# Understanding two similar definitions: Fréchet-Urysohn space and sequential space

Here are the definitions:

Fréchet-Urysohn space: A topological space $X$ where for every $A \subseteq X$ and every $x \in \text{cl}(A)$, there exists a sequence $(x_{n})_{n \in \mathbb{N}}$ in $A$ converging to $x$.

Sequential space: A topological space $X$ where a set $A \subseteq X$ is closed iff $A$ contains the limit points of every sequence contained in it.

As the title explains, I would like to know the difference between them. Thanks for any help.

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Consider the following operation on a subset $A$ of a space $X$, defining a new subset of $X$: $$\mbox{s-cl}(A) = \{ x \in X \mid \mbox{ there exists a sequence } (x_n)_n \mbox{ from } A \mbox{ such that } x_n \rightarrow x \}\mbox{.}$$ This set, the sequential closure of $A$, contains $A$ (take constant sequences) and in all spaces $X$ it will be a subset of the $\mbox{cl}(A)$, the closure of $A$ in $X$.

We can define $\mbox{s-cl}^{0}(A) = A$ and for ordinals $\alpha > 0$ we define $\mbox{s-cl}^\alpha(A) = \mbox{s-cl}(\cup_{\beta < \alpha} \mbox{s-cl}^\beta(A))$, the so-called iterated sequential closure.

A space is Fréchet-Urysohn when $\mbox{s-cl}(A) = \mbox{cl}(A)$ for all subsets $A$ of $X$, so the first iteration of the sequential closure is the closure.

A space is sequential if some iteration $\mbox{s-cl}^\alpha(A)$ equals the $\mbox{cl}(A)$, for all subsets $A$.

So basically by taking sequence limits we can reach all points of the closure eventually in a sequential space, but in a Fréchet-Urysohn space we are done after one step already.

For more on the differences and the "canonical" example of a sequential non-Fréchet-Urysohn space (the Arens space), see this nice topology blog, and the links therein.

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a space is a Sequential space space iff every sequencially closed set is closed. a space is frechet-urysohn if sequencial closure of a set and usual closure coincide

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a space is a Sequential space space iff every sequencially closed set is closed.But there can be closed sets which are not sequencially closed. a space is frechet-urysohn if sequencial closure of a set and usual closure coincide

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That's awfully wrong! Any closed subset $A$ is sequentially closed since a limit point of a sequence in $A$ is always an adherence point (or limit point) of the set $A$. –  Stefan Hamcke Jan 2 '13 at 17:45

Both Frechet-Urysohn and sequential spaces are related to first-countable spaces. In fact, first-countable $\Rightarrow$ Frechet-Urysohn $\Rightarrow$ sequential.

• Frechet-Urysohn gives a characterisation of what it means for a point to belong to the closure of a set: $x \in \overline{A}$ iff there is a sequence in $A$ converging to $x$.
• Sequentiality gives a characterisation of the closed subsets of a space: A set is closed exactly when it contains the limits of all its convergent sequences.

An example of a sequential space which is not Frechet-Urysohn is as follows (this is essentially taken from Engelking's text with some added details):

For $i \geq 1$ define $X_i = \left\{ \frac 1i \right\} \cup \left\{ \frac 1i + \frac 1{i^2 + k} : k \geq 0 \right\}$, and let $X = \{ 0 \} \cup \bigcup_{i=1}^\infty X_i$. (Note that $X_i \cap X_j = \emptyset$ for $i \neq j$.) We topologise $X$ as follows:

• all points of the form $\frac 1i + \frac 1{i^2+k}$ are isolated;
• the basic open neighbourhoods of $\frac 1i$ are of the cofinite subsets of $X_i$ containing $\frac 1i$; and
• the basic open neighbourhoods of $0$ are of the form $\{ 0 \} \cup \bigcup_{i=1}^\infty Y_i$ where $Y_i \subseteq X_i$ for each $i$, and $Y_i \neq \emptyset$ for all but finitely many $i$, and if $Y_i \neq \emptyset$, then $\frac 1i \in Y_i$ and $Y_i$ is a cofinite subset of $X_i$.

It is easy to see that $0 \in \overline{ X \setminus \left\{ 0 , \frac 11 , \frac 12 , \frac 13 , \ldots \right\} }$, but no sequence in this set converges to $0$: If $\{ x_j \}_{j=1}^\infty$ is any sequence in this set, note that if $X_i \cap \{ x_j : j \geq 1 \}$ is infinite for only finitely many $i$, then we can easily form a neighbourhood of $0$ containing no points of this sequence. If $X_i \cap \{ x_j : j \geq 1 \}$ is infinite for infinitely many $i$, enumerate them as $\{ i_k : k \geq 1 \}$. Inductively pick a sequence $\{ j_{k} \}_{k=1}^\infty$ so that $j_{k+1}$ is the least $j > j_k$ such that $x_j \in X_{i_k}$. Note that $X \setminus \{ x_{j_{k}} : k \geq 1 \}$ is a neighbourhood of $0$ which does not include a tail of the sequence.

Nevertheless, $X$ is sequential. Suppose that $A \subseteq X$ contains the limits of all convergent sequences of its points. If $x \in \overline{A}$, note that if $x \neq 0$ then $x$ has a countable neighbourhood base it follows that there is a sequence in $A$ converging to $x$, meaning that $x \in A$. If $x = 0$, then assume that $0 \notin A$. Note that there must be a subsequence $\{ x_j \}_{j=1}^\infty$ of $\{ \frac 1 i \}_{i=1}^\infty$ such that every neighbourhood of each $x_j$ intersects $A$ (otherwise we could form a neighbourhood of $0$ which is disjoint from $A$). Then each $x_j \in A$ (since $x_j \in \overline{A}$, and we have observed above that for these points we can build a sequence in $A$ converging to $x_j$), and it follows that $\lim_j x_j = 0$ (every neighbourhood of $0$ contains all but finitely many points of the form $\frac 1i$).

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