Every subspace of a Frechet-Urysohn space is a Frechet- Urysohn space and hence also a sequential space

Question

I'm trying to understand this statement:

Every subspace of a Frechet-Urysohn space is a Frechet- Urysohn space and hence also a sequential space.

Definition of Frechet-Urysohn space:

A topological space $(X, \tau)$ is said to be a Frechet-Urysohn space if for every subset $S$ of $(X, \tau)$ and every $a$ in the closure, $\overline{S}$, of $S$ there is a sequence $s_n \to a$, for $s_n \in S$, $n \in \mathbb{N}$.

I wonder how is singleton set subspace a Frechet-Urysohn space, according to the definition?

Answer 1

I suspect you are somehow misunderstanding the statement or the definition of Frechet-Urysohn, since the case of a singleton is really quite trivial. Hopefully the following explanation will help you find your misunderstanding.

If $X=\{x\}$ is a singleton, there are only two subsets of $X$: either $S=\emptyset$ or $S=\{x\}$. If $S=\emptyset$ then $\overline{S}=\emptyset$ and there does not exist any element $a\in \overline{S}$, so there's nothing to check. If $S=\{x\}$ then $\overline{S}=\{x\}$ and the only element $a\in\overline{S}$ is $x$, which is the limit of the constant sequence $s_n=x$ in $S$.

(So, in fact, you don't even need the ambient space to be Frechet-Urysohn to conclude that any singleton subspace is Frechet-Urysohn.)

Answer 2

The main thing to note is that if $A$ is a subspace of $X$ and $x \in \overline{S}$ (in the subspace topology on $A$) for some $S \subseteq A$ and $x \in A$, then $x \in \overline{S}$ in $X$ as well. So immediately Fréchet-Urysohn applies in $X$ etc.