How to show that every Suslin tree is Frechet-Urysohn

Question

A topological space $X$, is Frechet-Urysohn, if, given $x \in \overline A \subset X$, there exists a sequence of points in $A$ which converges to $x$.

I am trying to prove that every Suslin tree, is Frechet-Urysohn. (I have read in an article, that this claim is folklore).

Any help?

Thank you!

Answer 1

Given a tree $\langle T , \leq_T \rangle$, the tree topology on $T$ is generated by the basis consisting of all sets of the form

• $( \leftarrow , t ] = \{ x \in T : x \leq_T t \}$, for $t$ in $T$;
• $( s , t ] = \{ x \in T : s <_T x \leq_T t \}$, for $s < t$ in $T$.

Note that if $t \in T$ has countable height, then there is a countable neighbourhood basis for $t$: take all the intervals $( s , t ]$ where $s <_T t$; since $\mathrm{ht}_T ( t ) = \mathrm{ot} ( \langle \{ x \in T : x <_T t \} , \leq_T \rangle)$ is countable, there can only be countably many different such sets. (Any $t \in T$ of height $0$ has a smallest open neighbourhood, which is $ ( \leftarrow , t ] = \{ t \}$.)

From this it follows that any tree of height $\leq \omega_1$ is first-countable in the tree-topology, and is therefore Fréchet-Urysohn. (Souslin, and more generally Aronszajn, trees, have height $\omega_1$.)

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Somewhat more can be said:

• $t \in T$ has a countable neighbourhood base in the tree topology iff $\mathrm{ht}_T (t)$ is $0$, a successor ordinal, or has cofinality $\omega$.

• For essentially the same reason that no ordinal space $[0,\alpha)$ with $\alpha > \omega_1$ is Fréchet-Urysohn (no sequence in $[ 0 , \omega_1 )$ can converge to $\omega_1$), no tree of height $> \omega_1$ can be Fréchet-Urysohn in the tree topology.