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My problem: Let $X, Y$ be metric space and $f:X \rightarrow Y$. Prove that the following statements are equivalent

(a) $f$ is continuous on $X$.

(b) $f(\overline{A})\subset \overline{f(A)},\forall A\subset X.$

(c) $f^{-1}(\operatorname{Int}(B))\subset \operatorname{Int}(f^{-1}(B)), \forall B\subset Y.$

I have proven $(a)\Rightarrow (b)$ and $(c)\Rightarrow (a)$, but I still get stuck in proving $(b)\Rightarrow (c)$. I tried setting A in (b) by $f^{-1}(B)$ or $X\setminus f^{-1}(B)$,...or something like that, then use (b) and some relations I've known like $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$, $Y\setminus \overline{f(A)}=\operatorname{Int}(Y\setminus f(A))$,... to transform to (c), but looks like I went wrong. So it would be great if I get suggestions from you!

Jaclyn
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Nguyễn Hữu Nam
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3 Answers3

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Your idea is good, if you also use the relation

\begin{equation}\tag{1} \operatorname{Int}(B) = Y \setminus (\overline{Y \setminus B}) \end{equation}

you should be able to make your idea work.

Let $B \in Y$, by (b) you have that \begin{align} f(\overline{f^{-1}(B)}) \subseteq \overline{f(f^{-1}(B))} \subseteq \overline{B},\end{align} which implies that $\overline{f^{-1}(B)}\subseteq f^{-1}(\overline{B})$. Applying this last expression to $Y \setminus B$ we obtain \begin{align}\tag{2} \overline{f^{-1}(Y \setminus B}) \subseteq f^{-1}(\overline{Y \setminus B}),\end{align} which together with (1) gives you \begin{align} f^{-1}(\operatorname{Int}(B)) &= f^{-1}(Y\setminus{\overline{Y\setminus B}}) \\ &= X \setminus {f^{-1}}(\overline{Y\setminus B)} \\ &\subseteq X \setminus \overline{f^{-1}(Y \setminus B)} \\ &= X \setminus {\overline{X \setminus f^{-1}(B)}} \\ &= \operatorname{Int}(f^{-1}(B)).\end{align} (We used (1) in lines 1 and 5, (2) in line 3.)

noam.szyfer
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Your question should hold in topology and is not restricted to metric spaces. In topology, (a) is essentially defined as: the preimage of an open set is an open set.

The following lemma is easy. I hope you can prove it by yourself.

Lemma 1. In a topology space $X$ and any subset $S \subset X$, $X\backslash \overline{S} = Int(X\backslash S)$ and $X\backslash Int(S) = \overline{X\backslash S}$.

Now prove $(b) \Rightarrow (c)$ in your question: $\forall B \subset Y$,

$$ X \backslash Int(f^{-1}(B)) = \overline{X \backslash f^{-1}B} = \overline{f^{-1}(Y \backslash B)} \subset f^{-1}(\overline{f(f^{-1}(Y \backslash B)) }) \\ \subset f^{-1}(\overline{Y \backslash B}) = f^{-1}(Y \backslash Int(B)) = X \backslash(f^{-1}(Int(B))) $$ The inclusion comes from (b). So (c) holds.

Functor
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There are two better answers above. I did this before. But, let me start again, $$\begin{array} .f^{-1}(B^\circ)&\stackrel{\star}{=}f^{-1}(Y-\overline{Y-B})\\ &\stackrel{S}{=}f^{-1}(Y)-f^{-1}(\overline{Y-B})\\ &\stackrel{S}{=}X-f^{-1}(\overline{Y-B})\\ &\stackrel{(b)}{\subseteq} X-\overline{f^{-1}(Y-B)}\\ &\stackrel{\star\star}{=}(X-f^{-1}(Y-B))^\circ\\ &\stackrel{S}{=}(f^{-1}(B))^\circ \end{array}$$

$\star:$ The interior is the complement of the closure of the complement.

$\star\star:$ The complement of he closure is the interior of the complement.

$(b):$ Explained by Noam in detail

$S:$ Set theory

Bob Dobbs
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