# Does the expressibility of a set $A$ imply the expressibility of the singleton sets $\{a\}, a\in A$?

A set $A$ is expressible if there exists a first order logic formula with a single free variable $\alpha(x)$ such that $V_{I,v_{x=t}}(\alpha(x))=1 \iff t \in A$, that is, $A$ contains the values I should assign to the single free variable to make $\alpha(x)$ true.

Does the expressibility of a set $A$ imply the expressibility of the singleton sets $\{a\}, a\in A$ ?

E: Let $v:VAR \to U_I$ be a valuation (with $U_I$ being the universe of discourse of the interpretation $I$).

Then $v_{x=t}$ denotes the valuation that sastisfies: $v_{x=t}(x)=t, v_{x=t}(y)=v(y), \forall y\neq x$.

If $\alpha(x)$ is a formula with a single variable $x$, then after assigning some value to $x$, the truth value of $\alpha$ under the interpretation $I$ is independent of the valuation chosen.

• What first-order language are you working in - the usual language of first-order set theory? It is not clear whether your notion of "expressible" is different from the standard notion of "definable": it would help if you explained your notation $V_{I,v_{x=t}}(\alpha(x))=1$. – Rob Arthan Oct 11 '15 at 20:33
• I'm sorry Rob, I'm translating from my spanish notes and these terms seem to use very different words, could you tell me what the usual notion of "definable" is? – YoTengoUnLCD Oct 11 '15 at 20:45
• Let ${\cal M} = (U, \ldots)$ be astructure for a first-order language (with universe $U$), A subset $A \subset U$ is definable iff there is a formula $\phi(x) \in \cal L$, such that for every $a \in U$, $a \in A$ iff ${\cal M} \models \phi(a)$ (where $\phi(a)$ means the result of interpreting $\phi$ with $x$ interpreted as $a$). – Rob Arthan Oct 11 '15 at 20:52
• Yes! That's what I meant to say. Thanks a lot, Rob. – YoTengoUnLCD Oct 11 '15 at 20:54
• So now we need to know what theory or class of models you are working in. E.g., in the theory of a linear order in the language with signature $(U, <)$ the only definable sets are the empty set and $U$ and so no singleton sets are definable if $U$ has more than one element. – Rob Arthan Oct 11 '15 at 20:59

From the comments, you are interested in examples like the first-order theory of the additive group of real numbers. In $(\Bbb{R}, +)$, you can define the set $A = \Bbb{R}$, using the formula $\phi(x) \equiv x= x$, but the only singleton subset of $A$ that you can define is $\{0\}$ (because a definable subset has to be invariant under the automorphism that sends $x$ to $-x$).