Let us recall that a family $\mathcal N$ of subsets of a topological space $X$ is called a *$cs$-network* if for any sequence $\{x_n\}_{n\in\omega}\subset X$ that converges to a point $x\in X$ and any neighborhood $U\subset X$ of $x$ there exists a set $N\in\mathcal N$ such that $x\in N\subset U$ and $x_n\in N$ for all but finitely many numbers $n\in\omega$.

The following theorem answers the problem posed by Sam Nolen. I thank Martin Sleziak for his comment containing the reference to the paper of Franklin and Rajagopalan.

**Theorem.** *The smallest class of topological spaces that contains $\mathbb R$ and is closed under taking subspaces and quotient spaces coincides with the class of subspaces of sequential spaces that have cardinality $\le \mathfrak c$ and possess a countable $cs$-network.*

*Proof.* We shall prove that $\mathcal R=S\mathcal K=SQSQS\mathbb R$, where

$\bullet$ $\mathcal R$ is the smallest class of topological spaces, which contains the real line and is closed under taking subspaces and quotient spaces;

$\bullet$ $\mathcal K$ is the class of sequential spaces of cardinality $\le\mathfrak c$ that possess a countable $cs$-network;

$\bullet$ $S\mathcal K$ is the class of subspaces of spaces in the class $\mathcal K$.

$\bullet$ $S\mathbb R$ is the family of subspaces of the real line;

$\bullet$ $QS\mathbb R$ is the class of quotient spaces of spaces in the class $S\mathbb R$;

$\bullet$ $SQS\mathbb R$ is the class of subspaces of spaces in the class $QS\mathbb R$;

$\bullet$ $QSQS\mathbb R$ is the class of quotient spaces of spaces in the class $SQS\mathbb R$;

$\bullet$ $SQSQS\mathbb R$ is the class of subspaces of spaces in the class $QSQS\mathbb R$;

The proof of the equality $\mathcal R=S\mathcal K=SQSQS\mathbb R$ is divided into six lemmas.

The first lemma was proved in the answer of Will Brian.

**Lemma 1.** The class $QS\mathbb R\subset\mathcal R$ contains all separable metrizable spaces.

**Lemma 2.** For any zero-dimensional Polish space $Z$ and any anti-discrete space $A$ of cardinality continuum the product $Z\times A$ belongs to the class $QS\mathbb R\subset\mathcal R$.

*Proof.* The anti-discrete space $A$ can be identified with the topological group $\mathbb R/\mathbb Q$ endowed with the quotient topology (which is anti-discrete). Then $Z\times A$ belongs to $QS\mathbb R$, being a quotient space of the separable metrizable space $Z\times \mathbb R$, which belongs to $QS\mathbb R$ by Lemma 1.

**Lemma 3.** $\mathcal K\subset QSQS\mathbb R\subset \mathcal R$.

*Proof.* Let $X$ be a sequential space of cardinality $\le\mathfrak c$ that has a countable $cs$-network $\mathcal N$.

Without loss of generality, we can assume that $\mathcal N$ is closed under finite unions and intersections. Endow the set $\mathcal N$ with the discrete topology and consider the zero-dimensional Polish space $\mathcal N^\omega$.
Let $X_a$ be the set $X$ endowed with the anti-discrete topology $\{\emptyset, X\}$.

In the space $\mathcal N^\omega\times\ X_a$ consider the subspace
$Z$ consisting of pairs $((N_k)_{k\in\omega},x)\in\mathcal N^\omega\times X_a$ such that for any open neighborhood $U\subset X$ of $x$ there exists $n\in\omega$ such that $x\in N_{k+1}\subset N_k\subset U$ for all $k\ge n$.
By Lemma 2, the space $Z$ belongs to the class $SQS\mathbb R$.

Let $q:Z\to X$, $q:((N_k)_{k\in\omega},x)\mapsto x$, be the projection on the second factor. It is easy to see that the map $q$ is surjective and continuous. To see that $q$ is quotient, we need to check that a set $F\subset X$ is closed if its preimage $q^{-1}(F)$ is closed in $Z$. Assuming that $F$ is not closed in the sequential space $X$, we can find a sequence $\{x_n\}_{n\in\omega}\subset F$ that converges to some point $x\in X\setminus F$. Let $\mathcal N'$ be the family of all sets $N\in\mathcal N$ such that $x\in N$ and $x_n\in X$ for all but finitely many numbers $n$. Let $(N_i')_{i\in\omega}$ be an enumeration of the countable set $\mathcal N'$. For every $k\in\omega$ let $N_k=\bigcap_{i\le k}N_i'$.

Taking into account that the $cs$-network $\mathcal N$ is closed under finite unions, we can show that $z:=((N_k)_{k\in\omega},x)\in Z$. It is clear that $z\notin q^{-1}(F)$. It can be shown that the pair $z$ belongs to the closure of $q^{-1}(F)$, which is not possible as $q^{-1}(F)$ is closed in $Z$.
This contradiction shows that $X$ is a quotient space of $Z$ and hence $X\in QSQS\mathbb R$.

Lemma 3 implies

**Lemma 4.** $S\mathcal K\subset SQSQS\mathbb R\subset \mathcal R$.

**Lemma 5.** The class $S\mathcal K$ is closed under quotient spaces.

*Proof.* Let $S$ be a subspace of a space $K\in\mathcal K$ and $q:S\to X$ be a quotient map of $S$ onto some topological space $X$. The map $q$ determines the equivalence relation $$E=\{(x,y)\in K\times K:x=y\}\cup\{(x,y)\in X\times X:q(x)=q(y)\}.$$ In Proposition 3.2 of this paper, Franklin and Rajagopalan prove that $X$ can be identified with a subspace of the quotient space $K/E$.
The following lemma ensures that the quotient space $K/E$ belongs to the class $\mathcal K$.

**Lemma 6.** Let $f:X\to Y$ be a quotient map. If $X$ is a sequential space with countable $cs$-network, then so is the space $Y$.

*Proof.* The sequentiality of the quotient space $Y$ is a well-known fact (that can be found in Engelking, I hope). So, it remains to prove that the space $Y$ has a countable $cs$-network.

Let $\mathcal N$ be a countable $cs$-network for the space $X$. Without loss of generality, we can assume that the family $\mathcal N$ is closed under finite unions.

Consider the family $\mathcal M:=\{q(N):N\in\mathcal N\}$ and for every $M\in\mathcal M$ let $\ddot M$ be the intersection of all open sets in $Y$ than contain $M$. We claim that the countable family $\ddot{\mathcal M}=\{\ddot M:M\in\mathcal M\}$ is a $cs$-network for the space $Y$. Since the family $\mathcal N$ is closed under finite unions, so are the families $\mathcal M$ and $\ddot{\mathcal M}$.

Fix a sequence $\{y_n\}_{n\in\omega}\subset Y$, convergent to a point $y\in Y$ and let $U\subset Y$ be a neighborhood of $y$ in $Y$. Let $\mathcal M'=\{M\in\ddot{\mathcal M}: M\subset U\}$. We claim that the family $\mathcal M'$ contains a set $M$ with $y\in M$. To find such set $M$, take any point $x\in q^{-1}(y)$ and find a set $N\in\mathcal N$ such that $x\in N\subset q^{-1}(U)$. Then $M=q(N)$ contains $y$ and the set $\ddot M$ contains $y$ and belongs to the family $\mathcal M'$.

So, we can choose an enumeration $\{\ddot M_k\}_{k\in\omega}$ of the countable family $\mathcal M'$ such that $y\in \ddot M_0$. We claim that for some $k\in\omega$ the set $\bigcup_{i\le k}\ddot M_i$ contains all but finitely many points $y_n$. Assuming that this is not true, we can construct an increasing number sequence $(n_k)_{k\in\omega}$ such that $y_{n_k}\notin \bigcup_{i\le k}\ddot M_i$. Taking into account that the set $\ddot M_0$ contains $y$, but $\ddot M_0$ does not contain the points $y_{n_k}$, we conclude that $y$ does not belong to the closure $\overline{\{y_{n_k}\}}$ of the singleton $\{y_{n_k}\}$ for all $k\in\omega$. Consequently, the set $B=(X\setminus U)\cup\bigcup_{k\in\omega}\overline{\{y_{n_k}\}}$ does not contain its accumulation point $y$ and hence is not closed in $Y$.

Since the map $q$ is quotient, the preimage $q^{-1}(B)$ is not closed in $X$. By the sequentiality of $X$, there exists a sequence $(x_m)_{m\in\omega}\in q_E^{-1}(B)$ that converges to some point $x\notin q^{-1}(B)$. It follows from $X\setminus q^{-1}(U)=q^{-1}(Y\setminus U)\subset q^{-1}(B)$ that $x\in q^{-1}(U)$.
By the definition of a $cs$-network, there exists a set $N\in\mathcal N$ such that $x\in N\subset q^{-1}(U)$ and $N$ contains all but finitely many points $x_m$. Find $k\in\omega$ such that $q(N)=M_k$. Since the closed set $F_k:=\bigcup_{i\le k}q^{-1}(\overline{\{y_{n_i}\}})$ does not contain $x$, it does not contain the points $x_m$ for all sufficiently large numbers $m$. So, we can find $m\in\omega$ so large that $x_m\notin F_k$. Then $x_m\in q^{-1}(\overline{\{y_{n_i}\}})$ for some $i>k$ and hence $q(x_m)\in \overline{\{y_{n_i}\}}$. Now observe that $q(x_m)\in q(N)=M_k$ and hence $y_{n_i}\in\ddot M_k\subset \bigcup_{j\le i}\ddot M_j$, which contradicts the choice of $y_{n_i}$.

`$[0,1]^{\aleph_0}$`

. $\endgroup$ – Stephen S Mar 8 '11 at 21:17