Is Gromov-Hausdorff distance realized when one space is compact? The Gromov-Hausdorff distance between two complete$^1$ metric spaces $M,N$ is defined as the infimum, over all pairs $f,g$ of embeddings of $M,N$ into some third metric space $U$, of the Hausdorff distance in $U$ between the images $f[M]$ and $g[N]$. In general, this infimum need not be achieved by any specific pair of embeddings; however, if $M$ and $N$ are each compact, then a standard argument shows that the infimum is achieve. See here.
However, unless I'm having a silly moment, that argument seems to break down if only one of the spaces involved is compact. So my question is:

Suppose $M,N$ are complete metric spaces, $M$ is compact, and $N$ is bounded. Must there be isometric embeddings $f,g$ of $M,N$ into some metric space $U$ such that $d_{GH}(M,N)=d_{H}^U(f[M],g[N])$?

(Here $d_{GH}$ is the Gromov-Hausdorff distance, and $d_H^U$ is the Hausdorff distance within $U$.)

$^1$There is nothing gained by considering non-complete spaces here, so I'm restricting attention to complete spaces for simplicity.
 A: The answer is no even when the compact space has two points. Let $M$ be the two point space whose points are a distance $3$ apart, and let $N$ be the metric space whose underlying set is $\mathbb{N}$ with the metric $d(x,y) = (1-2^{-\max(x,y)})[x\neq y]$. (This is a metric because any $[\frac{1}{2},1]$-valued function $d(x,y)$ satisfying $d(x,x) = 0$ and $d(x,y) = d(y,x)$ is a metric.)
In order to compute $d_{GH}(M,N)$ exactly, it will be useful to consider the definition of $d_{GH}$ in terms of correspondences and distortions. A correspondence $R$ between $M$ and $N$ is basically just two non-empty subsets $A,B \subseteq N$ satisfying $A\cup B = N$, and the distortion of the correspondence is just $\mathrm{dis}(A,B):=\sup\{|d(a,b)-3|:a\in A,~b \in B\}$. Since all distances in $M$ are less than $1$, this is the same as $3-\inf\{d(a,b):a\in A,~b \in B\}\geq 2$. Now we have
$$d_{GH}(M,N) = \frac{1}{2}\inf\{\mathrm{dis}(A,B):A,B\subseteq N\text{ non-empty},~A\cup B = N\}\geq \frac{1}{2}2 = 1.$$
To show that $d_{GH}(M,N) = 1$, let $A = N \setminus\{x\}$ and $B = \{x\}$ with $x > 0$. We clearly have that $\mathrm{dis}(A,B) = 3-(1-2^{-x})=2+2^{-x}$. Since we can do this for any $x > 0$, we have that $d_{GH}(M,N) \leq 1$, wherefore $d_{GH}(M,N) = 1$.
Now let $f: M \to U$ and $g : N \to U$ be isometric embeddings. Let $\alpha$ and $\beta$ be the two elements of $M$. We may assume without loss of generality that the underlying set of $U$ is just $M \sqcup N$ and that $f$ and $g$ are just the inclusion maps. Assume for the sake of contradiction that $d_H^U(M,N) = 1$. Let $\alpha$ and $\beta$ be the two elements of $M$. Since $M$ is finite, we must have for each $x \in N$ that either $d(\alpha,x) \leq 1$ or $d(\beta,x) \leq 1$. Therefore, without loss of generality, we may assume that $d(\alpha,x) \leq 1$ for infinitely many $x \in N$. Note that if $d(\alpha,x) \leq 1$, then $d(\beta,x) \geq 2$, so there must be some $y \in N$ such that $d(\alpha,y) > 1$. Let $y$ be the least such and note that $d(\beta,y) \leq 1$. Find $x > y$ such that $d(\alpha,x) \leq 1$. Now we have that
$$ d(\alpha,\beta) \leq d(\alpha,x) + d(x,y) + d(y,\beta) \leq 1 + 1-2^{-x} + 1 < 3,$$
which is absurd. Therefore no such embedding can exist.
