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Good morning, I'm doing Problem III.2.12 from textbook Analysis I by Amann.

enter image description here 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.

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    $\begingroup$ Your reasoning looks sound to me! $\endgroup$ – Math1000 Sep 7 at 6:41
  • $\begingroup$ Thank you so much for your help @Math1000 :)))))) $\endgroup$ – Abstract Analysis Sep 7 at 6:45
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$f[B]=f[f^{-1}[A]] \subseteq A$, 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.)

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  • $\begingroup$ 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 $f[\overline A] \subseteq \overline{f[A]}$. $\endgroup$ – Abstract Analysis Sep 7 at 8:31
  • $\begingroup$ 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. $\endgroup$ – Abstract Analysis Sep 7 at 8:34
  • $\begingroup$ Thank you so much for correcting my critical logical mistake! $\endgroup$ – Abstract Analysis Sep 7 at 8:38
  • $\begingroup$ @LeAnhDung yes, that’s right. I assumed the intermediate step. $\endgroup$ – Henno Brandsma Sep 7 at 8:47

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