# Problem III.2.7 in textbook Analysis I by Amann: $f : X \to Y$ is continuous $\iff f(\overline{A}) \subseteq \overline{f(A)}$ for all $A \subseteq X$

Good morning, I'm doing Problem III.2.12 from textbook Analysis I by Amann. Here $$\overline A$$ is the set of all accumulation points of $$A$$ or equivalently the closure of $$A$$.

Could you please verify if my proof look fine or contains logical gaps/errors? Any suggestion is greatly appreciated.

My attempt:

$$\Longrightarrow$$ (Indirect proof): Assume the contrary that $$f(\overline A) \not \subseteq \overline{f(A)}$$. Then there is $$y \in f(\overline A)$$ such that $$y \notin \overline{f(A)}$$. Because $$y \in f(\overline A)$$, there is $$x \in \overline A$$ such that $$f(x) =y$$. Because $$y \notin \overline{f(A)}$$, there is a neighborhood $$U$$ of $$y$$ such that $$U \cap f(A) = \emptyset$$. Because $$f$$ is continuous, there is a neighborhood $$U'$$ of $$x$$ such that $$f(U') \subseteq U$$. As such, $$f(U') \cap f(A) = \emptyset$$ and thus $$U' \cap A = \emptyset$$, which contradicts $$x \in \overline A$$.

$$\Longrightarrow$$ (Direct proof): For $$y \in f(\overline A)$$, there is $$x \in \overline A$$ such that $$f(x) = y$$. Because $$f$$ is continuous, if $$U$$ is a neighborhood of $$y$$ then there is a neighborhood $$U'$$ of $$x$$ such that $$f(U') \subseteq U$$. It follows from $$x \in \overline A$$ that $$U' \cap A \neq \emptyset$$, so $$f(U') \cap f(A) \neq \emptyset$$. As such, $$U \cap f(A) \neq \emptyset$$ and hence $$y \in \overline{f(A)}$$.

$$\Longleftarrow$$: Let $$A \subseteq Y$$ be closed in $$Y$$ and $$B = f^{-1}(A) \subseteq X$$. Assume the contrary that $$B$$ is not closed in $$X$$, then $$C = \overline B - B \neq \emptyset$$ and $$C \cap B = \emptyset$$. We have $$f(C \cup B) = f(\overline B) \subseteq \overline {f(B)} = \overline A = A$$, so $$(C \cup B) \subseteq f^{-1}(A) = B$$, which contradicts $$C \neq \emptyset$$ and $$C \cap B = \emptyset$$. As such, $$B$$ is closed and thus $$f$$ is continuous.

• Your reasoning looks sound to me! – Math1000 Sep 7 at 6:41
• Thank you so much for your help @Math1000 :)))))) – Abstract Analysis Sep 7 at 6:45

$$f[B]=f[f^{-1}[A]] \subseteq A$$, not $$=A$$ (unless $$f$$ is surjective), but that's a minor thing; you only need the inclusion anyway.
The first part can be done easier: As $$\overline{f[A]}$$ is closed, $$f^{-1}[\overline{f[A]}]$$ is closed by continuity and contains $$A$$, i.e. $$A \subseteq f^{-1}[f[A]] \subseteq f^{-1}[\overline{f[A]}]$$
so $$f[\overline{A}] \subseteq \overline{f[A]}$$ is immediate. This does require you know that $$f$$ continuous means that the inverse images of closed sets are closed. (As for open sets.)
• Please check if my understanding is correct! It follows from $A \subseteq f^{-1}[f[A]] \subseteq f^{-1}[\overline{f[A]}]$ that $\overline{A} \subseteq \overline{f^{-1}[\overline{f[A]}]}$. Because $f^{-1}[\overline{f[A]}]$ is closed, $\overline{f^{-1}[\overline{f[A]}]} = f^{-1}[\overline{f[A]}]$. As such, $\overline A \subseteq f^{-1}[\overline{f[A]}]$ and thus $f[\overline A] \subseteq \overline{f[A]}$. – Abstract Analysis Sep 7 at 8:31
• To remove ambiguity, I prefer $f[A]$ than $f(A)$ for the induced set-value function from $f$ as you did. But not many people use this notation and it makes the notation complicated in some cases. Your approach is much more elegant beautiful than mine. – Abstract Analysis Sep 7 at 8:34