# Proof of Upward Lowenheim-Skolem Theorem

Currently I am reading Enderton's A Mathematical Introduction to Logic. In page $154,$ he stated the following:

Lowenheim-Skolem Tarski Theorem Let $\Gamma$ be a set of formulas in a language of cardinality $\lambda,$ and assume that $\Gamma$ satisfies in some infinite structure. Then for every cardinal $\kappa\geq \lambda,$ there is a structure of cardinality $\kappa$ in which $\Gamma$ is satisfiable.

The proof is as follows:

Proof: Let $\mathfrak{A}$ be the infinite structure in which $\Gamma$ is satisfiable. Expand the language by adding a set $C$ of $\kappa$ new constant symbols. Let $$\Sigma=\{c_1\neq c_2|c_1,c_2 \text{ distinct members of }C\}.$$ Then every finite subset of $\Sigma\cup\Gamma$ is satisfiable in the structure $\mathfrak{A},$ expanded to assign distinct objects to the finitely many new constant symbols in the subset. (Since $\mathfrak{A}$ is infinite, there is room to accommodate any finite number of these.) So by Compactness $\Sigma\cup\Gamma$ is satisfiable, and by the Downward Lowenheim-Skolem Theorem it is satisfiable in a structure $\mathfrak{B}$ of cardinality $\leq \kappa.$ (The expanded language has cardinality $\lambda+\kappa=\kappa.$) But any model of $\Sigma$ clearly has cardinality $\geq \kappa.$ So $\mathfrak{B}$ has cardinality $\kappa;$ restrict $\mathfrak{B}$ to the original language.

I have two questions:

Questions $(1):$ Why is every finite subset of $\Sigma\cup\Gamma$ satisfiable in the structure $\mathfrak{A}?$ If the finite subset is in $\Gamma$ only, then by assumption, the finite subset is satisfiable in $\mathfrak{A}.$ However, when the finite subset contains some elements from $\Sigma,$ I fail to understand why it is satisfiable in $\mathfrak{A}.$

$(2):$ Why is every model of $\Sigma$ has cardinality $\geq \kappa?$

Any hint would be appreciated.

• (2) Because $\Sigma$ has $\kappa$ many new distinct constants. Commented Mar 10, 2018 at 11:42
• @MauroALLEGRANZA: So whenever we add $\kappa$ new symbols, we are assuming that those symbols are all distinct? Commented Mar 10, 2018 at 11:43
• The constants $c_i, c_j$ are distinct and the corresponding "axiom" $c_i \ne c_j$ of $\Sigma$ forces us to interpret them with distict objects. Commented Mar 10, 2018 at 11:52
• So the theory $\Sigma$ is satisfied by an interpretation whose domain has $\kappa$ many objects. Commented Mar 10, 2018 at 12:09
• No; $\mathfrak A$ is what it is. It has a domain with infinite many elements. We "expand" the language adding constants (symbols) to it and we "expand" the theory $\Gamma$ adding the "axioms" of $\Sigma$. Commented Mar 10, 2018 at 14:01

We start from a theory $\Gamma$ in a language $\mathcal L$.

Then we expand the language adding infinite many new individual constant $c_i \in C$, where $C$ has cardinalty $\kappa$.

Then we consider the set of sentence $\Sigma = \{ c_1 \ne c_2 \mid c_1, c2 \text { distinct elements of } C \}$.

Then we consider the set of sentences: $\Gamma \cup \Sigma$ in the expanded language.

Lastly, we assume that $\Gamma$ has an infinite model $\mathfrak A$.

(2) Due to the fact that the expanded language has $\kappa$ many distinct constants, the "axiom" $c_i \ne c_j$ of $Σ$ forces us to interpret them with distict objects.

And (1) due to the fact that $\mathfrak A$ has an infinite domain, it has enough objects to satisfy a finite number of new "axioms" from $\Sigma$.

Thus, by compactness, there is model of $Γ \cup Σ$ and by Downward L-S there is a model $\mathfrak B$ (different from $\mathfrak A$) with the required cardinality.

Obviously, being a model of $Γ \cup Σ$, $\mathfrak B$ is also a model of $Γ$.