# open is dual to closed : continuity

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 $$A\subset X$$, $$f(int(A)) \supset int(f(A))$$

• "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...
– Surb
Jan 4, 2020 at 13:54

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.