Is the boundary of a rooted tree with finite valence compact? Given an infinite rooted tree $T$ with root $v$ it is possible do define the boundary $\partial T$ of $T$ to be the set of ininite paths emanating from $v$. If $T$ has bounded valence, then the natural tooplogy on $\partial T$ makes it compact. I know that $\partial T$ is no longer compact if $T$ has vertices of infinite valence, but I am not sure what is the case when the vertices of $T$ all have finite valence, but such that the valences are globally unbounded.

As long as the valences are finite, the boundary is still compact, even if the valences aren't bounded. Proof: For each natural number $n$, let $D_n$ be the set of vertices at distance $n$ from the root $v$. Because all the valences are finite, each of these sets $D_n$ is finite. So each $D_n$, when equipped with the discrete topology, is a compact space. Therefore so is the product $\prod_nD_n$. The boundary of your tree is a closed subset of this product, so it's also compact.