Proving these equivalent conditions for an open map using boundary of a set

Question

Let $X,Y$ be topological spaces.

Prove the following statements are equivalent.

$(1)$ $f\colon X\to Y$ is an open map.

$(2)$ For all $x\in X$ and open set $U \ni x$ there exists open set $V$ such that $f(x)\in V \subset f(U)$.

$(3)$ $f^{-1}(\text{Bd}(B))\subset \text{Bd}(f^{-1}(B))$ for all $B\subset Y$. (Here, $\text{Bd}(B)$ is the boundary of a set $B$.)

(The question is attached by image file.)

https://i.stack.imgur.com/PrWw4.png

I completely proved $(1) \implies (2)$, but couldn't find any idea to solve $(2) \implies (3)$ and $(3) \implies (1)$. I'm trying to solve using $\text{Cl}(B)=\text{Int}(B)-\text{Bd}(B)$, but failed.

*** $\text{Cl}(B)$ is the closure of a set $B$, $\text{Int}(B)$ is the interior of a set $B$ and $\text{Bd(B)}$ means boundary of $B$.

Please give me some hints.

Answer 1

I don't think you need to include the image, you have written in $\LaTeX$ well enough the three problems. I also changed the boundary notation from $\text{b}()$ to $\text{Bd}()$ since it is more standard. Lastly, you have the closure identity wrong. The identity is that $\text{Bd}(A) = \text{Cl}(A) - \text{Int}(B)$.

Here's a way you could show $(3) \implies (1)$: Take the contrapositive approach and suppose $f$ is not an open map. Then there exists an open set $U' \subset X$ such that $f(U') = B$ is not open. This means $B$ has a boundary point $y$ that is not in $B$. (You can show this via the closure identity I mentioned above). We now have that $f^{-1}(y) \in f^{-1}\left(\text{Bd}(B)\right)$. However since $y \notin B$ then $f^{-1}(y) \notin f^{-1}(B)$. Can you take the last step to show that $f^{-1}(y) \notin \text{Bd}\left(f^{-1}(B)\right)$? If so, $f^{-1}(y)$ is the element that demonstrates that $$f^{-1}\left(\text{Bd}(B)\right)\not\subseteq\text{Bd}\left(f^{-1}(B)\right)$$ You'll have to flush out some parts of the proof a bit more but I think it will hold. Please let me know if it doesn't and I'll delete/edit my post.

As for showing $(2) \implies (3)$, you may also have luck approaching with contrapositive. You can use $$f^{-1}\left(\text{Bd}(B)\right)\not\subseteq\text{Bd}\left(f^{-1}(B)\right)$$ to find another $f^{-1}(x)$ element that is in $f^{-1}\left(\text{Bd}(B)\right)$ but not in $\text{Bd}\left(f^{-1}(B)\right)$. I suspect if you push for it you can reveal that $x$ will have the property that: $$x \in X \text{ (**This one is a freebie**)} \tag{1}$$ $$\text{There exists an open set } U \text{ such that } x\in U \tag{2}$$ $$\text{There does not exist any open set } V \subset Y \text{ where } f(x) \in V \subseteq f(U) \tag{3}$$

Answer 2

Suppose (2) holds. We want to show (3), so pick $x \in f^{-1}[b(B)]$, where $b(B)$ is the boundary of $B$ and $B \subseteq Y$. We want to show that $x \in b(f^{-1}[B])$.

To this end, we pick an open set $O$ that contains $x$, and we need to show that $O$ intersects both $f^{-1}[B]$ and its complement (this is what being in the boundary of a set means).

We know that $x \in f^{-1}[b(B)]$, so $f(x) \in b(B)$. We know that $f(x) \in f[O]$ and by condition (2) we have that there exists $V$ open, with $f(x) \in V$ and $V \subseteq f[O]$. But $f(x)$ in $b(B)$ implies that $V$ intersects both $B$ and $Y \setminus B$. So any $y \in V \cap B$ is also in $f[O]$ (from $V \subseteq f[O]$) , so there exists $y' \in O$ with $f(y') = y$ and so for $y' \in O \cap f^{-1}[B]$. Also for the same reason, any $z \in V \cap (Y \setminus B)$ has $z' \in O$ with $f(z') = z \notin B$, so that $z' \in O \cap (X \setminus f^{-1}[B])$.

So using (2) we have that $O$ intersects both $f^{-1}[B]$ and its complement, so that $x \in b(f^{-1}[B])$, as required.

Now, to see that (3) implies (1), we assume the condition (3) holds for $f$, and we need to show that $f$ is an open map. So let $O \subseteq X$ be open, and we need to somehow show that $f[O]$ is open. How to connect this to boundaries?

Well, a set $U$ is open (in any space) iff $U \cap b(U) = \emptyset$ (check this, e.g. it follows from a set being closed iff it contains its boundary). So we can try to show that $b(f[O]) \cap f[O] = \emptyset$. So suppose not, and we have $x \in O$ with $f(x) \in f[O]$ and $f(x) \in b(f[O])$.

Set $B = f[O]$. Then $f^{-1}[b(B)] \subseteq b(f^{-1}[B])$ by condition (3). Then $x$ from before is in $f^{-1}[b(B)]$ (as $f(x) \in b(f[O]) = b(B)$), so $x \in b(f^{-1}[B])$. As $x \in O$, $O$ intersects both $f^{-1}[B]$ and $X \setminus f^{-1}[B]$. But if $x' \in (X \setminus f^{-1}[B]) \cap O$, then $f(x') \in f[O] = B$, but $f(x') \notin B$ at the same time. So said $x$ cannot exist, and so $b(f[O]) \cap f[O]$ is empty and so $f[O]$ is open. Hence $f$ is an open map. This shows (3) implies (1).

The proof writes itself once you keep the conditions in mind, and keep track of what you already know etc.