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I'd like to prove following proposition by the compactness theorem:

Suppose $\Sigma$ is a set of $\mathcal{L}$-sentences such that at least one sentence from $\Sigma$ is true in each $\mathcal{L}$-structure. Show that the disjunction of some finitely many sentences from $\Sigma$ is logically valid.

My proof:

Let $\Sigma = \{\sigma_1, ..., \sigma_n\}$ be a set of $\mathcal{L}$-sentences. According to the problem's assumption (at least one sentence from $\Sigma$ in each $\mathcal{L}$-structure), assume a set of $\mathcal{L}$-structures such that each one models at at least one sentence from $\Sigma$:

$\mathcal{A_1} \models \sigma_{1}$

$\mathcal{A_2} \models \sigma_{2}$

...

$\mathcal{A_n} \models \sigma_{n}$

To summary, $\mathcal{A_i} \models \sigma_{i}$ for $1\le i \le n~~~~~~~(**)$.

Equation above could be written as $\mathcal{A_i} \models \{\sigma_{i}\}$ for $1\le i \le n$. Now, assume $\Sigma^{*}$ be the disjunction of some finitely many sentences from $\Sigma$, as following:

$\Sigma^{*} = \bigvee_{m \le i \le n} \sigma_i$. (By Compactness theorem) since (according to $(**)$) every finite subset ${\sigma_i} \subseteq \Sigma^{*}$ has a model, then $\Sigma^{*}$ has a model. Hence $\Sigma^{*}$ is logically valid. Q.E.D.

Is my argument valid?

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    Two comments: (1) "such that at least one sentence from $\Sigma$ in each $\mathcal{L}$-structure" lacks a verb. (2) What would be the model of $\{\sigma_1,\sigma_2\}$? – 2017-02-19
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    @FabioSomenzi: (1) Thanks, fixed. (2): I'm not sure. Maybe I can consider an $\mathcal{L}$-structure for each combination of $\sigma_i$s. However, It doesn't sound to be promising... – 2017-02-19

2 Answers 2

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Suppose for finite $\Gamma \subseteq \Sigma$, $ \bigvee_{\sigma \in \Gamma} \sigma$ is not valid. Then $\bigwedge_{\sigma \in \Gamma} \neg \sigma$ is satisfiable. (This is just De Morgan's law.) If $\bigwedge_{\sigma \in \Gamma} \neg \sigma$ is satisfiable, then $\{\neg \sigma \mid \sigma \in \Gamma \}$ is a satisfiable finite set of sentences.

If this holds for all finite $\Gamma \subseteq \Sigma$, compactness says that $\{\neg \sigma \mid \sigma \in \Sigma\}$ is satisfiable. That is, $\{ \neg\sigma \mid \sigma \in \Sigma\}$ is a set of true sentences in some $\mathcal{L}$-structure $\mathcal{A}$. Now, this structure must be a model of all the negations of sentences in $\Sigma$; that is, $\mathcal{A} \models \neg \sigma$ for all $\sigma \in \Sigma$.

However, this contradicts the assumption that in every $\mathcal{L}$-structure, including $\mathcal{A}$, at least one sentence $\sigma \in \Sigma$ is true. We have to reject the assumption that there is no finite subset of $\Sigma$ such that the disjunction of all its members is valid.

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If there is at least one sentence $\phi \in \Sigma$ true in all structures, then that very sentence is a disjunction of some sentences from $\Sigma$ that is logically valid. (disjunctions can have one disjunct). No Compactness Theorem needed. In fact, this is of course just trivial ... So probably not really what you tried to prove?

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    I think @Roboticist means that for every $\mathcal{L}$-structure at least one sentence in $\Sigma$ is true. – 2017-02-19
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    @FabioSomenzi Ah! The good old Universal Existential swap! yeah, that must be it. – 2017-02-19
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    Yes, that sentence is ambiguous, but as you point out, one interpretation leads to triviality. – 2017-02-19
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    @FabioSomenzi Wait, you didn't just upvote me, did you?! – 2017-02-19
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    What's wrong with that? Your answer is useful because it highlights the ambiguity of the proposition. – 2017-02-19
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    Well, I won't complain ... But @Roboticist please accept Fabio's answer!! – 2017-02-19