I know open and closed are in somehow dual relations.

Now I am applying this concept in continuous map.

What I know is following

$f$ is continuous iff $\overline{f^{-1}(B)} \subset f^{-1}(\overline{B})$ for all $B\subset Y$

and its dual version also holds. i.e.,

$f$ is continuous iff $\operatorname{int}(f^{-1}(B)) \supset f^{-1}(\operatorname{int}(B))$ for all $B\subset Y$

Furthermore, many textbooks cover this as well

$f : X \rightarrow Y$ is continuous iff for each $A \subset X$, $f(\bar{A}) \subset \overline{f(A)}$.

But I don't see its dual version at any textbook.

Is its dual version works?

I mean

iff for each $A\subset X$, $f(int(A)) \supset int(f(A))$

  • 3
    $\begingroup$ "open is dual to close" has no real sense... but we can understand what you mean. I didn't think to your problem so much, but In your two first version (with $f^{-1}$), you can easily pass from the version with open set to the version with closed set using the fact that $f^{-1}$ is stable with intersection, i.e. $f^{-1}(A\cap B)=f^{-1}(A\cap B)$. Unfortunately, this is not true true with $f$, so I wouldn't be surprise that such a dual version doesn't exist... $\endgroup$
    – Surb
    Jan 4, 2020 at 13:54

2 Answers 2


The first two characterisations (using $f^{-1}$) are derivable from each other : in any space $X$ we have the complements duality between interior and closure

$$\operatorname{int}(A)=X\setminus \overline{X\setminus A}$$

and $$\overline{A}= X\setminus \operatorname{int}(X\setminus A)$$

and moreover, $f^{-1}$ plays nice with complements, i.e.

$$f^{-1}[Y\setminus A]=X\setminus f^{-1}[A]$$

and complements reverse inclusions.

But we don't have the complement property for forward images (we do for injective functions, but generally $f[X\setminus A]\neq f[X]\setminus f[A]$ etc.)

The property that $f[\operatorname{int}(A)] \subseteq \operatorname{int}(f[A])$ characterises open maps, as can be easily seen. But the reverse

$$f[\operatorname{int}(A)] \supseteq \operatorname{int}(f[A])$$

does not characterise continuity, as can be seen from the examples here, (both implications with continuity fail) which I won't repeat here.


No. It's false in both directions. Interior of a set depends on the ambient space. So it's easy to produce sets with empty interiors. For example, let $X = \mathbb{R}$ (with the usual topology) and $Y = \{\ast\}$ be a singleton and let $f : X \to Y$ be the constant map. Let $A = \{0\} \subset X$. Then $\mathrm{int}(A) = \emptyset$, but $\mathrm{int}(f(A)) = Y$ which shows that the only if statement is false. For the other direction, let both $X$ and $Y$ be $\mathbb{R}$. Define $f : X \to Y$ by sending $(-\infty, 0) \to \{-1\}$ and $[0, \infty) \to \{1\}$. The interior of the image is empty but the map is discontinuous.


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