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Let $f$ be a mapping. Prove that the following three statements are equivalent.

  1. $f$ is continuous;
  2. $\forall A \subseteq X: f(\overline{A}) \subset \overline{f(A)}$;
  3. $\forall B \subseteq Y: \overline{f^{-1}(B)} \subset f^{-1}(\overline{B})$.

I finally succeeded in proving these equivalences.

I want to go a bit further: To find (counter)examples to show $f(\overline{A}) \neq \overline{f(A)}$ in (2) and $\overline{f^{-1}(B)} \neq f^{-1}(\overline{B})$ in (3). However I failed.

My attempt: When I prove $(1) \implies (2)$, I use the fact of $f^{-1}(f(A)) \supset A$ and I know that $f^{-1}(f(A)) = A$ if $f$ is injective; Similarly, in $(2) \implies (3)$, I use the fact of $f(f^{-1}(B)) \subset B$ and I know that $f(f^{-1}(B)) = B$ if $f$ is surjective.

Thus, in my opinion, the counterexamples for (2) [or (3)] are likely to involve a mapping $f$ that is not injective [or surjective].

My question is:

What are the (counter)examples for $f(\overline{A}) = \overline{f(A)}$ in (2) and $\overline{f^{-1}(B)} = f^{-1}(\overline{B})$ in (3)?

P.S. Related posts without counterexamples:

1. Show that the statements are equivalents

2. For any $A \subseteq X, f(\overline{A}) \subset \overline{f(A)}$ , if and only if $f: X \to Y$ is continuous.

EDIT:

Continuous mappings $f$ are more preferable because they are continuous in (1). Non-continuous ones are also welcome.

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hengxin
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2 Answers2

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Let $X$ be the set $\{a,b\}$, and consider the map $$f:X_d=(X,\ \mathcal P(X))\longrightarrow X_i=(X,\{\emptyset,X\}) \\ f(a)=a,\quad f(b)=b$$ from the discrete space on $X$ to the indiscrete space on $X$. This $f$ is continuous and bijective, but $f\left(\overline{\{a\}}\right)=\{a\}$ while $\overline{f\left(\{a\}\right)}=X$. For the other inequality, let $B=\{b\}\subset X_i$.

Stefan Hamcke
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Consider the function

$f(x)=\begin{cases}0&x<0\\1&x\geq0\end{cases}.$

Let $A=(-\infty,0)$. $f(\bar{A})=\{0,1\}$, but $\overline{f(A)}=\{0\}$.

More generally, if $f$ is not continuous, there is some sequence $(x_n)$ such that $x_n\to x$, but $f(x_n)$ does not approach $f(x)$. Define $A$ to be $(x_n)$, and this will give you your desired counter-example, as long as $f(x)\neq f(x_n)$ for any $n$.

Trevor J Richards
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