# Exercise 11, Section 30 of Munkres’ Topology

alternative-proofgeneral-topologylindelof-spacesproof-writingsolution-verification

Let $$f \colon X \rightarrow Y$$ be continuous. Show that if $$X$$ is Lindelof, or if $$X$$ has a countable dense subset, then $$f(X)$$ satisfies the same condition.

My attempt:

Approach(1): It’s easy to check, $$A$$ is lindelof $$\iff$$ Every open cover of $$A$$ in $$X$$ has countable subcover. Proof is very similar to lemma 26.1 of Munkres’ topology. Exercise 9, Section 30 of Munkres’ Topology. Let $$U=\{ U_\alpha \in \mathcal{T}_Y| \alpha \in J\}$$ be an open cover of $$f(X)$$ in $$Y$$. $$f(X)\subseteq \bigcup_{\alpha \in J}U_\alpha$$. So $$X\subseteq f^{-1}(f(X))\subseteq f^{-1}(\bigcup_{\alpha \in J} U_\alpha)=\bigcup_{\alpha \in J} f^{-1}(U_\alpha)$$. Thus $$X= \bigcup_{\alpha \in J} f^{-1}(U_\alpha)$$. Since $$U_\alpha \in \mathcal{T}_Y$$, $$\forall \alpha \in J$$ and $$f$$ is continuous, $$f^{-1}(U_\alpha) \in \mathcal{T}_X$$, $$\forall \alpha \in J$$. So $$V=\{ f^{-1}(U_\alpha)| \alpha \in J\}$$ is an open cover of $$X$$. Since $$X$$ is lindelof, $$\exists \{f^{-1}(U_{\alpha_n})|n\in \Bbb{N}\}$$ countable subcover of $$V$$. $$X= \bigcup_{n\in \Bbb{N}} f^{-1}(U_{\alpha_n})$$. So $$f(X)=f( \bigcup_{n\in \Bbb{N}} f^{-1}(U_{\alpha_n}))=\bigcup_{n\in \Bbb{N}} f(f^{-1}(U_{\alpha_n}))\subseteq \bigcup_{n\in \Bbb{N}} U_{\alpha_n}$$. Hence $$\{U_{\alpha_n}|n\in \Bbb{N}\}$$ is countable subcover of $$U$$.

Since $$X$$ is separable, $$\exists D\subseteq X$$ such that $$\overline{D}=X$$ and $$D$$ is countable. By Theorem 18.1 of Munkres’ Topology, $$f(X)=f(\overline{D})\subseteq \overline{f(D)}$$. Thus $$f(X)=(\ \overline{f(D)})_{f(X)}$$. Since $$D$$ is countable, $$D=\{x_n \in X| n\in \Bbb{N}\}$$. So $$f(D)=\{ f(x_n)| n\in \Bbb{N}\}$$ is countable. Hence $$f(D)$$ is countable dense subset of $$f(X)$$. Is this proof correct?

Approach(2): It’s easy to check, $$X$$ is lindelof $$\iff$$ If $$\{ A_\alpha \subseteq X| A_\alpha$$ is closed in $$X$$, $$\alpha \in J\}$$ have countable intersection property, then $$\bigcap_{\alpha \in J} A_\alpha \neq \emptyset$$. Proof is very similar to theorem 26.9 of Munkres’ topology or proposition 2.4 page no. 20. Let $$U=\{ U_\alpha |\alpha \in J\}$$ be an indexed family of closed subset of $$f(X)$$ such that $$\bigcap_{i \in I}U_i \neq \emptyset$$ for any countable subset $$I\subseteq J$$. By theorem 17.2, $$U_\alpha = f(X) \cap C_\alpha$$, where $$C_\alpha$$ is closed in $$Y$$, $$\forall \alpha \in J$$. So $$\emptyset \neq \bigcap_{i \in I}U_i = \bigcap_{i \in I}(f(X) \cap C_i)=f(X)\cap (\bigcap_{i \in I}C_i)$$, for any countable subset $$I\subseteq J$$. Since $$f(X)\cap (\bigcap_{i \in I}C_i) \neq \emptyset$$ and $$f(X)\cap (\bigcap_{i \in I}C_i)\subseteq f(X)$$, we have $$\emptyset \neq f^{-1}(f(X)\cap (\bigcap_{i\in I}C_i)) =X\cap (\bigcap_{i\in I} f^{-1}(C_i)) = \bigcap_{i\in I} f^{-1}(C_i)$$, for any countable subset $$I\subseteq J$$. Since $$f$$ is continuous, $$f^{-1}(C_\alpha)$$ is closed in $$X$$, $$\forall \alpha \in J$$. Since $$X$$ is lindelof, $$\bigcap_{\alpha \in J} f^{-1}(C_\alpha) \neq \emptyset$$. So $$\emptyset \neq f(\bigcap_{\alpha \in J}f^{-1}(C_\alpha)) \subseteq \bigcap_{\alpha \in J}f(f^{-1}(C_\alpha))\subseteq \bigcap_{\alpha \in J} C_\alpha$$. Since $$f(\bigcap_{\alpha \in J}f^{-1}(C_\alpha)) \subseteq f(X)$$, we have $$\emptyset \neq f(\bigcap_{\alpha \in J}f^{-1}(C_\alpha)) = f(\bigcap_{\alpha \in J}f^{-1}(C_\alpha)) \cap f(X) \subseteq (\bigcap_{\alpha \in J} C_\alpha) \cap f(X)=\bigcap_{\alpha \in J}U_\alpha$$. Hence $$\bigcap_{\alpha \in J}U_\alpha \neq \emptyset$$. Our desired result. Is this proof correct?

Let $$f: X \rightarrow Y$$ be continuous, where $$X$$ is Lindelöf. Let $$\mathcal{A}$$ be an open cover of $$f(X)$$; then $$\left\{f^{-1}(A)\right\}_{A \in \mathcal{A}}$$ is an open cover of $$X$$. Let $$f^{-1}\left(A_{1}\right), f^{-1}\left(A_{2}\right), \cdots$$ be a countable subcover of $$X$$. Then $$A_{1}, A_{2}, \cdots$$ is a countable subcover of $$f(X)$$. Since $$\mathcal{A}$$ was arbitrary, we conclude that $$f(X)$$ is Lindelöf.
Now let $$f: X \rightarrow Y$$ be continuous, where $$X$$ has a countable dense subset $$A$$. Let $$g: X \rightarrow f(X)$$ be the function obtained from $$f$$ by restricting its range. Then $$g(A)$$ is a countable subset of $$f(X)$$; we claim that it is dense in $$f(X)$$. Let $$f(x) \in f(X)$$, and let $$V$$ be a neighborhood in $$f(X)$$ of $$f(x)$$. Then, since $$f$$ is continuous, $$g$$ is as well, so $$U=g^{-1}(V)$$ is open. Since $$g(x)=f(x) \in V$$, we conclude that $$x \in U$$, and hence $$U$$ is a neighborhood of $$x$$. Since $$A$$ is dense in $$X$$, there exists some $$a \in A \cap U$$. Then $$g(a) \in g(A) \cap V$$; since $$x$$ and $$V$$ were arbitrary, we conclude that $$g(A)$$ is dense in $$f(X)$$, as claimed.