Is $x \mapsto d(x, A)$ a quotient map?

Question

Consider a metric space $(X, d)$ and a nonempty closed set $A \subset X$. Is the map $d_A : X \to \mathbb{R}, x \mapsto d(x, A)$ a quotient map when restricted to its image? Note $d(x, A) = \inf\{ d(x, a) : a \in A \}$.

If this is not true, can a counterexample exist in $\mathbb{R}^2$? Some partial results follow.

If $X$ is compact, then $d_{A}$ is closed. This follows since, more generally, continuous maps from compact spaces to Hausdorff spaces are closed.

If $X$ has the Heine-Borel property and $A$ is compact then $d_A$ is closed. For $B \subset X$ closed and $y \in \overline{d_A(B)}$ there exist $x_n \in B$ such that $d_{A}(x_n) \to y$. Exercise 27.2b of Munkres 2ed states that since $A$ is compact, $d(x, A) = d(x, a)$ for some $a \in A$. So there exist $a_n \in A$ such that $d(x_n, a_n) \to y$. Since $A$ is compact there exists a subsequence $a_{i_n}$ of $a_n$ such that $a_{i_n} \to a$. Then $d(x_{i_n}, a) \le d(x_{i_n}, a_{i_n}) + d(a_{i_n}, a)$ for all $n$, and the right-hand side of this inequality is eventually less than $y + 1$. So $x_{i_n}$ is eventually contained in a closed bounded closed set, and by the Heine-Borel property there is a subsequence $x_{j_n}$ of $x_{i_n}$ such that $x_{j_n} \to x$. Since $B$ is closed, $x \in B$. Then continuity and the Hausdorff property imply $d_{A}(x) = y$, implying $y \in d_{A}(B)$. So the claim follows. In the general case, is $d_A$ always a closed map when $A$ is compact?

$d_A$ is not generally a closed map, even when $A$ is closed. Consider the case where $X = \mathbb{R}^2$ and $A$ is the $x$-axis. Then $d_A$ maps the graph of the exponential function to $(0, \infty)$.

Answer 1

This can fail if $X$ is not connected but the image of $d_A$ is. For example consider $X = (-1, 0] \cup [1, \infty)$ and $A = \{0\}$, with the Euclidean metric $d_A(x) = | x |$. Then $d_A(X) = [0, \infty)$, which has one component, but no pair of points in distinct components of $X$ has been identified.

In the case $X$ is connected, I don’t see an obvious answer.

Answer 2

For your second question. $A$ is compact doesn't imply $d_A$ is a closed map. Let $X$ be the space of irrational numbers and $A$ be $\sqrt 2$. Then $B=[0,\sqrt 2]\cap X$ is closed in $X$. But $\sqrt 2\notin d_A (B),$ although it belongs to $d_A(X)$. It's easy to see $\sqrt 2\in \overline{d_A(B)}$, hence $d_A(B)$ is not closed.

Answer 3

I stumbled across the following paper and think it adds to the answers given so far.

Which Connected Spaces Have a Quotient Homeomorphic to an Arc

It gives additional counterexamples. Example 9 is a counterexample in the case that $X$ is connected and locally compact, and $A$ is a point.

There are also some propositions characterizing when certain maps will be quotients. Theorem 5 states

If $X$ is a connected, locally connected space, $f : X \to [0, 1]$ is a continuous surjection, then $f$ is a quotient.

In particular, this implies $\frac{d_{A}}{d_A + d_B}$ will be a quotient in the case that $X$ is connected, locally connected, and $A, B$ are disjoint closed sets. This gives an Urysohn map from $A$ to $B$ which is a quotient. I'm not immediately sure if $d_A$ will be a quotient but it seems likely.