Reflection of the chain-condition: clarifying a suspected typo In  the 2002 paper Reflecting Lindelöfness in Topology and its Applications, J. Baumgartner and F. Tall state that "by an easy  Löwenheim–Skolem argument," an uncountable (say $T_2$ or $T_3$) ccc topological space has a ccc subspace of cardinality $\aleph_1$. 
My question is: Is this (reflection statement) actually true?
(Of course, I can see for example that indeed, "by an easy (ramification) argument," this is true for the linearly orderable topological spaces). 
 A: The reflection statement is dramatically false at least for Hausdorff spaces.
Theorem 1. For any infinite cardinal $\kappa$ there is a Hausdorff ccc space $X$ with no uncountable ccc subspace of cardinality less than or equal than $\kappa$. 
Proof. Let $X=2^\kappa$ with the topology $\tau$ generated by sets of the form $P \cap C$ where $P$ is basic clopen on the product topology and $C\subseteq X$ with $|X \setminus C|<2^\kappa$. Note that the generating set is closed under finite intersections so it is in fact a base for $\tau$. 
Clearly $\tau$ contains the product topology on $X$ so $(X,\tau)$ is a Hausdorff space. Moreover since any clopen in the product topology has size $2^\kappa$ we have that for any two basic open sets $P_1 \cap C_1$ and $P_2 \cap C_2$, $$(P_1 \cap C_1) \cap (P_2 \cap C_2) \neq \emptyset \iff P_1 \cap P_2 \neq \emptyset.$$
It follows that $(X,\tau)$ is ccc (since the product topology is ccc).
Finally if $Y$ is a subspace of $X$ with $|Y|<2^\kappa$ (in particular if $|Y|\leq \kappa$) then $Y$ is discrete. To see this just note that given $y \in Y$ we can take $C=(X \setminus Y) \cup \{y\}$ and since $|X \setminus C|=|Y \setminus \{y\}|<2^\kappa$, we have that $\{y\}=Y \cap (X \cap C)$ is open in $Y$. Hence $Y$ is not ccc unless $Y$ is countable. $\square$
However there are some positive results.
Theorem 2. If $(X,\tau)$ is an uncountable ccc topological space and $\chi(X) \leq \aleph_1$ then $X$ contains a ccc subspace of size $\aleph_1$.
Proof. Fix $M$ an elementary submodel of $H_\theta$ (with $\theta$ large enough) such that $\omega_1 \subseteq M$, $(X,\tau) \in M$ and $|M|=\aleph_1$. By elementarity the space $(X \cap M, \tau \cap M)$ is ccc and $|X \cap M|=\aleph_1$ (because $\omega_1 \subset M$). But since $\chi(X) \leq \aleph_1$, again by elementarity and the fact that $\omega_1 \subset M$, we have that $\tau \cap M$ generates the subspace topology in $X \cap M$. $\square$
Theorem 3. If $(X,\tau)$ is an uncountable ccc compact Hausdorff topological space then $X$ contains a ccc subspace of size $\aleph_1$.
Proof. Fix $M$ an elementary submodel of $H_\theta$ (with $\theta$ large enough) such that $\omega_1 \subseteq M$, $(X,\tau) \in M$ and $|M|=\aleph_1$. Define a relation on $X$ by $x\sim y$ if and only if $f(x)=f(y)$ for all $f \in C(X)\cap M$. It is fairly well known that this is an equivalence relation for which the quotient space $X/M$ is Hausdorff (in fact $X/M$ is Tychonoff whenever $X$ is Tychonoff, even if $X$ is not compact), the quotient map $\rho:X \to X/M$ is inyective when restricted to $X \cap M$ (so $|X/M| \geq \aleph_1$) and $\rho(X \cap M)$ is dense in $X/M$ (so $d(X/M) \leq \aleph_1$).
Fix a closed $Z \subseteq X$ such that the restriction $\rho:Z \to X/M$ is suryective and irreducible. Since $X/M$ is ccc being a continuous image of the ccc space $X$, we get by irreducibility that $Z$ is also ccc. Since $d(X/M) \leq \aleph_1$ we can find $Y \subseteq X$ with $|Y|=\aleph_1$ such that $\rho(Y)$ is dense in $X/M$. By irreducibility $Y$ is dense in $Z$ and therefore $Y$ is the ccc space we were looking for. $\square$
A combination of the ideas in the proofs of Theorems 2 and 3 can be used to prove that the reflection statement is true for $T_3$ spaces of pointwise countable type ($h(X) \leq \aleph_1$ is enough). In particular it is true for locally compact Hausdorff spaces.
I don´t know what happens for $T_3$ spaces in general (note that the space described in the proof of Theorem 1 is not regular).
A: This is meant to flesh out a claim made in a previous comment and clear up a point of confusion about what the topology on $X \cap M$ is normally assumed to be when trying to reflect properties of $X$ using elementary submodels.
Proposition:  Assume $(X,\tau)$ is Tychonoff. If $M\prec H(\theta)$ is an elementary sub-model with $X,\tau \in M$. Then, $(X_M, \tau_M)$ where $X_M = X \cap M$ and $\tau_M$ is the topology generated by $\{ U \cap X_M: U \in \tau \cap M \}$, is homeomorphic to a dense subset of a continuous image of $X$. 
In particular, if $(X,\tau)$ is c.c.c., then so is $(X_M, \tau_M)$
Proof: Without loss of generality assume $X \subset [0,1]^{\lambda}$ ($\lambda$ a cardinal) with the topology on $X$ taken to be the topology induced by $[0,1]^{\lambda}$. 
Let $\mathcal{B} = \cup \{ \{ [0, q), (q,p), (p,1] \}: q < p \text{ and } p,q\in (0,1)\cap \mathbb{Q}\}$ and $\mathbb{P} = \mathsf{Fn}(\lambda, \mathcal{B})$ (the set of finite partial functions $p\subset \lambda\times \mathcal{B}$); moreover for each $p \in \mathbb{P}$ define $$[p] = \{ a \in [0,1]^{\lambda}: (\forall \gamma \in dom(p))(a(\gamma) \in p(\gamma))\}.$$  Noting that $\{ [p]:p \in \mathbb{P} \}$ is a base for the product topology on $[0,1]^{\lambda}$, we readily have that $\mathcal{B}_X=\{ [p] \cap X: p \in \mathbb{P} \}$ is a base for the relative topology on $X$.
Claim: For any elementary submodel $M\prec H(\theta)$:  


*

*$\mathcal{B} \in M$, $\mathcal{B} \subset M$, and $M \cap \mathbb{P} = \mathsf{Fn}(\lambda \cap M, \mathcal{B})$.

*Assuming $X \in M$: the set $Y_0 = \{ x_0 \cap M: x_0 \in X \cap M \} \subset [0,1]^{\lambda \cap M}$ is a dense subset of the space $Y=\pi_M[X]$ (the image of $X$ under the projection $\pi_M:[0,1]^{\lambda} \rightarrow [0,1]^{\lambda \cap M}$.)
Proof of Claim: For the first portion, simply note that $\mathcal{B}$ is a definable and countable family of subsets of $[0,1]$; in addition to see that $p \in \mathbb{P} \cap M \iff p \in \mathsf{Fn}(\lambda \cap M, \mathcal{B})$. First note that the set $dom(p)$ is definable from $p$ by a $\Delta_0$ formula. As such, it follows that $p \in M $ implies $dom(p) \in M$ which, since $p$ is finite, also entails $dom(p) \subset M$. For the second portion, it is enough to observe that by elementarity, for every $p \in \mathbb{P} \cap M$, if $[p] \cap X \neq\emptyset$ then, there is some $x \in [p] \cap X \cap M$ serving as a witness. This is enough to establish the second portion (and the proposition) since for each relevant $p$, we have $dom(p) \subset M \cap \lambda$ and the projection $\pi_M$ does not effect indices laying within $M$. 
Edit:
Theorem: Assume $(X,\tau)$ is Tychonoff and c.c.c. and $M \prec H(\theta)$ with $X \in M$, then  $X_M = X \cap M$ (as a subspace of $X$) is c.c.c.
Proof: As before assume $X \subset [0,1]^\lambda$ and $M\prec H(\theta)$ are such that $X\in M$. Next, suppose $X$ has the c.c.c., then


*

*$Y = cl(X)\in M$ is compact and c.c.c.,

*The compact space $Z = cl(Y_M)$ contains $X_M$, and

*The restriction of the projection $\pi_M(x)=x\vert_M$ to $Z$ is irreducible and maps $Z$ onto $\pi_M[ Y ]\subset [0,1]^{\lambda \cap M}$ (this follows from the proof of the previous claim, combined with the observation that for any proper closed  $Z_0\subset Z$, we have $\emptyset \neq Y_M \cap (Z\backslash Z_0) \subset M$)


With that setup, for any $p \in \mathbb{P}$, define $U_p = \pi_M[Z] \backslash \pi_M[Z \backslash [p]] \subset [0,1]^{\lambda \cap M}$. Then each $U_p \subset \pi_M[Z] \subset [0,1]^{\lambda \cap M}$ is (relatively) open. Moreover, for any $p, q \in \mathbb{P}$, $$X_M \cap [p] \cap [q] = \emptyset \implies U_p \cap U_q \cap \pi_M[X_M] = \emptyset. $$
It follows that if $X_M$ is not a c.c.c. subspace of $X$, then $\pi_M[X_M]$ is not a c.c.c. subspace of $\pi_M[Z]$; however, noting that $\pi_M[X_M]$ is a dense subset of $\pi_M[X]$, which is in turn a dense subset of $\pi_M[Z] = \pi_M[Y]$, we have that $\pi_M[Y]$ is not c.c.c. (a contradiction.)
Remark: I guess I should have been more clear in the Proposition, the space $(X_M, \tau_M)$ is homeomorphic to $\pi_M[X_M] \subset \pi_M[X]$, which is a dense subset of $\pi_M[X]$. Moreover, since the original Proposition applies to all Tychonoff spaces, it applies to any compact space within which $X$ embeds as a dense subset.
