(a) Show that every metrizable space with a countable dense subset has a countable basis.
(b) Show that every metrizable Lindelof space has a countable basis.
My attempt:
(a) Since $X$ is separable, $\exists Y\subseteq X$ such that $Y$ is countable and $\overline{Y}=X$. Let $\mathcal{B}=\{ B_d (y, \frac{1}{n})|y\in Y , n\in \Bbb{N}\}$. By theorem 2.13 of Baby Rudin, $\mathcal{B}$ is countable. (1) Clearly $\mathcal{B}\subseteq \mathcal{B}_X \subseteq \mathcal{T}_X$. (2) Let $U\in \mathcal{T}_X$ and $x\in U$. Since $X$ is metrizable, $\exists \epsilon \gt 0$ such that $B_d (x,\epsilon) \subseteq U$. By theorem 1.20(a) of Baby Rudin, $\exists n\in \Bbb{N}$ such that $\frac{1}{n}\lt \frac{\epsilon} {2}$. Since $x\in \overline{Y} =X$ and $B_d (x, \frac{1}{n}) \in \mathcal{N}_x$, we have $B_d (x, \frac{1}{n}) \cap Y \neq \emptyset$. So $\exists z\in B_d (x,\frac{1}{n}) \cap Y$. $z\in Y$ and $d(x,z)\lt \frac{1}{n} \lt \frac{\epsilon}{2}$. Claim: $x\in B_d (z, \frac{1}{n})\subseteq B_d(x, \epsilon)$. Proof: since $d(x,z)\lt \frac{1}{n}$, we have $x\in B_d(z,\frac{1}{n})$. Let $p\in B_d(z,\frac{1}{n})$. Then $d(p,z) \lt \frac{\epsilon}{2}$. $d(p,x)\leq d(p,z)+d(z,x) \lt \frac{1}{n} + \frac{1}{n} =\frac{2}{n}\lt \epsilon$. So $p\in B_d(x, \epsilon)$. Thus $B_d (z, \frac{1}{n})\subseteq B_d(x, \epsilon)$. Hence $\exists B_d (z, \frac{1}{n}) \in \mathcal{B}$ such that $x\in B_d (z, \frac{1}{n})\subseteq B_d(x, \epsilon) \subseteq U$. By lemma 13.2, $\mathcal{B}$ is a basis of $\mathcal{T}_X$. Is this proof correct?
(b) $B_n =\{B(x, \frac{1}{n})|x\in X\}$ is an open cover of $X$, $\forall n\in \Bbb{N}$. Since $X$ is lindelof, $\exists A_n= \{ B(x_{n,i},\frac{1}{n})| i\in \Bbb{N}\}$ countable subcover of $B_n$, $\forall n\in \Bbb{N}$. By elementary set theory, $\bigcup_{n\in \Bbb{N}}A_n =\{ B(x_{n,i},\frac{1}{n})| i\in \Bbb{N}, n\in \Bbb{N}\}$. Let $\mathcal{B}= \bigcup_{n\in \Bbb{N}}A_n$. By theorem 2.12 of Baby Rudin, $\mathcal{B}$ is countable. (1) Clearly $\mathcal{B}\subseteq \mathcal{B}_X \subseteq \mathcal{T}_X$, where $\mathcal{B}_X =\{ B(x,r)| x\in X, r\in \Bbb{R}^+\}$. (2) Let $U\in \mathcal{T}_X$ and $x\in U$. Since $X$ is metrizable, $\exists \epsilon \gt 0$ such that $B(x,\epsilon)\subseteq U$. By theorem 1.20(a) of Baby Rudin, $\exists n\in \Bbb{N}$ such that $\frac{1}{n}\lt \frac{\epsilon} {2}$. Since $x\in X= \bigcup A_n= \bigcup_{i\in \Bbb{N}} B(x_{n,i}, \frac{1}{n})$, $\exists m\in \Bbb{N}$ such that $x\in B(x_{n,m}, \frac{1}{n})$. Now we show $B(x_{n,m}, \frac{1}{n}) \subseteq B(x,\epsilon)$. Let $p\in B(x_{n,m}, \frac{1}{n})$. Then $d(x_{n,m}, p)\lt \frac{1}{n}$. $d(p,x)\leq d(p, x_{n,m}) +d(x_{n,m},x) \lt \frac{2}{n} \lt \epsilon$. So $p\in B(x,\epsilon)$. Thus $B(x_{n,m}, \frac{1}{n}) \subseteq B(x,\epsilon)$. Hence $\exists B(x_{n,m}, \frac{1}{n}) \in \mathcal{B}$ such that $x\in B (x_{n,m}, \frac{1}{n}) \subseteq B(x,\epsilon)\subseteq U$. By lemma 13.2, $\mathcal{B}$ is basis of $\mathcal{T}_X$. Is this proof correct?
In proof of exercise 5(b) section 30, we use hint of Exercise 4, Section 30 of Munkres’ Topology. My first approach without using hint was: $\forall x\in X, \exists B_x \in \mathcal{B}_X$ such that $x\in B_x$. $B=\{ B_x |x\in X\}$ is an open cover of $X$. Since $X$ is lindelof, $\exists \{B_{x_n}| n\in \Bbb{N} \}$ countable subcover of $B$. I hypothesis $\mathcal{B}=\{B_{x_n} |n\in \Bbb{N} \}$ is basis of $\mathcal{T}_X$.