# Image of a Topologically Regular Set?

This question has come up as a small question in my research and I think I'm a little too thick in the weeds with extraneous details to see it cleanly. If necessary, I can add some more conditions on the spaces and maps that follow---just ask!

Suppose $$(X, d_x, \mu_x)$$ and $$(Y, d_y, \mu_y)$$ are metric measure spaces, with metrics $$d_x$$ and $$d_y$$ and Borel measures $$\mu_x$$ and $$\mu_y$$ respectively. Let $$f:X \to Y$$ be continuous (and hence measureable). Generally assume that $$X$$ and $$Y$$ are "nice" (e.g., manifolds) but $$f$$ is "messy."

Definition: A set $$U$$ is called topologically regular (or a regular closed set) if it is the closure of its interior, i.e., $$\overline{\mathrm{int } U} = U$$.

Theorem: If $$A \subseteq X$$ is compact and $$f$$ is continous, then $$f(A) \subseteq Y$$ is compact.

Suppose for everything that follows that $$A \subseteq X$$ is non-empty, compact, and topologically regular. My aim is to establish the weakest possible condition on $$f$$ such that $$f(A) \subseteq Y$$ is also topologically regular.

Baby Question: If $$f:X \to Y$$ is continuous, is $$f(A) \subseteq Y$$ topologically regular?

Counter Example: Consider $$f:\mathbb{R}^2 \to \mathbb{R}^2$$ with the Euclidean metrics and Lebesgue measures. Let $$f(x,y):=(x, 0)$$; observe that $$f$$ is clearly continuous. However, $$[0,1]^2$$ is topologically regular but $$f([0,1]^2) = [0,1] \times \{0\}$$ which has empty interior and is hence not topologically regular. $$\blacksquare$$

Clearly we need a stronger condition on $$f$$ in order to guarantee that the image is also topologically regular. The following would be sufficient, but is a stronger condition than I could ever possibly dream of in my context.

Theorem: If $$f:X \to Y$$ is a homeomorphism, then $$f(A)$$ is topologically regular.

Proof: Clearly $$f( \mathrm{int}~A) = \mathrm{int}~ f(A)$$ as $$f$$ is a homeomorphism; since $$A$$ is topologically regular and non-empty, we have that $$\mathrm{int}~A \neq \emptyset$$ and hence $$\mathrm{int}~f(A) \neq \emptyset$$.

Let $$y \in \partial f(A)$$ and $$\epsilon > 0$$ be given. We seek to show that the ball $$B(y, \epsilon)$$ has non-trivial intersection with $$\mathrm{int}~f(A)$$. Since $$f(A)$$ is compact (and hence closed), we have that $$y \in f(A)$$ and hence $$x:= f^{-1}(y) \in A$$. Since $$A$$ is topologically regular, we have that $$f^{-1}(B(y, \epsilon)) \cap \mathrm{int}~A \neq \emptyset$$ and therefore $$B(y, \epsilon) \cap f(\mathrm{int}~A) = B(y, \epsilon) \cap \mathrm{int}~f(A) \neq \emptyset.$$ Hence $$y \in \overline{\mathrm{int}~f(A)}$$ and thus $$f(A) = \overline{\mathrm{int}~f(A)}$$ as desired. $$\blacksquare$$

## The Question

In my specific context, $$f$$ will land somewhere on the spectrum between being continuous (trivially easy) and being a homeomorphism (provably impossible). What are the weakest possible conditions on $$f$$ to guarantee that $$f(A)$$ is topologically regular? Do any (or all?) of the following suffice? If they fail, what additional (if any) hypotheses on $$f$$ or the spaces $$X$$ and $$Y$$ are necessary?

1. $$f:X \to Y$$ is continuous and bounded-to-one, i.e., there exists $$M \in \mathbb{N}$$ such that $$\mathrm{Card } f^{-1}(y) \leq M$$ for all $$y \in Y$$.

2. $$f:X \to Y$$ is continuous and almost everywhere constant-to-one, i.e., there exists $$M \in \mathbb{N}$$ such that $$\mathrm{Card}~f^{-1}(y) = M$$ almost everywhere.

3. $$f:X \to Y$$ is continuous and almost everywhere injective.

4. $$f: X \to Y$$ is Lipschitz continuous. Edit: See above counter example

5. $$f: X \to Y$$ is $$\alpha$$-Holder continuous. Edit: See above counter example

## Edit

Per comments from @WilliamElliot and @HennoBrandsma, $$f$$ being continuous and an open (or closed) mapping suffices. The proof is similar to that of $$f$$ being a homeomorphism as above, but with some care taken as to inclusions instead of inequalities. In my specific context, this actually solves my problem as I can show that $$f$$ is a closed map. I still think there's some interesting analysis to be done regarding the five conditions I've listed above, so I'll leave the question open for now.

## Edit 2: Open Mapping

There seems to be some confusion in the comments. I give a proof below that $$f$$ is continuous and an open mapping suffices. I also give proof that my Counter-Example is neither an open mapping nor a closed mapping.

Proposition E1: The map $$f:\mathbb{R}^2 \to \mathbb{R}^2$$ given by $$f(x,y):= (x,0)$$ is not an open mapping.

Proof: Given open $$U \subseteq \mathbb{R}^2$$, we have that $$f(U)= A \times \{0\}$$ for some $$A \subseteq \mathbb{R}$$, which is not open in $$\mathbb{R}^2$$. $$\blacksquare$$

Proposition E2: The map $$f:\mathbb{R}^2 \to \mathbb{R}^2$$ given by $$f(x,y):= (x,0)$$ is not a closed mapping.

Proof. Define $$U:=\{(x,y) \in \mathbb{R}^2 \mid y \geq 1/x \text{ and } x > 0\}$$. We have that $$U$$ is closed, but $$f(U) = (0, \infty)\times \{0\}$$ which is not closed. $$\blacksquare$$

Proposition E3: Suppose $$f:X \to Y$$ is continuous. If $$f$$ is an open mapping and $$A \subseteq X$$ is topologically regular, then $$f(A)$$ is topologically regular.

Proof. Let topologically regular $$A \subseteq X$$ be given. We begin by noting that if $$A=\emptyset$$, then the proposition holds vacuously. Therefore, assume that $$A \neq \emptyset$$ and thus $$\mathrm{int} A \neq \emptyset$$. Since $$\mathrm{int} A \subseteq A$$, we have that $$f(\mathrm{int} A) \subseteq \mathrm{int} f(A)$$ and in particular is non-empty.

Let $$y \in f(A)$$ and open neighborhood $$U\ni y$$ be given. Since $$f$$ is continuous, $$f^{-1}(U)$$ is open in $$X$$ and in particular $$f^{-1}(U) \cap A$$ is non-empty. As $$\overline{\mathrm{int} A} = A$$, we have that $$f^{-1}(U) \cap \mathrm{int} A \neq \emptyset$$. Therefore $$U \cap \mathrm{int} f(A) \supseteq U \cap f(\mathrm{int} A) \supseteq f \left(f^{-1}(U) \cap \mathrm{int} A \right) \neq \emptyset$$ and thus $$y \in \overline{\mathrm{int} f(A)}$$ as desired. $$\blacksquare$$

## Edit 3: Closed Mapping

Having $$f$$ be continuous and a closed mapping is insufficient to guarantee that the image is topologically regular.

Counter-Example: Consider $$f:[0,1]^2 \to [0,1]^2$$ (with the standard topologies) defined by $$f(x,y):=(x,0)$$. We can clearly see that $$f$$ is continuous and we will show that $$f$$ is a closed mapping. Let a closed subset $$A \subseteq [0,1]^2$$. As $$A$$ is a closed and bounded subset of $$\mathbb{R}^2$$, we have that $$A$$ is compact. Since the image of a compact set is compact (and thus closed), we have that $$f(A)$$ is closed as well and thus $$f$$ is a closed mapping.

However, $$f(A) \subseteq [0,1] \times \{0\}$$ and thus $$\mathrm{int}~f(A) = \emptyset$$ in $$[0,1]^2$$. Therefore, given any non-empty $$A$$ we have that $$\overline{\mathrm{int} f(A)} = \emptyset \neq f(A)$$ and thus $$f(A)$$ cannot be topologically regular. $$\blacksquare$$

• The first theorem is false unless f is continuous. Commented Aug 17, 2019 at 22:42
• @WilliamElliot edited to make more prominent. Commented Aug 17, 2019 at 22:49
• The counterexample uses a function which is not closed. Perhaps you want to require the function to be closed. Are you familiar with equivalent condition for a function to be closed (preserves closeness)? Commented Aug 17, 2019 at 22:57
• Isn't $f$ an open map sufficient too? Commented Aug 17, 2019 at 23:08
• @HennoBrandsma A counter example of an open projection that is not a regular closed map was given. Commented Aug 18, 2019 at 6:21

We give a counter-example which shows that none of the following conditions are sufficient to guarantee that the image of a regular set is regular. More conditions on $$X,Y, f$$ could presumably be added so as to give sufficiency, but this vague enough as to not be a well-posed question.

1. $$f:X \to Y$$ is continuous and bounded-to-one, i.e., there exists $$M \in \mathbb{N}$$ such that $$\mathrm{Card } f^{-1}(y) \leq M$$ for all $$y \in Y$$.

2. $$f:X \to Y$$ is continuous and almost everywhere constant-to-one, i.e., there exists $$M \in \mathbb{N}$$ such that $$\mathrm{Card}~f^{-1}(y) = M$$ almost everywhere.

3. $$f:X \to Y$$ is continuous and almost everywhere injective.

4. $$f: X \to Y$$ is continuous and injective.

Counter-Example: Let $$f: \{0\} \to \mathbb{R}$$ be defined by $$f(0):=0$$. Equip $$\mathbb{R}$$ with the standard topology, Euclidean metric, and Lebesgue measure $$\mu$$. Equip $$\{0\}$$ with the discrete topology (which is the same as the trivial topology in this case), the discrete metric, and the discrete measure $$\lambda$$. Observe the following:

• The Lebesgue measure $$\mu$$ is a Borel measure on $$\mathbb{R}$$.
• $$\lambda$$ is a Borel measure on $$\{0\}$$: the only open sets are $$\emptyset$$ and $$\{0\}$$, which have measures $$0$$ and $$1$$ respectively. Hence every open set is measurable and thus $$\lambda$$ is a Borel measure.
• $$f$$ is continuous: Let open $$U \subseteq \mathbb{R}$$ be given. Either $$U$$ contains $$0$$ or it doesn't. If $$0 \in U$$, then $$f^{-1}(U) = \{0\}$$ which is open. If $$0 \not \in U$$, then $$f^{-1}(U) = \emptyset$$ which is also open. Therefore the preimage of any open set is open and thus $$f$$ is continuous.
• $$f$$ is Lipschitz continuous and Holder continuous: trivial.
• $$f$$ is injective: suppose $$f(x) = f(x')$$. Since $$\{0\}$$ contains a single point, we have that $$x=0=x'$$ and thus $$f$$ is injective.
• $$f$$ is almost everywhere injective: this follows immediately from $$f$$ being injective. Explicitly, let $$N \subseteq \{0\}$$ be the set on which $$f$$ is not injective. Then $$N=\emptyset$$ so $$\lambda(N) = \lambda(\emptyset) = 0$$ as desired.
• $$f$$ is bounded-to-one: given $$y \in \mathbb{R}$$, we have that $$\mathrm{Card}~f^{-1}(y)$$ is either $$0$$ or $$1$$. Hence $$f$$ is bounded-to-one.
• $$f$$ is almost everywhere constant-to-one: this follows immediately from $$f$$ being injective or from $$f$$ being almost everywhere injective.
• $$f$$ is a closed mapping: we have that $$f(\emptyset) = \emptyset$$ and $$f(\{0\}) = \{0\}$$, which are both closed in $$\mathbb{R}$$. Hence the image of every closed subset of $$\{0\}$$ is closed in $$\mathbb{R}$$.
• $$f$$ is NOT an open mapping: $$\{0\}$$ is an open subset of $$\{0\}$$, but $$f(\{0\}) = \{0\}$$ is not open in $$\mathbb{R}$$.

Observe that $$\{0\}$$ is a regular closed set in $$\{0\}$$. Explicitly, $$\{0\}$$ is equipped with the discrete topology, so $$\{0\}$$ is open and thus $$\mathrm{int}~\{0\} = \{0\}$$. Furthermore, $$\{0\}$$ is closed and thus $$\overline{\mathrm{int}~\{0\}} = \{0\}$$.

However, $$f(\{0\}) = \{0\} \subseteq \mathbb{R}$$ has empty interior and thus $$\overline{\mathrm{int}~f(\{0\})} = \emptyset \neq \{0\} = f(\{0\})$$. Therefore $$f$$ does not have the property of mapping regular closed sets to regular closed sets as desired. $$\blacksquare$$