# How to express the closure under countable union in formal language?

I'm trying to express the closure under countable union by using formal language, as I'm praticing the language. The following is the statement that I found on proofwiki. $$\forall x_n\in\Sigma:n=1,2,\cdots:\bigcup_{n=1}^{\infty}x_n\in\Sigma$$ It seems that it's written in formal language, but strictly speaking, the sentence has some grammatical errors, somewhat informal and implicit (e.g., an alphabet after a quantifier must be a variable, but $x_n$ is implicitly given as a function, and dots in $n=1,2,\cdots$ is not an alphabet of formal language.)

I think the sentence should be in $L^{\mathrm{II}}$, which means it's second-order logic. What I've done is the following but am not sure. $$\forall x:((\exists f:f:S\leftrightarrow\mathbb{N}\wedge Sx\wedge \Sigma x)\implies \Sigma\bigcup S)$$ ($f:S\leftrightarrow\mathbb{N}$ means a bijection from a set $S$ to the set of natural numbers $\mathbb N$, $S$ and $\Sigma$ are unary relations.)

• If by "closure under countable union" you mean under all countable unions, then shouldn't you have $\forall f$ rather than $\exists f$ ? – DanielWainfleet Sep 30 '15 at 7:03
• @user254665 Umm, I don't think so. Since a set $S$ is free variable, it can be any set that satisfies the statement. – Shin Kim Sep 30 '15 at 7:24
• You don’t want $f$ to be a bijection, since countable includes finite. And I frankly think that even with that corrected, your style obfuscates the definition. I would simply define cardinality (or at least countable cardinality) first and write $$\forall S\subseteq\Sigma\left(|S|\le\omega\to\bigcup S\in\Sigma\right)\;.$$ – Brian M. Scott Sep 30 '15 at 15:58
• I was going to ask about the notation of the Q. I didn't know whether it was just a standard that I'm not familiar with so I waited for someone else to comment on it. – DanielWainfleet Sep 30 '15 at 16:04
• @BrianM.Scott Ah, it makes everything clear. Yeah, after that I formulated another one, $\forall S((S\subseteq \Sigma \wedge \exists f f:S\hookrightarrow \mathbb N)\rightarrow\Sigma\bigcup S)$, but I prefer Your formula way more than mine. Really appreciated. – Shin Kim Oct 1 '15 at 7:54