 Solution

In a neutral Moran model, a population has $N$ labelled individuals; reproduction copies one individual's type to a uniformly chosen other individual, keeping population size fixed. One convenient time convention puts independent rate-$1/2$ Poisson arrows on each ordered pair of distinct individuals. Backwards in time, two sampled ancestral lineages merge whenever one copies the other, at total rate $1$ per unordered pair. In the convention where each individual reproduces at rate $1$, this pair rate is $2/(N-1)$ if the parent cannot replace itself, and time must be rescaled accordingly.

Kingman's coalescent on $n$ labels is the continuous-time Markov chain on partitions of $\{1,\dots ,n\}$, initially all singleton blocks, in which each pair of distinct blocks merges at rate $1$. With $j$ blocks, the waiting time is exponential of rate $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$, and the merging pair is uniform. Kingman's infinite coalescent is the consistent partition process on $\mathbb{N}$ whose restriction to each finite label set has this law; consistency and the extension theorem construct it. These time conventions are part of the definition.

Let $B_t^{(n)}$ be the number of blocks in the first $n$ labels. The expected time to reach at most $m$ blocks isMarkov inequality gives $\mathbb{P}(B_t^{(n)}>m)\le 2/(mt)$. As $n$ increases these counts increase to the total number of infinite-coalescent blocks, so the same bound holds for that number. Letting $m\to \infty$ shows it is finite almost surely for each $t>0$. Apply this simultaneously to positive rational times; monotonicity of the block count then covers every positive real time. Hence Kingman's infinite coalescent comes down from infinity almost surely.