All spaces are assumed to be Tychonoff. Given a realcompact space X, we denote by $\mathsf {Exp}(X)$ the smallest infinite cardinal $\kappa $ such that X is homeomorphic to a closed subspace of $\mathbb {R}^\kappa $. Our main result shows that, given a cardinal $\kappa $, the following conditions are equivalent:

• • There exists a countable crowded space X such that $\mathsf {Exp}(X)=\kappa $.

• • $\mathfrak {p}\leq \kappa \leq \mathfrak {c}$.

$\mathfrak {d}\leq \kappa \leq \mathfrak {c}$

$2^\kappa $

$\kappa $

$\kappa $

$\mathsf {Exp}(X)$

In a recent NJL article (Enger 2013), Hans-Olav Enger argues against some analyses of gender and ‘pancake sentences’, in particular against Josefsson (2009). In this short contribution, I will discuss what I take to be misunderstandings in Enger (2013). In addition I will discuss some data not included in Enger's (2013) analysis, which I will show to be crucial for the comparison between the different analyses proposed.

The relative rank $\rank(S:A)$ of a subset $A$ of a semigroup $S$ is the minimum cardinality of a set $B$ such that $\langle A\cup B\rangle=S$. It follows from a result of Sierpiński that, if $X$ is infinite, the relative rank of a subset of the full transformation semigroup $\mathcal{T}_{X}$ is either uncountable or at most $2$. A similar result holds for the semigroup $\mathcal{B}_{X}$ of binary relations on $X$.

A subset $S$ of $\mathcal{T}_{\mathbb{N}}$ is dominated (by $U$) if there exists a countable subset $U$ of $\mathcal{T}_{\mathbb{N}}$ with the property that for each $\sigma$ in $S$ there exists $\mu$ in $U$ such that $i\sigma\le i\mu$ for all $i$ in $\mathbb{N}$. It is shown that every dominated subset of $\mathcal{T}_{\mathbb{N}}$ is of uncountable relative rank. As a consequence, the monoid of all contractions in $\mathcal{T}_{\mathbb{N}}$ (mappings $\alpha$ with the property that $|i\alpha-j\alpha|\le|i-j|$ for all $i$ and $j$) is of uncountable relative rank.

It is shown (among other results) that $\rank(\mathcal{B}_{X}:\mathcal{T}_{X})=1$ and that $\rank(\mathcal{B}_{X}:\mathcal{I}_{X})=1$ (where $\mathcal{I}_{X}$ is the symmetric inverse semigroup on $X$). By contrast, if $\mathcal{S}_{X}$ is the symmetric group, $\rank(\mathcal{B}_{X}:\mathcal{S}_{X})=2$.

AMS 2000 Mathematics subject classification: Primary 20M20