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I read two different textbooks and saw two different definitions.

$\mathbf{Definition\ 1}$: Let $X$ be an arbitratry set. A non-empty class $\mathcal{S}$ of subsets of $X$ is called a countably additive algebra ($\sigma$-algebra) if the following conditions are satisfied:

(1) If $A_i\in\mathcal{S}$ ($i=1,2,\ldots)$ then $\bigcup\limits_{i=1}^{\infty} A_{i}\in\mathcal{S}$,

(2) if $A,B\in\mathcal{S}$ then $A\backslash B\in\mathcal{S}$.

$\mathbf{Definition\ 2}$: Consider $X$ a set. A $\sigma-$algebra $\mathcal{S}$ of subsets of $X$ is a collection of subsets of $X$ satisfying the following conditions:

(a) $\emptyset\in\mathcal{S}$,

(b) If $B\in\mathcal{S}$ then its complement $B^c\in \mathcal{S}$,

(c) If $B_1,B_2,\ldots$ is a countable collection of sets in $\mathcal{S}$ then $\bigcup\limits_{i=1}^{\infty} B_{i}\in\mathcal{S}$.

My attempt is to prove that those two definitions are equivalent. This is what I did.

Suppose that $\mathcal{S}\subset 2^X$, and satisfies the conditions (a), (b), (c) in the $\mathbf{Definition\ 2}$. Then it satisfies the conditions (1) and (2) in the $\mathbf{Definition\ 1}$. Indeed, condition (1) is clearly satisfied. For condition (2), we have that $A\backslash B=A\cap B^c=(A^c\cup B)^c=(A^c\cup B\cup B\cup B\ldots)^c\in\mathcal{S}$, by (c) and (b). So, (2) is also satisfied.

Now, I want to prove that if $\mathcal{S}$ satisfies the conditions $(1)$ and $(2)$ in the $\mathbf{Definition\ 1}$ then it also satisfies the condition (a), (b), (c) in the $\mathbf{Definition\ 2}$. For (a), since $\mathcal{S}$ is nonempty, there exists $A\in\mathcal{S}$. By (2), $A\backslash A=\emptyset\in\mathcal{S}$. (c) is satisfied since it's the same as (1). The only problem is, how to show that $\mathcal{S}$ satisfies (b) if it satisfies (1) and (2).

I am stuck and I am not really sure if it can be proved or not. Need help.

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    Well, let $\mathcal{S} = \{\varnothing\}$. That satisfies (1) and (2). One would need to demand $X\in \mathcal{S}$. – 2017-01-04
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    Thank you. So, the two textbooks that I read have their own $\sigma-$algebra. Am I right? – 2017-01-04
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    I'm not sure whether it's intentional or a mistake. The structures satisfying definition 1 are usually called $\sigma$-rings. Definition 2 is the standard one. – 2017-01-04
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    Why don't you identify the two books that you copied those definitions from? Quotations should be attributed. Also, some people might like to know what textbook gives that wrong definition of $\sigma$-algebra. – 2017-01-05
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    The first textbook is "An Introduction to the Theory of Real Functions by Stanislaw Lojasiewicz". That's the one which uses the definition 1. The Definition 2 is standard one, according to what Daniel said. – 2017-01-05

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