. Let $\mathcal{B}=\{ B_n : n \in \mathbb{N} \}$ be a countable basis and take any open cover $U = \{ U_\alpha : \alpha \in I\}$ of $X$. For each $n \in \mathbb{N}$, let $U’ = \{ A_n \in U : B_n \subseteq A_n \in U\}$. If we take any $x \in X$, then $\exists U_\alpha$ s.t. $x \in U_\alpha$. Then, $\exists n \in \mathbb{N}$ s.t. $x \in B_n \subseteq U_\alpha$, i.e., $ U_\alpha = U_{\alpha, n} \in U’$. Hence, $X = \bigcup_{A \in U’} A$. Note that $U’$ is countable by definition, which concludes the proof.
2. For each $n \in \mathbb{N}$, choose $x_n \in B_n \ne \emptyset$. Let $A := \{ x_n : n \in \mathbb{N}\}$. NTS: $\overline{A} = X$.
   Since $\overline{A} \subseteq X$ is clear, take any $x \in X$ and any basis $B_n \ni x$. If $x_n \ne x$, then $x_n \in (B_n \setminus \{x \})$. Thus, $(B_n \setminus \{x \}) \cap X \ne \emptyset$. If $x_n = x$, then $x=x_n \in A$. Therefore, $x \in \overline{A}$.

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\[\text{For T.S. }X, \text{ the followings hold:} \nonumber\]

1. If $X$ is 1st or 2nd countable, then $Y \subseteq X$ as a subspace also has the same axioms.
2. If $X_i$’s are 1st or 2nd countable for $i \in I$ (all $X_i$ share the same axiom), then $\prod_i X_i$ is also has the same axiom.

Proof

Consider the 2nd countable case, then the proof of the 1st countable case would be similar.

1. Let $\mathcal{B}$ be a countable basis for $X$. Then, $\{B \cap A : B \in \mathcal{B} \}$ is a countable basis for the subspace $A$ of $X$.
2. Let $\mathcal{B}_i$ be a countable basis for the space $X_i$. Then, $\{ \prod_i U_i : U_i \in \mathcal{B}_i \}$ for finitely many $i$ and $U_i = X_i$ for other values of $i$, is a countable basis for $\prod_i X_i$.

Topologies of function space

Def. For a given T.S. $(Y,T)$, let $Y^X = \{ f : X \to Y\}$ be a set of functions. Let $S(x \in Y, U \in T) := \{f \in Y^X : f(x) \in U \}$ be a subbasis of $Y^X$. Then, $\mathcal{B} = S(x_1, U_1) \cap \cdots \cap S(x_n, U_n)$ becomes a basis and $T_{\mathcal{B}}$ is called pointwise-convergence topology of $Y^X$.

NB7; $X_1 \times X_2 = \{ (x_1, x_2) : x_1 \in X_1, x_2 \in X_2\} = \{ f: \{1,2\} \to \cup_{i=1,2} X_i : f(1) \in X_1, f(2) \in X_2\}$, i.e., $\prod_{\alpha \in I}X_\alpha = \{ f \mid f:I \to \cup_{\alpha \in I} X_\alpha, f(\alpha) \in X_{\alpha}, \forall \alpha \in I \}=Y^X$.
NB8; Given $(Y^X, T_{pc}), (f_n) : X \to Y$, $f_n \to f$ in ptwise-convergence is equal to $\forall x \in X, \lim_{n\to \infty} f_n(x) = f(x)$.
NB9; Let $X = I = [0, 1]$ and $f_n(x) = x^n \to f(x) = \begin{cases} 0 &\text{ if } x <1 \\ 1 &\text{ if } x = 1\end{cases}$. Then, for the collection $C(I, \mathbb{R})$ of continuous functions $f: I \to \mathbb{R}$, it holds that $(f_n) \subseteq C(I, \mathbb{R}) \lneq (\mathbb{R}^{[0, 1]}, T_{pc})$. Then, $f_n \to f \not\in C(I, \mathbb{R})$, i.e., $C(I, \mathbb{R})$ is not closed.

Def. Let $X$ be a set and $Y$ be a metric space with $d$. Then, we say $f_n$ converges uniformly to $f$, ${f_n}^\to_\to f$iff $\forall \epsilon > 0, \exists N \in \mathbb{N} $ s.t. $\forall n \geq N, \forall x \in X$, $d(f_n(x), f(x)) < \epsilon$.

Def. Given T.S. $(X,T)$ and metric space $(Y, d)$, define $B_C(f, \epsilon) := \{ g \in Y^X : \sup_{x\in C } d(f(x), g(x)) < \epsilon\}$ for each compact set $C$ in $X$, $f:X\to Y$ and $\epsilon > 0$. Then, $\mathcal{B} = \{ B_C(f, \epsilon)\}$ is a basis of $Y^X$ and we call $T_{\mathcal{B}}$ a compact convergence topology on $Y^X$.

NB10; $(f_n) \subseteq (Y^X, T_{cct})$, $f_n \to f$ in cct sense is equal to $\forall$ cpt $C$, $f_n {\mid_{C}}^\to_\to f\mid_C$.
NB11; For metric space $Y$, $T_{cct} = T_{\text{compact open topology}}$.

Propositions

\[\text{Given } (X,T), (Y, d), \text{ if } (f_n)\subseteq C(X,Y) \text{ s.t. } {f_n}^\to_\to f, \text{ then } f\in C(X,Y), \text{ i.e., } f \text{ is cts.} \nonumber\]

Proof

Fix $x_0 \in X$ and $\epsilon >0$ and consider a ball $B_{\epsilon}(f(x_0))$. By uniform convergence, for any $x\in X$, $\exists N \in \mathbb{N}$ s.t. $d(f(x), f_N(x) ) < \frac{\epsilon}{3}$. By continuity, $\exists$ nbhd $U$ of $x_0$ s.t. $d(f_N(x), f_N(x_0)) < \frac{\epsilon}{3}$. By uniform convergence, $\exists N \in \mathbb{N}$ s.t. $d(f_N(x_0), f(x_0)) < \frac{\epsilon}{3}$. Then, $d(f(x), f(x_0)) \leq d(f(x), f_N(x) ) + d(f_N(x), f_N(x_0)) + d(f_N(x_0), f(x_0)) < \epsilon$.

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\[(\mathbb{R}^w , T_{\text{uniformly convergence topolgy} }) \text{ is 1st countable but not 2nd countable.} \nonumber\]

Proof

• For a metric space $Y$ and index set $J$, consider a function space $Y^J$. Then, $x,y : J \to Y \, ( \alpha \mapsto x_{\alpha}, y_{\alpha})$. Let $\bar{d}(x_\alpha, y_\alpha) := \max (d(x_\alpha, y_\alpha),1)$ and $\tilde{d}(x_\alpha, y_\alpha) := \sup_{\alpha \in J}\bar{d}(x_\alpha, y_\alpha) \, (\leq 1)$. Then, $(Y^J, \tilde{d})$ is called uniformly convergence topology.
  1. If $\mathbb{R}^w$ is 2nd countable, then $\exists$ countable basis $\mathcal{B}$. Then, for any discrete subspace $D$ of $\mathbb{R}^w$, $D$ must be at most countable since one-point set is open. Precisely, define $\Phi: D \to \mathcal{B}\, (a \mapsto B_a)$ s.t. $B_a \cap D = \{ a \}$. If $a \ne b$ then, $\Phi(a) \ne \Phi(b)$, i.e., $\Phi$ is injective. Thus, $D$ is at most countable.
  2. Define $F:= \{ a \in \mathbb{R}^w : \text{each entry of }a \text{ is either } 0 \text{ or }1 \}$ which is uncountable. Then, $\forall a,b \in F, \tilde{d}(a,b) = \begin{cases} 1 \quad a \ne b \\ 0 \quad a = b\end{cases}$, i.e., $F$ is discrete subspace of $\mathbb{R}^w$. This implies that $F$ is at most countable, which is contradicts to the definition of $F$.