Question
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> For which pairs $M,N$ of compact metric spaces does there exist a metric space $K$ along with isometric embeddings $i:M \to K$ and $j:N \to K$ so that the Hausdorff distance between $i(M)$ and $j(N)$ in $K$ equals the Gromov-Hausdorff distance between $M$ and $N$ exactly? 
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We always have $d\_\text{GH}(M,N) \leq d\_\text{H}(i(M),j(N))$ by definition, so it suffices to decide if there exist any constraints intrinsic to $M$ and $N$ which force the reverse inequality to also hold for some judicious choice of $i,j,K$.


Update
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Bill Johnson's nice answer appears to settle the question for compact $M$ and $N$ affirmatively: the Gromov-Hausdorff distance can always be realized by an ultraproduct construction. In light of that answer -- and also in light of the fact that I did not intend to assume compactness in the original question but stupidly wrote it down anyway -- here is a modified question:



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> Can the GH-distance be similarly achieved if we only assume that $M$ and $N$ are *bounded*, rather than compact?
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In general, I am interested in the weakest hypotheses on $M$ and $N$ which are known to guarantee embeddings into a common target space $K$ so that the Gromov-Hausdorff distance between $M$ and $N$ equals the Hausdorff distance of their isometric images in $K$. It is not clear to me that a modification of Bill's argument works when we drop compactness.


Background
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Given a compact metric space $M$ and two subspaces $A, B \subset M$, their *Hausdorff distance* -- denoted $d\_\text{H}(A,B)$ -- is defined to be the smallest $\epsilon > 0$ so that $B$ is contained in $A$ thickened by $\epsilon$ and vice-versa. 


The *Gromov-Hausdorff* distance $d\_\text{GH}(M,N)$ between two compact metric spaces $M$ and $N$ is defined to be the infimum over triples $(i,j,K)$ of $d\_\text{H}(i(M),j(N))$ where $K$ is a metric space and $i,j$ are isometric embeddings of $M, N$ into $K$. The question asks when this infimum can be explicitly realized.



For which metric spaces is Gromov-Hausdorff distance actually achieved?
Unless I am missing something, the distance is achieved for compact metric spaces. Take isometries $i\_n :M \to K\_n$ and $j\_n : N \to K\_n$ that give you the Gromov-Hausdorff distance up to $1/n$. Let $K$ be an ultraproduct of the $K\_n$ and $i$, $j$ the isometries from $M$, $N$ into $K$ induced by $i\_n$, $j\_n$. This achieves the distance, I think, when $M$ and $N$ are compact. To see that, given $x$ in $M$ take $y\_n$ in $N$ s.t. the distance of $i\_n(x) $ to $j\_n(y\_n)$ is less than the Hausdorff distance from $i\_n(M)$ to $j\_n(N)$. Let $y$ be the ultralimit of $y\_n$ in $N$ (this is where compactness is used). Then the distance from $i(x)$ to $j(y)$ is at most the Gromov-Hausdorff distance from $M$ to $N$. 


This is a pretty obvious argument, so I guess it is either written somewhere or is complete nonsense.


