Proof verification- $ \sigma $ - Algebra, Algebra, Ring

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I want to prove following for a set $ X \neq \emptyset $ : $ M \subset P(X) $, where $ P(X) $ is the power set.

1) Any Ring is an Algebra

2) Any Algebra is a Ring

3) Any Ring is a $ \sigma $ -Algebra

4) Any Algebra is a $ \sigma $- Algebra

5) Any $ \sigma $- Algebra is an Algebra

For a set $ X \neq \emptyset $, $M$ is

a) A Ring if

  • $ \emptyset \in M$

  • $ A,B \in M \rightarrow A \cup B \in M $

  • $ A,B \in M \rightarrow B \backslash A \in M $

b) An Algebra, if $ M$ Ring , and $ X \in M $

c) A $ \sigma $-Algebra if

  • $ X \in M $

  • $ A \in M \rightarrow X \backslash A \in M $

  • $ (A_i)_{i \in \mathbb{N}} \in M \rightarrow \cup_{i=0}^{ \infty} A_i \in M $


for 1) For $ A \subset X $ is M:={ \emptyset , A} a Ring, but for $ A \neq X$ not an Algebra.

2) let be $ A,B \in M $

$ A \backslash B = ( A^c \cup B )^c \in M $ So M is also a ring.

3) I guess I can use the same example as for 1)?

4) let be $ X= \mathbb{N} , M:= \{ A \subset X : A $ or $ A^c $ finite $\}$ is this a right example for an Alegbra , which is not a s.Algeba?

5) $A,B \in M $

$ X= X \backslash \emptyset \in M $

$ A\backslash B = A \cap B^c = ( A^c \cup B)^c \in M $

any union you can discribe as : $ A \cup B = A \cup B \cup \emptyset \cup ...\emptyset \in M $

so any s. Algebra is also an Algebra.

are my arguments correct and formally right? any adjustments? Appreciate any of your help !