Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-155/3/solution

The Bourgain embedding theorem states that every -point metric space embeds into a Hilbert space with distortion at most ; the random-subset construction may be taken to have dimension .
We use the following finite Frechet embedding lemma. For every integer , an -point metric space has subsets such that
and, for every , some satisfies
Here is the probabilistic proof. Put . For every level , independently form
random subsets by including each point with probability .
Fix and put . Form open balls by taking to have radius and center for even and for odd . Consecutive balls are disjoint because their radii sum to at most . Partition the cardinality interval into
Among the cardinalities , either two consecutive ones lie in one interval, or some consecutive pair decreases. In either case there are such that, after orienting the pair appropriately,
At level , a random set hits with probability at least and misses the disjoint with probability at least . These events are independent, and when both occur the two distance-to-set values differ by at least . Thus one sample succeeds with probability at least . All samples at that level fail with probability at most
A union bound over the fewer than unordered pairs leaves a simultaneous successful choice. There are coordinates, proving the lemma.
Define
Every distance-to-set coordinate is one-Lipschitz, while the lemma supplies the lower bound. Hence
proving the Low-dimensional Frechet embedding into linfinity.
Take . Then , , and the distortion into is . Since
the identity has distortion . Composition gives
Finally let a -regular expander embed into with distortion at most . For , the identity
has distortion at most . Therefore the assumed lower bound gives
For , choose , so and
Adjusting the constant handles bounded , and exponentiation proves the Dimension lower bound for an expander embedded in linfinity
where depends only on and .

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