There is a family $K_X:X\subseteq\mathfrak{c}\}$ of separable compact
zero-dimensional spaces such that there is a continuous surjection of $K_X$ 
onto $K_Y$ if and only if $X\subseteq Y$.

These spaces are continuous images of $\omega^*$, the Stone space of 
$\mathcal{P}(\omega)/\mathit{fin}$.
So Stone duality then yields a family 
$\{B_X:X\subseteq\mathfrak{c}\}$ of subalgebras of
$\mathcal{P}(\omega)/\mathit{fin}$ such that $B_Y$ embeds into $B_X$ if and 
only if $X\subseteq Y$. 

We can make such spaces using variations on Alexandroff's double-arrow space,
which is the set 
$D=([0,1]\times\{0,1\})\setminus\{\langle0,0\rangle,\langle0,1\rangle\}$ 
ordered lexicographically and endowed with the order topology;
The resulting space will be denoted $\mathbb{A}$. 
(We drop the points $\langle0,0\rangle$ and $\langle0,1\rangle$ because they
would be (the only) isolated points of $\mathbb{A}$.)

Pictorially we have taken the unit interval $[0,1]$ and split all points
$x$ of the open interval $(0,1)$ into two copies:
The space $\mathbb{A}$ is compact and separable, hence a continuous image
of $\omega^*$.

The variations will be obtained by specifying a subset $X$ of $(0,1)$ and taking
$\mathbb{A}_X=\{\langle x,i\rangle\in D: x\in X \to i=0\}$; that is, 
by splitting the points of $(0,1)\setminus X$ only.
Thus we can write $\mathbb{A}=\mathbb{A}_\emptyset$, 
and $[0,1]=\mathbb{A}_{(0,1)}$.

One direction in the if and only if will be immediate by the following
observation: if $X\subseteq Y$ then there is a natural continuous map $h_{X,Y}$ from
$\mathbb{A}_X$ onto $\mathbb{A}_Y$; it sends
the points $\langle x,i\rangle$ to $\langle x,i\rangle$ if $x\notin Y$;
the points $\langle x,i\rangle$ to $\langle x,0\rangle$ if $x\in Y\setminus X$;
and $\langle x,0\rangle$ to $\langle x,0\rangle$ if $x\in X$.

We shall construct a family $\{A_\alpha:\alpha\in\mathfrak{c}\}$ of subsets
of $(0,1)$ (all disjoint from $\mathbb{Q}$)
and put 
$S_X=\mathbb{Q}\cup\bigcup_{\alpha\in X}A_\alpha$ for $X\subseteq\mathfrak{c}$.
Then $K_X$ will be $\mathbb{A}_{S_X}$.
It is readily seen that $h_{S_X,S_Z}=h_{S_Y,S_Z}\circ h_{S_X,S_Y}$ 
whenever $X\subseteq Y\subseteq Z$, so the maps dual to the surjections
$h_{S_\emptyset,S_X}$ yields an embedding of $\mathcal{P}(\mathfrak{c})$
into the family of subalgebras of the clopen algebra of $K_\emptyset$
ordered by the subset relation and so also into the family of
subalgebras of $\mathcal{P}(\omega)/\mathit{fin}$.


To this end we let $\mathcal{F}$ be the set of all maps $f$ that satisfy:
$\operatorname{dom}f$ is a co-countable subset of $[0,1]$ and
$f:\operatorname{dom}f\to[0,1]$ is continuous.
For every $f\in\mathcal{F}$ we let 
$S(f)=\{x\in\operatorname{dom}f:f(x)\neq x\}$ and 
$E(f)=\operatorname{dom}f\setminus S(f)$.
We choose a subset $C(f)$ of $\operatorname{dom}f$ such that the 
restriction $f:C(f)\to f[S(f)]$ is a bijection.

We apply Theorem 2.0 of [*A method for constructing ordered continua*, 
Topology and its Applications, 21 (1985), 35--49][1]
to this family and obtain a pairwise disjoint family 
$\{V\}\cup\{A_\alpha:\alpha\in\mathfrak{c}\}$ 
of Bernstein sets in $(0,1)$ with the following properties:
all are disjoint from $\mathbb{Q}$ and 
for every $f\in\mathcal{F}$ such that $f[S(f)]$ (and hence $C(f)$) has 
cardinality $\mathfrak{c}$ the intersections
$C(f)\cap A_\alpha$ and $f[C(f)\cap A_\alpha]\cap V$ have 
cardinality $\mathfrak{c}$, for all $\alpha$.

For later use we note that for a given function $f$ in $\mathcal{F}$ the 
sets $E(f)$ and $S(f)$ are closed and open in $\operatorname{dom}f$ respectively and 
hence completely metrizable, and the set $f[Sf)]$ is analytic.
Therefore for each of the three sets we know that either it is countable
or it contains a copy of the Cantor set.
Thus, to show that one is countable it suffices to show that its intersection
with some Bernstein set is countable.

The following claim establishes the other direction in our if and only if.

Let $\alpha\in\mathfrak{c}$ and let $X$ and $Y$ be subsets of $(0,1)$ 
such that $A_\alpha\subseteq X$ and $(A_\alpha\cup V)\cap Y=\emptyset$.
We claim that there is no continuous surjection from $\mathbb{A}_X$
onto $\mathbb{A}_Y$.

To why this claim suffices assume that $U$ and $V$ are subsets 
of $\mathfrak{c}$ and that $U\not\subseteq V$.
Take $\alpha\in U\setminus V$ and note that $A_\alpha\subseteq S_U$
and that $(A_\alpha\cup V)\cap S_V=\emptyset$, so that
there is no continuous surjection from $\mathbb{A}_{S_U}$ 
onto $\mathbb{A}_{S_V}$.

For brevity we write $A=A_\alpha$.
We first observe that the points of $A$ are not split in $\mathbb{A}_X$,
so its subspace topology is the same as its subspace topology in $[0,1]$.

Now let $s:A_X\to A_Y$ be continuous and let $t:A_Y\to[0,1]$ be the natural
map that identifies $\langle y,0\rangle$ and $\langle y,1\rangle$ for all 
$y$ outside $Y$.

The composition $t\circ s$ is continuous and we let $g$ denote its restriction
to $A$.
Since $A$ is dense in $\mathbb{A}_X$ we can recover $t\circ s$ from $g$.
If $x\in X$ then $(t\circ s)(x,0)=\lim_{y\to x}g(y)$, and 
if $x\notin X$ then $(t\circ s)(x,0)=\lim_{y\uparrow x}g(y)$ and
$(t\circ s)(x,0)=\lim_{y\downarrow x}g(y)$, where in each case $y$ runs 
through $A$.

By one half of Lavrentieff's theorem 
(Theorem 4.3.20 in Engelking's *General Topology*)
we can find a $G_\delta$-set $G$ that contains~$A$ and a continuous 
map $f:G\to[0,1]$ that extends $g$.
The complement of $G$ in $[0,1]$ is a countable union of closed sets, 
each of which is countable because they are closed and disjoint from 
the Bernstein set $A$.
It follows that $f$ belongs to the family $\mathcal{F}$.

The arguments given above yield information about the locations of
$s(x,0)$ and $s(x,1)$ whenever $x\in G$.
Indeed, if $x\in G\setminus X$ then $(t\circ s)(x,0)=f(x)=(t\circ s)(x,1)$
and so in all cases $s(x,i)\in\{\langle f(x),0\rangle,\langle f(x),1\rangle\}$.

To summarize: divide $\mathbb{A}_X$ into two sets $P$, the points with
first coordinate in $G$, and $Q$, the points with first coordinate outside $G$.
The set $Q$ is countable and once we show that $f[G]$ is countable we can 
conclude that $s[\mathbb{A}_X]$ is countable and that $s$ is definitely not
surjective.

Our first claim is that $E(f)$ is countable.
We show this by showing that its intersection with the Bernstein set $A$ 
is countable.

Indeed, let $x\in E\cap A$, then $f(x)=x$ and hence 
$s(x)=\langle x,0\rangle$ or $s(x)=\langle x,1\rangle$;
remember: the points of $A$ are split in $\mathbb{A}_Y$.
So $E\cap A$ is the union of two sets
$E_0=\{x\in E\cap A:s(x)=\langle x,0\rangle\}$
and $E_1=\{x\in E\cap A:s(x)=\langle x,1\rangle\}$.

By continuity of $s$ we have for every $x\in E_0$ an interval $(p_x,q_x)$
with rational end points that is mapped by $s$ into the interval 
$[0,\langle x,0\rangle]$ in $\mathbb{A}_Y$.
It follows that if $x<y$ in $E_0$ then $y\in(p_y,q_y)\setminus (p_x,q_x)$.
We see that $x\mapsto(p_x,q_x)$ is injective and $E_0$ is countable. 
Likewise $E_1$ is countable and we conclude that $E(f)\cap A$ is countable
and hence so is $E(f)$ itself.
 
The second claim is that $f[S(f)]$ is countable.
We show that $f[S(f)]$ is countable by showing that 
$f[C(f)\cap A]\cap V$ is countable. 
The contrapositive of the properties of our family of Bernstein sets
then shows that $C(f)$ and hence $f[S(f)]$ is countable.

As above: let $x\in C(f)\cap A$ and assume $f(x)\in V$.
Then $s(x)=\langle f(x),0\rangle$ or $s(x)=\langle f(x),1\rangle$.
As above we split $C(f)\cap A$ into two pieces:
$C_0=\{x:s(x)=\langle f(x),0\rangle\}$ and  
$C_1=\{x:s(x)=\langle f(x),1\rangle\}$.

If $x\in C_0$ then we take a rational interval $(p_x,q_x)$ that gets mapped
into $[0,\langle f(x),0\rangle]$ by $s$.

If $x,y\in C_0$ and $f(x)<f(y)$ then $y\in(p_y,q_y)\setminus (p_x,q_x)$, 
and we conclude that $x\mapsto(p_x,q_x)$ is injective and $C_0$ is countable.
Likewise $C_1$ is countable.

It follows that $f[G]=E(f)\cup f[S(f)]$ is countable.


  [1]: http://dx.doi.org/10.1016/0166-8641(85)90056-2