# What spaces can be obtained from $\mathbb{R}^{n}$ by taking quotient spaces and subspaces?

Is there a good characterization of the smallest collection of topological spaces which contains $\mathbb{R}^{n}$ for each $n$, and is closed under taking subspaces and quotient spaces?

A bit of motivation: A friend of mine asked me to give an argument why the definition of a topological space is "right" or "natural", considered perhaps as a generalization of manifolds or cell complexes. While trying to answer him, I briefly wondered whether the collection of topological spaces is the closure of $\{ \mathbb{R}^{n} \}_{n \geq 0}$ under certain operations, say taking subspaces and quotient spaces. I quickly realized that this is false in general, though (there are counterexamples which have very large cardinality or don't satisfy first or second countability).

• Your motivation question seems much more interesting! Related questions have been discussed on MO before, e.g. mathoverflow.net/questions/19152/… . The answers that most convinced me were the ones involving logic and computability. They suggest that the definition of a topology is useful because it is absurdly general, hence general enough to include nice things. But it is not necessarily geometrically natural. I think Grothendieck once expressed an opinion that the definition is "wrong" e.g. for homotopy theory? – Qiaochu Yuan Mar 8 '11 at 10:40
• If you used quotients and topological sums, you would get sequential spaces. Using subspaces, quotiens and sums, you would get subsequential spaces. (S. P. Franklin, M. Rajagopalan: On subsequential spaces, Topology. and its Applications 35 (1990), 1–19) Your class will definiely be a subclass of the class of subsequential spaces. I am not sure about the precise characterization. – Martin Sleziak Mar 8 '11 at 11:25
• This way you can get only separable spaces with finite dimension and I guess you can get all of them (?) – Anton Petrunin Mar 8 '11 at 15:48
• @Anton Petrunin: Finite dimension can't be right, since every compact metrizable space is a quotient of the Cantor set, and that includes things like $[0,1]^{\aleph_0}$. – Stephen S Mar 8 '11 at 21:17
• Concerning your friends question: Take a look at Bill Lawvere's answer in mathoverflow.net/questions/127841/… "My paper about Volterra's functionals [...] discusses the unsuitability for functional analysis (as well as for homotopy theory) of the attempt to characterize continuity or cohesion using open sets or other contravariant structure." – Gerrit Begher Feb 4 '14 at 17:38

These are exactly the spaces which are countable unions of dyadic spaces.

By definition, a dyadic space is the continuous quotient of the Cantor set. (This class includes all compact, metrizable spaces.)

Let $C\subset [0,1]$ denote the Cantor set, and for $k\in\mathbb{Z}$ let $C+k\subset[k,k+1]$ denote the shifted Cantor set as a subset of $\mathbb{R}$. If $X$ is the union of dyadic spaces $X_1$, $X_2$, ..., then $X$ is a continuous image of the space $\bigcup_k (C+2k)$, which is a closed subset of $\mathbb{R}$.

Conversely, assume $X$ is a quotient of a closed subset $Y\subset\mathbb{R}^n$. Then $Y$ is a countable union of compact subsets $Y_1$, $Y_2$, ... of $\mathbb{R}^n$. Since each $Y_k$ is compact and metrizable, its image in $X$ is dyadic.

This also shows that one obtains the same class by considering just quotients of closed subsets of $\mathbb{R}$.

• I don't think this is right. A "dyadic space" typically means a space that is (up to homeomorphism) dense in a subspace of a continuous image of $2^\kappa$ for any cardinal $\kappa$ (not just for $\kappa = \omega$, where we get the Cantor space). . . . But even if we redefine "dyadic" to mean what you say it means, your argument doesn't quite work, because you're only considering the operation of taking closed subsets. The OP asked about taking arbitrary subsets. (link: en.wikipedia.org/wiki/Dyadic_space) – Will Brian Jan 30 '18 at 17:06
• @WillBrian: OK, I guess that is a matter of definition. The wikipedia article you linked also sais: "However, many authors use the term dyadic space with the same meaning as dyadic compactum." I don't really have a strong opinion about that. When I wrote my answer I must have thought that "dyadic space" means quotient of the Cantor set. But you are right that I only address closed subspaces. – Hannes Thiel Feb 1 '18 at 20:16

The class of spaces you're describing contains all second countable spaces of cardinality $\leq\!\mathfrak{c}$.

For $T_3$ spaces, "second countable spaces and of cardinality $\leq\!\mathfrak{c}$" is the same as separable and metrizable, so in particular every separable metrizable space can be obtained in this way. Taras Banakh points out in the comments that there are other spaces that can be obtained in this way too, for example the Frechet-Urysohn fan.

Let me prove first the assertion about $T_3$ spaces. Every second countable $T_3$ space is metrizable (this is Urysoh's metrization theorem, the theorem for which Urysohn's famous lemma was proved). Every second countable metrizable space is homeomorphic to a subspace of $[0,1]^\omega$. And $[0,1]^\omega$ is a continuous image of the Cantor space, which is a subspace of $\mathbb R$. (Proof that $[0,1]^\omega$ is a continuous image of the Cantor space: everyone knows that $[0,1]$ is a continuous image of $2^\omega$, but then $[0,1]^\omega$ is a continuous image of $(2^\omega)^\omega \approx 2^\omega$.)

Let's consider the $T_0$ case next, and show that every second countable $T_0$ space can be obtained in the way you describe. To see this, let's begin with a fascinating basic fact about $T_0$ spaces:

Fact: (possibly due to Sierpinski?) Let $S$ denote the topological space on two points, $0$ and $1$, where the open sets are precisely $\emptyset$, $\{1\}$, and $\{0,1\}$. (This is called the Sierpinski space.) Then every $T_0$ topological space is a subspace of $S^I$ for some sufficiently large set $I$.

The standard proof of this is as follows. Let $X$ be an arbitrary space, and let $I$ be the collection of all open subsets of $X$. Define a map from $e$ from $X$ into $S^I$ as follows. For every $U \in I$, let the value of $e(x)$ in the "$U$-coordinate" be equal to $0$ if $x \notin U$ and $1$ if $x \in U$. It is not too hard to check that $e$ is an embedding $X \rightarrow S^I$.

But notice that we can modify this standard proof by taking $I$ to be some basis for $X$, rather than the collection of all open sets. The proof still goes through, but it now shows:

Fact: Every second countable $T_0$ space is a subspace of $S^\omega$.

This fact, together with the following observation, finishes the proof for the $T_0$ case:

Observation: $S^\omega$ is a quotient of a quotient of a subspace of $\mathbb R$.

Proof: The Cantor space is a subspace of $\mathbb R$, and we already mentioned that $[0,1]^\omega$ is a quotient of the Cantor space. To prove the observation, then, it suffices to show that $S^\omega$ is a quotient of $[0,1]^\omega$. To prove this, it suffices to show that $S$ is a quotient of $[0,1]$. But this is obvious: take the quotient mapping that collapses $(0,1]$ to a point, and does nothing else. QED

Now let's consider the general case. Note that every topological space $X$ that is not $T_0$ has a quotient that is $T_0$, known as its Kolmogorov quotient. This quotient looks exactly like $X$, except that topologically indistinguishable sets of points have been collapsed.

Let $X$ be a second countable space of cardinality $\leq\!\mathfrak{c}$. By what we just saw in the $T_0$ case, the Kolmogorov quotient of $X$, let's called it $\tilde X$, is a subspace of $S^\omega$. But this means that $X$ is a subspace of $D \times S^\omega$, where $D$ is the indiscrete space of size $\mathfrak{c}$ (just embed $\tilde X$ in the second part, and then use $D$ to "undo" the quotient mapping $X \rightarrow \tilde X$).

Thus, to finish the argument, it suffices to show that

Observation: $D \times S^\omega$ is a quotient of a quotient of a subspace of $\mathbb R$.

As before, we know that $[0,1]^\omega \approx [0,1] \times [0,1]^\omega$ is a quotient of a subspace of $\mathbb R$. We already showed that $S^\omega$ is a quotient of $[0,1]^\omega$, so this shows that $[0,1] \times S^\omega$ is a quotient of $[0,1] \times [0,1]^\omega$. To finish the proof it suffices to show that $D$ is a quotient of $[0,1]$. For this, just take the quotient of $[0,1]$ using Vitali's famous equivalence relation, $x \sim y$ iff $x-y \in \mathbb Q$. This gives you $\mathfrak{c}$ equivalence classes, and they are topologically indistinguishable.

• Your first sentense is false: quotients of subsets (even closed) od $\mathbb R$ need not be second countable: just consider the Frechet-Urysohn fan, which is a quotient space of the convergence sequence times countable discrete space. Finite-dimensional countable CW-complexes need not be non-metrizable but are qoutient spaces of the topological sums of simplexes (which are subspaces of suitable $\mathbb R^n$) – Taras Banakh Feb 2 '18 at 7:02
• @TarasBanakh: You are right (and that is a mistake I shouldn't be making). I guess this means I've showed inclusion in the class of spaces asked about in the OP, but not equality. I will edit my post accordingly. Thanks for pointing this out. – Will Brian Feb 2 '18 at 13:40
• In general, products and quotients don't commute, so it's not immediately obvious that the product topology on $S^\omega$ will match the quotient topology coming from $[0, 1]^\omega$. – Todd Trimble Feb 2 '18 at 15:12
• @ToddTrimble: OK, I'm being a bit hasty there. After thinking about it for a few minutes, it seems to me that the two topologies do match in this case. I will think about it some more to be sure, and edit again later. Thanks for pointing this out. – Will Brian Feb 2 '18 at 15:41

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 $y_{n_i}$.