Does this strengthening of continuity have a characterization in terms of familiar concepts?

Question

Definition 0. Whenever $X$ is a metric space, $A \subseteq X$ is a subset, and $r \in \mathbb{R}_{>0}$ is a positive real number, define that $$A \oplus r = \bigcup_{a \in A} B_r(a).$$

Definition 1. Whenever $X$ is a metric space, and $A,A' \subseteq X$ are subsets, define that $A \lhd A'$ iff $A\oplus r \subseteq A'$ for some $r \in \mathbb{R}_{>0}$.

Question. Suppose $X$ and $Y$ are metric spaces and $f : X \rightarrow Y$ is a function. Is there a nice characterization of the following condition in terms of familiar concepts?

$(*)$ For all $B,B' \subseteq Y,$ we have the following implication.$$B \lhd B' \rightarrow f^{-1}(B) \lhd f^{-1}(B')$$

Discussion. Its certainly true that if $(*)$ holds, then $f$ is continuous. To see this, observe that a subset $A \subseteq X$ of a metric space $X$ is open iff for all $a \in A$ it holds that $a \lhd A$. Hence, we may argue as follows.

Proof. Consider an open set $B \subseteq Y$ and fix $a \in f^{-1}(B)$. We need to show that $a \lhd f^{-1}(B)$. Since $a \in f^{-1}(B)$, hence $f(a) \in B$. So $f(a) \lhd B$ because $B$ is open. So $f^{-1}(f(a)) \lhd f^{-1}(B)$ by (2). So $a \lhd f^{-1}(B).$

However, continuity is not enough to ensure $(*)$. An example is the function $f : (0,\infty) \rightarrow [-1,1]$ given by $f(x) = \sin(1/x)$. Consider $B = (-1/4,1/4)$ and $B' = (-1/2,1/2)$. Then $B \lhd B'$, but not $f^{-1}(B) \lhd f^{-1}(B')$.

Answer 1

Your property $(\ast)$ is precisely uniform continuity.

The proof that uniform continuity implies $(\ast)$ is easy: according to the definition of uniform continuity, for every $\varepsilon > 0$ there is a $\delta > 0$ with $d(x,y) < \delta \implies d(f(x),f(y)) < \varepsilon$, and from that follows that

$$f(A\oplus\delta) \subset f(A) \oplus \varepsilon$$

for every $A\subset X$, whence

$$f^{-1}(B) \oplus \delta \subset f^{-1}(B\oplus \varepsilon)$$

for all $B\subset Y$, and that gives you the desired

$$B \lhd B' \implies f^{-1}(B) \lhd f^{-1}(B').$$

For the other direction, suppose that $f\colon X\to Y$ is not uniformly continuous. Then there is an $\varepsilon > 0$ such that for every $n\in\mathbb{N}$ there are points $x_,z_n \in X$ with $d(x_n,z_n) < 2^{-n}$ but $d(f(x_n),f(z_n)) \geqslant \varepsilon$.

Now if there is a $y\in Y$ such that one of $M(y) := \{ n\in\mathbb{N} : d(f(x_n),y) < \varepsilon/4\}$ and $N(y) := \{ n\in\mathbb{N} : d(f(z_n),y) < \varepsilon/4\}$ is infinite, then $B = B_{\varepsilon/4}(y)$ violates $(\ast)$: $B' := B \oplus \varepsilon/4 \subset B_{\varepsilon/2}(y)$ by the triangle inequality, and for every $n\in\mathbb{N}$ we have

$$d(f(x_n),y) < \varepsilon/2 \implies d(f(z_n),y) \geqslant d(f(z_n),f(x_n)) - d(f(x_n),y) > \varepsilon - \varepsilon/2,$$

so $f(x_n) \in B' \implies f(z_n)\notin B'$. By symmetry, $f(z_n)\in B' \implies f(x_n)\notin B'$. In particular, $f(z_n) \notin B'$ for all $n\in M(y)$ and $f(x_n) \notin B'$ for all $n\in N(y)$. But if, say, $M(y)$ is infinite, then set $A = \{ x_n : n \in M(y)\}$. By construction, $A\subset f^{-1}(B)$. Since $d(x_n,z_n) < 2^{-n}$, every $A\oplus \delta$ contains $z_n$ for infinitely many $n\in M(y)$, and thus

$$A \not\lhd f^{-1}(B')$$

and a fortiori $f^{-1}(B) \not\lhd f^{-1}(B')$ although $B\lhd B'$. If $M(y)$ is finite and $N(y)$ infinite, swap the roles of $x_n$ and $z_n$.

So it remains to treat the case that for all $y\in Y$ the two sets $M(y)$ and $N(y)$ are finite. But then we can construct an infinite subset $K\subset\mathbb{N}$ such that $\{ f(x_k) : k\in K\}$ and $\{ f(z_k) : k\in K\}$ have positive distance: Let $n_0 = 0$, and define

$$n_{k+1} := 1 + \max \bigl(M(f(x_{n_k})) \cup M(f(z_{n_k})) \cup N(f(x_{n_k})) \cup M(f(z_{n_k}))\bigr)$$

for $k \geqslant 0$. In other words, $n_{k+1}$ is the smallest $n\in\mathbb{N}$ such that for all $m\geqslant n$ the distance of $f(x_m)$ and of $f(z_m)$ to each of $f(x_{n_k})$ and $f(z_{n_k})$ is at least $\varepsilon/4$. Let $K = \{ n_k : k\in\mathbb{N}\}$.

Then by construction, we have $d(f(x_k), f(z_j)) \geqslant \varepsilon/4$ for all $k,j\in K$, and we can take $B = \{ f(x_k) : k \in K\}$ and $B' = B \oplus \varepsilon/8$. For $A = \{ x_k : k\in K\}$ we have $A\subset f^{-1}(B)$, and every $A\oplus r$ contains infinitely many $z_k$ with $k\in K$, but $f^{-1}(B')$ contains no $z_k$ for $k\in K$, hence

$$A \not\lhd f^{-1}(B')$$

and a fortiori $f^{-1}(B) \not\lhd f^{-1}(B')$, although $B\lhd B'$.

Thus $(\ast)$ implies uniform continuity.

Answer 2

Uniform continuity implies $(*)$. In particular $(*)$ is equivalent with continuity if $X$ is compact.

Proof: Suppose $B,B'\subset Y$ are such that $B\oplus r\subset B'$ for some $r>0$. To show that $f^{-1}(B)\oplus s\subset f^{-1}(B')$ (for a suitable $s>0$), we need to show that for any $x\in f^{-1}(B)$ we have $B_s(x)\subset f^{-1}(B')$. By uniform continuity there is $s>0$ so that $f(B_s(x))\subset B_r(f(x))$ for all $x\in X$. Now if $z\in B_s(x)$, then $f(z)\in B_r(f(x))\subset B'$ and so $z\in f^{-1}(B')$.

A continuous function on a compact metric space is uniformly continuous. See eg. this old post for a proof. $\square$

I cannot come up with an example where $f$ satisfies $(*)$ but is not uniformly continuous. It is possible that $(*)$ is equivalent with uniform continuity, but I'm not sure.